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NCERT Exemplar Chapter 12 Electricity Class 10 Science Solutions

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NCERT Exemplar Class 10 Science Chapter 12 Electricity Solutions

NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity covers all the important questions and answers as well as advanced level questions. It helps in learning about the electric current, charge, conductors, insulators, electrons, convection, potential difference, ohms law, resistors and resistance.

The NCERT Exemplar solutions for class 10 science is very important for board exams. NCERT Exemplar Solutions for Class 10 Science Chapter 12 Electricity is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like electric fuse, short circuit, overloading, factors on which resistance of a conductor depends and safety devices.


Chapter Name

Chapter 12 Electricity

Book Title

NCERT Exemplar for Class 10 Science

Related Study

  • NCERT Solutions for Class 10 Science Chapter 12 Electricity
  • Revision Notes for Class 10 Science Chapter 12 Electricity
  • MCQ for Class 10 Science Chapter 12 Electricity
  • Important Questions for Class 10 Science Chapter 12 Electricity

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 12 Electricity Class 10 Science

Multiple Choice Questions

1. A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams of Figure12.1. The current recorded in the ammeter will be

(a) maximum in (i)
(b) maximum in (ii)
(c) maximum in (iii)
(d) the same in all the cases

Solution
(d) the same in all the cases

The current will be same in all the circuits as there are no changes in any of the circuits.


2. In the following circuits (Figure 12.2), heat produced in the resistor or combination of resistors connected to a 12 V battery will be

(a) same in all the cases
(b) minimum in case (i)
(c) maximum in case (ii) 
(d) maximum in case (iii) 

Solution
(c) maximum in case (ii) 

The two resistors are connected in series. The total resistance will be less than individual resistances in figure (iii) as they are connected parallel. Higher resistance produces more heat.


3. Electrical resistivity of a given metallic wire depends upon
(a) its length
(b) its thickness
(c) its shape
(d) nature of the material

Solution

(d) nature of the material 


4. A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly

(a) 1020
(b) 1016
(c) 1018
(d) 1023

Solution
(a) 1020

Therefore, n = 1020 electrons 
The number of electrons flowing is 1020 electrons  


5. Identify the circuit (Figure 12.3) in which the electrical components have been properly connected. 

(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Solution
(b) (ii) 

6. What is the maximum resistance which can be made using five resistors each of 1/5 Ω?
(a) 1/5 Ω
(b) 10 Ω
(c) 5 Ω
(d) 1 Ω
Solution
(d) 1 Ω
Maximum resistance is obtained when resistors are connected in series.

7. What is the minimum resistance which can be made using five resistors each of 1/5 Ω?
(a) 1/5 Ω
(b) 1/25 Ω
(c) 1/10 Ω
(d) 25 Ω
Solution
(b) 1/25 Ω
Minimum resistance is obtained when resistors are connected parallel 
1/R = 5 + 5 +5+ 5 + 5 = 25 Ω
R = 1/25 Ω

8. The proper representation of series combination of cells (Figure 12.4) obtaining maximum potential is
(a) (i) 
(b) (ii) 
(c) (iii) 
(d) (iv) 
Solution
(a) (i)
The positive terminal of next cell is adjacent to negative terminal of previous cell.

9. Which of the following represents voltage ? 

(a) Work done/(Current × Time) 
(b) Work done × Charge
(c) (Work done × Time)/Current 
(d) Work done × Charge × Time 

Solution

(a) Work done/(Current × Time) 


10. A cylindrical conductor of length l and uniform area of crossnsection A has resistance R. Another conductor of length 2l and resistance R of the same material has area of cross section
(a) A/2
(b) 3A/2
(c) 2A
(d) 3A

Solution

(c) 2A

⇒ A = 2A


11. A student carries out an experiment and plots the V-I graph of three samples of nichrome wire with resistances R1 , R2 and R3 respectively (Figure.12.5). Which of the following is true?

(a) R1 = R2 = R3
(b) R1 > R2 > R3
(c) R3 > R2 > R1
(d) R2 > R3 > R1

Solution
(c) R3 > R2 > R1

Current flow is inversely proportional to resistance. Highest resistance will show least amount of current.


12. If the current I through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be
(a) 100 %
(b) 200 %
(c) 300 %
(d) 400 %

Solution
(c) 300 %

Heat generated is directly proportional to square of current. Therefore, when current doubles, the dissipation of heat will multiply by 4.


13.  The resistivity does not change if
(a) the material is changed
(b) the temperature is changed
(c) the shape of the resistor is changed
(d) both material and temperature are changed

Solution
(c) the shape of the resistor is changed


14. In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?
(a) Brightness of all the bulbs will be the same
(b) Brightness of bulb A will be the maximum
(c) Brightness of bulb B will be more than that of A
(d) Brightness of bulb C will be less than that of B

Solution
(c) Brightness of bulb B will be more than that of A

As the three bulbs are connected in parallel, so the resistance of the combination would be less than arithmetic sum of individual resistances of all the bulbs. Therefore, there will be no negative effect on current flow. As a result, the bulbs will glow according to their wattage.


15. In an electrical circuit two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be
(a) 5 J
(b) 10 J
(c) 20 J
(d) 30 J

Solution
(c) 20 J

Equivalent resistance of the circuit is R = 4+2 = 6Ω

Therefore, heat dissipated by 4 ohm resistor is, H = IRt = 20J

16. An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?
(a) 1 A
(b) 2 A
(c) 4 A
(d) 5 A

Solution
(d) 5 A

P=V ×I

Or, 1000 W= 220 V×I


17. Two resistors of resistance 2 Ω and 4 Ω when connected to a battery will have
(a) same current flowing through them when connected in parallel
(b) same current flowing through them when connected in series
(c) same potential difference across them when connected in series
(d) different p

Solution
(b) same current flowing through them when connected in series

In series combination the resistors receive a common current as the current does not get divided into branches.


18. Unit of electric power may also be expressed as
(a) volt ampere
(b) kilowatt hour
(c) watt second
(d) joule second

Solution
(b) kilowatt hour

Volt-ampere(VA) is the unit for apparent power. Watt second(also watt-second, symbol W s or W.s) is a derived unit of energy and is equivalent to the joule. The joule-second is the unit used for Planck’s constant.


Short Answer Questions 

19. A child has drawn the electric circuit to study Ohm’s law as shown in Figure 12.6. His teacher told that the circuit diagram needs correction. Study the circuit diagram and redraw it after making all corrections.

Solution

20. Three 2 Ω resistors, A, B and C, are connected as shown in Figure 12.7. Each of them dissipates energy and can withstand a maximum power of 18W without melting. Find the maximum current that can flow through the three resistors?
Solution

Current P = I2R
⇒ 18W = I2 × 2Ω
⇒ I2 = 18W/2Ω = 9
⇒ I = 3A
This is the maximum current that can flow through the three resistors.


21. Should the resistance of an ammeter be low or high? Give reason.

Solution

The resistance of an ammeter should be close to zero. Ideally it should be zero ohm. If the resistance of the ammeter is non-zero then the true current will be affected substantially.


22. Draw a circuit diagram of an electric circuit containing a cell, a key, an ammeter, a resistor of 2 Ω in series with a combination of two resistors (4 Ω each) in parallel and a voltmeter across the parallel combination. Will the potential difference across the 2 Ω resistor be the same as that across the parallel combination of 4Ω resistors? Give reason.

Solution

Total resistance for parallel combination of 4Ω resistors can be calculated as follows:
1/R = 1/4 + 1/4 = 1/2 
⇒ R = 2Ω
This shows that the resistance of parallel combination is equal to resistance of the series resistor. Therefore, the potential difference across 2Ω resistance will be same as potential difference across the other two resistors which are connected in parallel.

23. How does use of a fuse wire protect electrical appliances?
Solution
The resistance of the fuse wire is greater than that of the main wiring. The fuse wire melts and breaks the circuit when there is significant increase in the electric current.. This prevents the damage of electrical appliance.

24. What is electrical resistivity? In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 5 A. The reading of the ammeter decreases to half when the length of the wire is doubled. Why?
Solution
Resistivity is the property of the conductor through which it resists the flow of electric current. Resistivity for any particular material is unique. Resistance is directly proportional to length of conductor and inversely proportional to current flow. As the length is doubled, the resistance also doubles but current flow reduces to half. This is the reason why the ammeter reading decreases.


25. What is the commercial unit of electrical energy? Represent it in terms of joules.

Solution
Commercial unit of electrical energy is killowatt-hour

1 kw/hr = 1 kW h
= 1000 W × 60 × 60s
= 3.6 × 106 J


26. A current of 1 ampere flows in a series circuit containing an electric lamp and a conductor of 5 Ω when connected to a 10 V battery. Calculate the resistance of the electric lamp. Now if a resistance of 10 Ω is connected in parallel with this series combination, what change (if any) in current flowing through 5 Ω conductor and potential difference across the lamp will take place? Give reason.

Solution

Let R be the resistance of the electric lamp.
In series total resistance = 5 + R
Therefore, R = 5 ohm
Now, potential drop across lamp + conductor = 10 V
Therefore, V across the lamp = I × R = 1 × 5 = 5 Volt


27. Why is parallel arrangement used in domestic wiring?

Solution

In domestic wiring, parallel arrangement is used as it provides the same potential difference across each electrical appliance.


28. B1 , B2 and B3 are three identical bulbs connected as shown in Figure 12.8. When all the three bulbs glow, a current of 3A is recorded by the ammeter A.

(i) What happens to the glow of the other two bulbs when the bulb B1 gets fused?
(ii) What happens to the reading of A1 , A2 , A3 and A when the bulb B2 gets fused?
(iii) How much power is dissipated in the circuit when all the three bulbs glow together?

Solution

(i) In parallel circuit, the potential difference does not get divided. Therefore, if one bulb is fused, then the glowing of other bulbs will not be affected.

(ii) Ammeter A shows a reading of 3A. This means each of the Al, A2, and A3 show 1 A reading.

Now, P= I2R
= (3A)2× 1.5 Ω
= 13.5 W


Long Answer Questions 

29. Three incandescent bulbs of 100 W each are connected in series in an electric circuit. In another circuit another set of three bulbs of the same wattage are connected in parallel to the same source.
(a) Will the bulb in the two circuits glow with the same brightness? Justify your answer.
(b) Now let one bulb in both the circuits get fused. Will the rest of the bulbs continue to glow in each circuit? Give reason. 

Solution

(a) In series connection, the resistance of the bulbs will be three times the resistance of single bulb. Therefore, the current in the series combination will be one-third of the current in each bulb in parallel combination. This in parallel combination the bulbs will glow more brightly.

(b) In series combination, the bulbs will stop glowing as the circuit is broken and therefore the current is zero. However, in parallel connection the bulbs will continue to glow with the same brightness.


30. State Ohm’s law? How can it be verified experimentally? Does it hold good under all conditions? Comment.

Solution

Ohm’s law states that at constant temperature, the potential difference (voltage) across an ideal conductor is proportional to the current through it.
V/I = R

Verification of Ohm’s law 
Steps to be followed in order to verify the Ohm’s law:
  • Set up a circuit as shown in figure. The circuit consists of a nichrome wire XY of length, say 0.5 m, an ammeter, a voltmeter and four cells of 1.5 V each.
  • First use only one cell as the source in the circuit and note the reading in the ammeter I, for the current and reading of the voltmeter V for the potential difference across the nichrome wire XY in the circuit. Tabulate them in the Table given
  • Next connect two cells in the circuit and again note down the respective readings of the ammeter and voltmeter.
  • Repeat the above steps using three cells and then four cells in the circuit separately.


31. What is electrical resistivity of a material? What is its unit? Describe an experiment to study the factors on which the resistance of conducting wire depends.

Solution

The inherent property of a conductor by virtue of which it resists the flow of electric current is known as resistivity. Resistivity of each material is unique.

The SI unit of resistivity is Ω m. 

Experiment to study the factors on which the resistance of conducting wire depends:

Set up:

  • Take a nichrome wire, a torch bulb, a 10 W bulb and an ammeter (0 –5 A range), a plug key and some connecting wires.
  • Connect four dry cells of 1.5 V each in series with the ammeter leaving a gap XY in the circuit, as shown in Fig. 12.4.
  • Complete the circuit by connecting the nichrome wire in the gap XY. Plug the key. Note down the ammeter reading. •Now, change the length of the nichrome wire and take the ammeter reading.
  • Next, change the thickness of nichrome wire and note down the ammeter reading.
  • After few steps, use copper wire for the experiment. Attach copper wire and note down the ammeter reading.
  • One by one change the length and thickness of the copper wire and observe the ammeter reading.
  • Repeat the steps with different materials. 

Observation:

It is observed that the resistance depends on:

  • Material of conductor.
  • Length of conductor.
  • Area of cross section.


32. How will you infer with the help of an experiment that the same current flows through every part of the circuit containing three resistances in series connected to a battery?

Solution

  • Connect the three resistors R1, R2, R3 in series to make the circuit.
  • Use ammeter to see the current flowing through the three resistors.
  • Remove R1 and take the reading of potential difference of R2 and R3
  • Remove R2 and take the reading of potential difference of R1 and R3
Observation:
Ammeter reading is same in each case. This shows same current flows across each resistors connected in series.

33. How will you conclude that the same potential difference (voltage) exists across three resistors connected in a parallel arrangement to a battery?
Solution
  • Take three resistors RI. ity. and R3 and connect them in parallel to make a circuit as shown in figure.
  • Use voltmeter to take reading of potential difference of three resistors in parallel combination.
  • Now, remove the resistor R1 and take the reacting of potential difference of remaining resistors combination.
  • Then, remove the resistor R, and take the reading of potential difference of remaining resistor.

Observation:

It is observed that in each case the voltmeter reading is the same. This shows that in parallel arrangement, same potential difference exists across each resistor.


34. What is Joule’s heating effect? How can it be demonstrated experimentally? List its four applications in daily life.

Solution

According to Joules heating effect, the heat produced in a resistor is:

  • Directly proportional to square of current.
  • Directly proportional to resistance.
  • Directly proportional to the time of current flowing through the resistor.
This can be mathematically expressed as,
H = I2 Rt
H is heating effect,
I is electric current,
R is resistance and
t is time

Experiment to demonstrate Joules law of heating 

  • Take an immersion rod and connect it to a socket which in turn is connected to a regulator.
  • It is important as the regulator controls the amount of current flowing through a device.
  • Keeping the pointer of regulator at minimum reading, measure the time taken by immersion rod to heat a certain amount of water.
  • Next, increase the pointer of regulator to the next level.
  • Measure the time taken by immersion rod to heat the same amount of water.
  • Repeat above step for higher levels on regulator to measure the time.
Observation:
It is observed that with increased electric current, the time required to heat the same amount of water decreases. This verifies the Joule’s Law of Heating.
Application:
The electric appliances like electric toaster, oven, electric kettle and electric heater work on the basis of heating effect of electric current.

35. Find out the following in the electric circuit given in Figure 12.9

(a) Effective resistance of two 8 Ω resistors in the combination
(b) Current flowing through 4 Ω resistor
(c) Potential difference across 4 Ω resistance
(d) Power dissipated in 4 Ω resistor
(e) Difference in ammeter readings, if any

Solution
(a) Since, two 8 Ω resistors are in parallel, then their effective resistance Rρ is given by

1/ Rρ = 1/R1 + 1/R2 = 1/8 + 1/8 = 1/4
= Rρ = 4 Ω
(b) Total resistance in the circuit
R = 4 Ω + Rρ = 4 Ω + 4 Ω = 8 Ω
Current through the circuit,
I = V/R = 8/8 = 1A
Thus, current through 4 Ω resistor is 1 A as 4 Ω and Rρ are in series and same current flows through them.
(c) Potential difference across 4 Ω resistor is potential drop by the 4 Ω resistor.
i.e. V = IR = 1 × 4 = 4V
(d) Power dissipated in 4 Ω resistor
P = I2R = I2 × 4 = 4
(e) There is no difference in the reading of ammeters A1 and A2 as same current flows through all elements in a series current.


NCERT Exemplar Chapter 13 Magnetic Effects of Electric Current Class 10 Science Solutions

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NCERT Exemplar Class 10 Science Chapter 13 Magnetic Effects of Electric Current Solutions

NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current covers all the important questions and answers as well as advanced level questions. It helps in learning about the magnetic field and field lines, Magnetic field due to a current-carrying conductor, right-hand thumb rule, circular loop, solenoid and Force on a current-carrying conductor in a magnetic field.

The NCERT Exemplar solutions for class 10 science is very important for board exams. NCERT Exemplar Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like electric motor, electromagnetic induction, Fleming's right-hand rule, electric generator, electric power and domestic electric circuits.


Chapter Name

Chapter 13 Magnetic Effects of Electric Current

Book Title

NCERT Exemplar for Class 10 Science

Related Study

  • NCERT Solutions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current
  • Revision Notes for Class 10 Science Chapter 13 Magnetic Effects of Electric Current
  • MCQ for Class 10 Science Chapter 13 Magnetic Effects of Electric Current
  • Important Questions for Class 10 Science Chapter 13 Magnetic Effects of Electric Current

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 13 Magnetic Effects of Electric Current Class 10 Science

Multiple Choice Questions

1. Choose the incorrect statement from the following regarding magnetic lines of field
(a) The direction of magnetic field at a point is taken to be the direction in which the north pole of a magnetic compass needle points
(b) Magnetic field lines are closed curves
(c) If magnetic field lines are parallel and equidistant, they represent zero field strength
(d) Relative strength of magnetic field is shown by the degree of closeness of the field lines

Solution

(c) If magnetic field lines are parallel and equidistant, they represent zero field strength

Parallel lines of magnetic field represent uniform magnetic field.


2. If the key in the arrangement (Figure 13.1) is taken out (the circuit is made open) and magnetic field lines are drawn over the horizontal plane ABCD, the lines are
(a) concentric circles
(b) elliptical in shape
(c) straight lines parallel to each other
(d) concentric circles near the point O but of elliptical shapes as we go away from it

Solution

(a) concentric circles

If key is taken out then current will stop and no magnetic field exists due to the conductor. Therefore, at the point O, there will be earth’s magnetic field and they will be straight lines parallel to each other.


3. A circular loop placed in a plane perpendicular to the plane of paper carries a current when the key is ON. The current as seen from points A and B (in the plane of paper and on the axis of the coil) is anti clockwise and clockwise respectively. The magnetic field lines point from B to A. The N-pole of the resultant magnet is on the face close to
(a) A
(b) B
(c) A if the current is small, and B if the current is large
(d) B if the current is small and A if the current is large

Solution

(a) A

The direction of magnetic field line is from south to north pole. As the field lines are pointing form B to A, point A is indicating the north pole.


4.  For a current in a long straight solenoid N- and S-poles are created at the two ends. Among the following statements, the incorrect statement is
(a) The field lines inside the solenoid are in the form of straight lines which indicates that the magnetic field is the same at all points inside the solenoid
(b) The strong magnetic field produced inside the solenoid can be used to magnetise a piece of magnetic material like soft iron, when placed inside the coil
(c) The pattern of the magnetic field associated with the solenoid is different from the pattern of the magnetic field around a bar magnet
(d) The N- and S-poles exchange position when the direction of current through the solenoid is reversed

Solution

(c) The pattern of the magnetic field associated with the solenoid is different from the pattern of the magnetic field around a bar magnet

A solenoid behaves like a bar magnet, because the pattern of the magnetic field lines associated with a solenoid are similar to the pattern of the magnetic field lines around a bar magnet.


5.  A uniform magnetic field exists in the plane of paper pointing from left to right as shown in Figure 13.3. In the field an electron and a proton move as shown. The electron and the proton experience
(a) forces both pointing into the plane of paper
(b) forces both pointing out of the plane of paper
(c) forces pointing into the plane of paper and out of the plane of paper, respectively
(d) force pointing opposite and along the direction of the uniform magnetic field respectively.

Solution

(a) forces both pointing into the plane of paper 

The direction of movement of electron is opposite to the direction of flow of electric current. This makes the current flow upwards. If the index finger shows the direction of magnetic field, ring finger shows the direction of current then the direction of thumb is in to the plane of paper.


6.  Commercial electric motors do not use
(a) an electromagnet to rotate the armature
(b) effectively large number of turns of conducting wire in the current carrying coil
(c) a permanent magnet to rotate the armature
(d) a soft iron core on which the coil is wound

Solution

(c) a permanent magnet to rotate the armature

In electric motors electromagnet is used instead of permanent magnet.


7. In the arrangement shown in Figure 13.4 there are two coils wound on a non-conducting cylindrical rod. Initially the key is not inserted. Then the key is inserted and later removed. Then
(a) the deflection in the galvanometer remains zero throughout
(b) there is a momentary deflection in the galvanometer but it dies out shortly and there is no effect when the key is removed
(c) there are momentary galvanometer deflections that die out shortly; the deflections are in the same direction
(d) there are momentary galvanometer deflections that die out shortly; the deflections are in opposite directions.

Solution

(d) there are momentary galvanometer deflections that die out shortly; the deflections are in opposite directions

When key is plugged in, the galvanometer shows deflection in one direction and the direction of deflection reverses if the key is unplugged.


8. Choose the incorrect statement
(a) Fleming’s right-hand rule is a simple rule to know the direction of induced current
(b) The right-hand thumb rule is used to find the direction of magnetic fields due to current carrying conductors
(c) The difference between the direct and alternating currents is that the direct current always flows in one direction, whereas the alternating current reverses its direction periodically
(d) In India, the AC changes direction after every 1/50 second

Solution

(d) In India, the AC changes direction after every 1/50 second

AC frequency in India is 50 Hz. As the direction changes twice in each cycle so that change of direction takes place after every 1/00 second.


9. A constant current flows in a horizontal wire in the plane of the paper from east to west as shown in Figure 13.5. The direction of magnetic field at a point will be North to South
(a) directly above the wire
(b) directly below the wire
(c) at a point located in the plane of the paper, on the north side of the wire
(d) at a point located in the plane of the paper, on the south side of the wire

Solution

(b) directly below the wire

Using the right hand thumb rule, it can be found that the direction of magnetic field is from North to South below the wire.


10. The strength of magnetic field inside a long current carrying straight solenoid is
(a) more at the ends than at the centre
(b) minimum in the middle
(c) same at all points
(d) found to increase from one end to the other

Solution

(c) same at all points

Magnetic field lines are straight inside the solenoid. This indicates strong magnetic field. Therefore, inside the solenoid the magnetic field is uniform at all points.


11. To convert an AC generator into DC generator
(a) split-ring type commutator must be used
(b) slip rings and brushes must be used
(c) a stronger magnetic field has to be used
(d) a rectangular wire loop has to be used

Solution

(a) split-ring type commutator must be used

After each half turn, the split-ring type commutator reverse the direction of current. This maintains a DC current.


12. The most important safety method used for protecting home appliances from short circuiting or overloading is
(a) earthing
(b) use of fuse
(c) use of stabilizers
(d) use of electric meter

Solution

(b) use of fuse

Fuse has a thin wire made of tin and lead in the ratio of 75: 25%. The fuse wire melts and breaks the circuit when current exceeds specified limit thereby protecting home appliances.


Short Answer Questions 

13. A magnetic compass needle is placed in the plane of paper near point A as shown in Figure 13.6. In which plane should a straight current carrying conductor be placed so that it passes through A and there is no change in the deflection of the compass? Under what condition is the deflection maximum and why?

Solution

The current carrying conductor is placed in the plane of the paper itself. This is because the axis of the compass is vertical and the field due to the conductor is also vertical. The deflection is maximum when the conductor passes through A in a plane perpendicular to the plane of paper. This is because the field is maximum in the plane of the paper.


14. Under what conditions permanent electromagnet is obtained if a current carrying solenoid is used? Support your answer with the help of a labelled circuit diagram.

Solution

Conditions in which a permanent electromagnet is obtained if a current carrying solenoid is used are:

  1. Circuit should be closed.
  2. (A core of soft iron should be used.


15. AB is a current carrying conductor in the plane of the paper as shown in Figure 13.7. What are the directions of magnetic fields produced by it at points P and Q? Given r1> r2, where will the strength of the magnetic field be larger?

Solution

At P into the plane of the paper and out of it at Q. The strength of the magnetic field is larger at Q as it is located closer.


16. A magnetic compass shows a deflection when placed near a current carrying wire. How will the deflection of the compass get affected if the current in the wire is increased? Support your answer with a reason.

Solution 

The deflection of the magnetic compass increases because the strength of magnetic field is directly proportional to the magnitude of current passing through the straight conductor.


17. It is established that an electric current through a metallic conductor produces a magnetic field around it. Is there a similar magnetic field produced around a thin beam of moving
(i) alpha particles,
(ii) neutrons? Justify your answer.

Solution

(i) As alpha particles are positively charged, they constitutes current in the direction of motion.

(ii) Neutrons do not conduct electricity as the have no charge on them.


18. What does the direction of thumb indicate in the right-hand thumb rule. In what way this rule is different from Fleming’s left-hand rule?

Solution

The thumb indicates the direction of current in the straight conductor held by curled fingers, whereas the Fleming’s left-hand rule gives the direction of force experienced by current carrying conductor placed in an external magnetic field.


19. Meena draws magnetic field lines of field close to the axis of a current carrying circular loop. As she moves away from the centre of the circular loop she observes that the lines keep on diverging. How will you explain her observation.

Solution

As the distance increases, the strength of the magnetic field falls. This is indicated by the decrease in degree of closeness of the lines of field.


20. What does the divergence of magnetic field lines near the ends of a current carrying straight solenoid indicate?

Solution

The divergence of magnetic field lines indicates the decrease in strength of magnetic field near and beyond the ends of the solenoid.


21. Name four appliances wherein an electric motor, a rotating device that converts electrical energy to mechanical energy, is used as an important component. In what respect motors are different from generators?

Solution

Electric fans, mixers, washing machines, computer drives, etc. are devices that convert electrical energy into mechanical energy.

Generators convert mechanical energy into electrical energy.


22. What is the role of the two conducting stationary brushes in a simple electric motor?

Solution

The brushes are connected to the battery and hence they draw current and supply it to the armature of the coil.


23. What is the difference between a direct current and an alternating current? How many times does AC used in India change direction in one second?

Solution

Direct current always flows in one direction whereas the alternating current reverses its direction periodically. The frequency of AC in India is 50 Hz and in each cycle it alters direction twice. Therefore AC changes direction 2×50 = 100 times in one second.


24. What is the role of fuse, used in series with any electrical appliance? Why should a fuse with defined rating not be replaced by one with a larger rating?

Solution

Fuse is used for protecting appliances during short-circuiting or overloading. The fuse is rated for a certain maximum current and the fuse wire melts when the current flowing through it exceeds the rated value. The appliance may get damaged, if a fuse is replaced by one with larger ratings because the fuse wire will not melt. Therefore, the practice of using fuse of improper rating should always be avoided.


Long Answer Questions 

25. Why does a magnetic compass needle pointing North and South in the absence of a nearby magnet get deflected when a bar magnet or a current carrying loop is brought near it. Describe some salient features of magnetic lines of field concept.

Solution

Current carrying loops act as bar magnets which have associated magnetic fields. These field lines change the existing earth’s magnetic field and cause a deflection.

Some salient features of magnetic lines of field are as follows:

  • Magnetic field has both direction and magnitude.
  • Magnetic field lines emerge from the north pole and enter the south pole.
  • The strength of a magnetic field is represented diagrammatically by the degree of closeness of the field lines.
  • At a single point, two values of net field cannot as the field lines cannot cross each other.
  • If the field lines are parallel and equi-spaced in some region, this means that the field is uniform.


26. With the help of a labelled circuit diagram illustrate the pattern of field lines of the magnetic field around a current carrying straight long conducting wire. How is the right hand thumb rule useful to find direction of magnetic field associated with a current carrying conductor?

Solution

Right hand thumb rule states that if a current carrying straight conductor is held in the right hand with the thumb pointing towards the direction of current, then the fingers will wrap around the conductor in the direction of the magnetic field line.


27. Explain with the help of a labelled diagram the distribution of magnetic field due to a current through a circular loop. Why is it that if a current carrying coil has n turns the field produced at any point is n times as large as that produced by a single turn?

Solution

Even in current carrying loop, the right hand thumb rule is obeyed. This shows that the magnetic field lines are present around the conducting wire. But the circular shape of conductor means that field lines at different points of the loop appear to be making ring around the periphery of the loop. This looks like small ring looping around big ring.

If the number of turns of the coils increases, the magnetic field increases. Therefore the strength of the magnetic field increases with number of turns in a coil.


28. Describe the activity that shows that a current-carrying conductor experiences a force perpendicular to its length and the external magnetic field. How does Fleming’s left-hand rule help us to find the direction of the force acting on the current carrying conductor?

Solution

Activity that shows that a current-carrying conductor experiences a force perpendicular to its length and the external magnetic field:

  • Take a small aluminium rod AB (of about 5 cm).
  • Suspend the rod horizontally from a stand, as shown in Fig.
  • Place a strong horse-shoe magnet in such a way that the rod lies between the two poles with the magnetic field directed upwards. 
  • Connect the aluminium rod in series with a battery, a key and a rheostat.
  • Now, allow a current to pass through the aluminium rod from end B to end A.

Observation:

It is observed that the rod is displaced towards the left. On reversing the direction of current flowing through the rod, the direction of displacement of the rod also changes.

According to Fleming’s left hand rule, stretch the thumb, forefinger and central finger of your left hand such that they are mutually perpendicular. If the fore finger points in the direction of magnetic field and the central in the direction of current, then the thumb will point in the direction of motion or force acting on the conductor.


29. Draw a labelled circuit diagram of a simple electric motor and explain its working. In what way these simple electric motors are diffferent from commercial motors?

Solution

An electric motor consists of a coil ABCD placed between the two poles of a permanent magnet. The coil is placed in such a way that the direction of current is perpendicular to the direction of magnetic field.

  • Arms of the coil are attached to a split ring. Arm AB is attached to the half P and arm CD is attached to the half Q.
  • The split rings are touching two static brushes X and Y which provide electric supply to the split rings. The split rings are insulated from inside. An axle is passing through the split rings.
  • A battery supplies electric current to the coil. The flow of current is from A to B and from C to D.

Working of electric motor:

When current flows from A to B, then the arm AB moves down according to the Flemings left hand rule. On the contrary, the arm CD of the coil moves up when current passes from C to D.

P of split ring touches brush Y when the coil completes half turn and the opposite happens with Q of split ring. Therefore, the direction of current gets reversed. That is the current now flows D to C and from B to A. This pushes the arm CD downwards and AB upwards. As a result the coil turns continuously.

Simple electric motor

Commercial electric motor

Permanent magnet is used

Electromagnet megnet is used

Turn number in coil is less

Turn number in coil is large

Iron core is not used

Soft Iron core is used


30. Explain the phenomenon of electromagnetic induction. Describe an experiment to show that a current is set up in a closed loop when an external magnetic field passing through the loop increases or decreases.

Solution

The phenomenon of generation of an electric current in a closed circuit by changing the magnetic field is known as electromagnetic induction. The current produced is called the induced current and the potential difference set up is called the induced emf.
Experiment to show that a current is set up in a closed loop when an external magnetic field passing through the loop increases or decreases:
  • Take two different coils of copper wire having large number of turns (say 50 and 100 turns respectively). 
  • Insert them over a non-conducting cylindrical roll, as shown.
  • Connect the coil-1having larger number of turns, in series with a battery and a plug key. 
  • Also connect the other coil-2 with a galvanometer as shown.
  • Plug in the key.
Observation:
  • When the key is plugged, the needle of the galvanometer shows deflection, indicating a momentary current in coil-2.
  • When the key is unplugged, the needle momentarily moves, but now to the opposite side. 
  • This shows that there is current in coil-2 whenever there is change in magnetic field.

31. Describe the working of an AC generator with the help of a labelled circuit diagram. What changes must be made in the arrangement to convert it to a DC generator?

Solution

A conductor coil ABCD is placed in a magnetic field. The direction of magnetic flux will be form N pole to S pole. The coil is connected to slip rings, and the load is connected through brushes resting on the slip rings.
Consider the case 1 from above figure. The coil is rotating clockwise, and thus the direction of induced current is along A-B-C-D according to the Fleming’s right hand rule. After half time period, the coil is rotating clockwise. Therefore, the direction of the induced current will be along D-C-B-A. This shows that the direction of the current changes after half of the time period. That means the induced current is an alternating current.
A split ring type commutator must be used in place of slip ring type commutator to get a direct current.

32. Draw an appropriate schematic diagram showing common domestic circuits and discuss the importance of fuse. Why is it that a burnt out fuse should be replaced by another fuse of identical rating?

Solution

Fuse is important for the following reasons:
  • It is a safety device used in household wiring and in electrical appliance.
  • It prevents the damage to the circuit and appliance due to overload.
  • It prevents accidental fire which may happen because of short circuit.
Fuse of a particular rating is used with a particular appliance. A fuse with lower rating will need to be changed frequently, hence it would result in lot of inconvenience. A fuse with higher rating will not melt if the current exceeds the safety limit, hence it would not serve the purpose. Therefore, a burnt out fuse should be replaced by another fuse of identical rating.

NCERT Exemplar Chapter 14 Sources of Energy Class 10 Science Solutions

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NCERT Exemplar Class 10 Science Chapter 14 Sources of Energy Solutions

NCERT Exemplar Solutions for Class 10 Science Chapter 14 Sources of Energy covers all the important questions and answers as well as advanced level questions. It helps in learning about the magnetic field and field lines, Magnetic field due to a current-carrying conductor, right-hand thumb rule, circular loop, solenoid and Force on a current-carrying conductor in a magnetic field.

The NCERT Exemplar solutions for class 10 science is very important for board exams. NCERT Exemplar Solutions for Class 10 Science Chapter 14 Sources of Energy is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like electric motor, electromagnetic induction, Fleming's right-hand rule, electric generator, electric power and domestic electric circuits.


Chapter Name

Chapter 14 Sources of Energy

Book Title

NCERT Exemplar for Class 10 Science

Related Study

  • NCERT Solutions for Class 10 Science Chapter 14 Sources of Energy
  • Revision Notes for Class 10 Science Chapter 14 Sources of Energy
  • MCQ for Class 10 Science Chapter 14 Sources of Energy
  • Important Questions for Class 10 Science Chapter 14 Sources of Energy

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 14 Sources of Energy Class 10 Science

Multiple Choice Questions

1. Which of the following is a non-renewable source of energy?
(a) Wood
(b) Sun
(c) Fossil fuels
(d) Wind

Solution
(b) Sun

Sources of energy which are present in limited amount and can’t be recycled once used are known as non-renewable sources of energy. Fossil fuels are non-renewable source of energy.


2. Acid rain happens because
(a) sun leads to heating of upper layer of atmosphere
(b) burning of fossil fuels release oxides of carbon, nitrogen and sulphur in the atmosphere
(c) electrical charges are produced due to friction amongst clouds
(d) earth atmosphere contains acids

Solution
(b) burning of fossil fuels release oxides of carbon, nitrogen and sulphur in the atmosphere

When rain water mixes with the oxides of carbon, nitrogen and sulphur in the atmosphere, acid rain takes place.


3. Fuel used in thermal power plants is
(a) water
(b) uranium
(c) biomass
(d) fossil fuels

Solution
(d) fossil fuels 

Coal is the major fossil fuel used in thermal power plants. Along with coal natural gas and petroleum are also used.


4. In a hydro power plant
(a) Potential energy possessed by stored water is converted into electricity
(b) Kinetic energy possessed by stored water is converted into potential energy
(c) Electricity is extracted from water
(d) Water is converted into steam to produce electricity

Solution
(a) Potential energy possessed by stored water is converted into electricity . 

Water in the dams possesses potential energy. This potential energy converts into kinetic energy when the water is released. This kinetic energy turns the turbines. Turbines in turn produce electricity.


5. Which is the ultimate source of energy?
(a) Water
(b) Sun
(c) Uranium
(d) Fossil fuels 

Solution
(b) Sun

Plants use solar energy to produce food. Fossil fuels are produced from dead plants. Hence directly or indirectly, sunlight is the source of all energy.


6. Which one of the following forms of energy leads to least environmental pollution in the process of its harnessing and utilisation?
(a) Nuclear energy
(b) Thermal energy
(c) Solar energy
(d) Geothermal energy

Solution
(c) Solar energy

In nuclear energy, metals like uranium are used which release harmful rays. In thermal energy, coal is used as a fuel. Burning of coal releases sulphur and carbon dioxide which causes air pollution. In geothermal energy, the energy is absorbed by the pressure which can sometimes be extremely hot and dangerous.


7. Ocean thermal energy is due to
(a) energy stored by waves in the ocean
(b) temperature difference at different levels in the ocean
(c) pressure difference at different levels in the ocean
(d) tides arising out in the ocean

Solution
(b) temperature difference at different levels in the ocean

Energy from the surface of the ocean is used to boil volatile liquids. Vapors from these are used to turn the turbine.


8. The major problem in harnessing nuclear energy is how to
(a) split nuclei?
(b) sustain the reaction?
(c) dispose off spent fuel safely?
(d) convert nuclear energy into electrical energy?

Solution
(c) dispose off spent fuel safely?

Nuclear plants produce highly radioactive materials. Thus disposing the nuclear waste in the environment will pose a major threat to the living organisms. Therefore, safe disposal of nuclear waste is a major problem in harnessing nuclear energy.


9. Which part of the solar cooker is responsible for green house effect?
(a) Coating with black colour inside the box
(b) Mirror
(c) Glass sheet
(d) Outer cover of the solar cooker

Solution
(c) Glass sheet 

The trapping of heat by the atmosphere leads to the greenhouse effect. Same thing happens in a solar cooker. Solar energy gets trapped which is then reflected by the glass sheet, this helps in cooking the food.


10. The main constituent of biogas is 
(a) methane 
(b) carbon dioxide 
(c) hydrogen 
(d) hydrogen sulphide 

Solution
(a) methane 

Methane gas is the major product of biogas. Around 70% of the gas is methane and remaining is ammonia and carbon dioxide.


11. The power generated in a windmill
(a) is more in rainy season since damp air would mean more air mass hitting the blades
(b) depends on the height of the tower
(c) depends on wind velocity
(d) can be increased by planting tall trees close to the tower

Solution
(c) depends on wind velocity

High speed winds will turn the turbine at a higher speed. This in turn will increase the production of electricity.


12. Choose the correct statement
(a) Sun can be taken as an inexhaustible source of energy
(b) There is infinite storage of fossil fuel inside the earth
(c) Hydro and wind energy plants are non polluting sources of energy
(d) Waste from a nuclear power plant can be easily disposed off

Solution
(a) Sun can be taken as an inexhaustible source of energy

Fossil fuels are exhaustible energy sources. Hydro and wind energy plants can change the environment and affect the land use, homes, and natural habitats there by having a large environmental impact. The disposal of waste from a nuclear power plant is the major issue faced by nuclear plant.


13. In a hydroelectric power plant more electrical power can be generated if water falls from a greater height because
(a) its temperature increases
(b) larger amount of potential energy is converted into kinetic energy
(c) the electricity content of water increases with height
(d) more water molecules dissociate into ions

Solution
(b) larger amount of potential energy is converted into kinetic energy


14. Choose the incorrect statement regarding wind power
(a) It is expected to harness wind power to minimum in open space
(b) The potential energy content of wind blowing at high altitudes is the source of wind power
(c) Wind hitting at the blades of a windmill causes them to rotate The rotation thus achieved can be utilised further
(d) One possible method of utilising the energy of rotational motion of the blades of a windmill is to run the turbine of an electric generator

Solution
(b) The potential energy content of wind blowing at high altitudes is the source of wind power

Kinetic energy of the wind runs the wind power.


15. Choose the incorrect statement
(a) We are encouraged to plant more trees so as to ensure clean environment and also provide bio-mass fuel
(b) Gobar-gas is produced when crops, vegetable wastes etc., decompose in the absence of oxygen (c) The main ingredient of bio-gas is ethane and it gives a lot of smoke and also produces a lot of residual ash
(d) Bio-mass is a renewable source of energy

Solution
(c) The main ingredient of bio-gas is ethane and it gives a lot of smoke and also produces a lot of residual ash

Methane is the main product of a biogas. Biogas burns with less smoke and does not produce residual ash.


Short Answer Questions 

16. Why is there a need to harness non-conventional sources of energy? Give two main reasons.

Solution
Following are the reasons to harness non-conventional sources of energy:

  • In the near future, the conventional sources will get exhausted.
  • Non-conventional source of energy do not cause environmental pollution.


17. Write two different ways of harnessing energy from ocean.

Solution
Different ways of harnessing energy form the ocean are:

  1. Tidal energy
  2. Ocean thermal energy


18. What steps would you suggest to minimise environmental pollution caused by burning of fossil fuels?

Solution
Steps to minimise pollution caused by burning of fossil fuels are as follows.

  • Use of non-conventional sources of energy.
  • Afforestion should be adopted to control air pollution.
  • Smokeless appliances should be used.


19. What is the role of a plane mirror and a glass sheet in a solar cooker?

Solution
Plane mirror increases the intensity of solar energy as it reflects the sunlight. Glass sheet prevents the escape of heat from the solar cooker.


20. Mention three advantages of a solar cell?

Solution
Advantages of a solar cell areas follows.

  • They are the source of renewable energy.
  • They do not cause pollution.
  • They require minimum maintenance.


21. What is biomass? What can be done to obtain bio-energy using biomass?

Solution
Biomass is the fuel obtained from organic material. The methods to obtain bio-energy using biomass are as follows:

  • Using firewood as a fuel.
  • Preparing cow dung cakes and using them as a fuel.
  • Preparation of gobar gas.


22. What are the limitations in obtaining energy from wind?

Solution

Limitations in obtaining energy from wind are as follows:

  • No continuous and consistent production.
  • Area required for installation of wind power mill is large.
  • Minimum wind speed should be 15 km/hour.


Long Answer Questions 


23. Which is the process used to harness nuclear energy these days? Explain it briefly.
Solution
Nuclear energy is harnessed in nuclear power plants. A nuclear power plant has a nuclear reactor, a boiler and a turbine.
  • Nuclear fission takes place in the nuclear reactor. This produces a large amount of energy.
  • Energy produced by nuclear fission is utilized to boil water, in order to generate steam.
  • Steam is then utilized to turn the turbines which produce electricity.

24. How can solar energy be harnessed? Mention any two limitations in using solar energy. How are these limitations overcome?
Solution
Solar energy can be harnessed by the following methods:
  1. Solar Cookers and solar water heaters:
    Black surfaces are good absorbers of heat as compared to a white or a reflecting surface. This is used in solar cookers and solar water heaters. Solar cookers are covered with a glass plate. Solar energy trapped within the solar cooker gets reflected by the glass plate and this helps in cooking the food.
  2. Solar Cell
    A typical cell when exposed to sunlight develops a voltage of 0.5–1 V and produces about 0.7 W of electricity. Solar panels are a large number of solar cells that deliver enough electricity for practical use.
Limitations of solar energy:
  • Cannot be harnessed beyond a certain latitude.
  • Cannot be harnessed during the night.
  • Can only be harnessed at places where there is plenty of sunlight.
  • It is an expensive source of energy.
Cheaper technologies must be discovered. Batteries can be used to store electrical energy form photovoltaic cells.

25. Make a list of conventional and non-conventional sources of energy. Give a brief description of harnessing one nonconventional source of energy.
Solution
Coal, petroleum, fire wood and tidal energy are conventional sources of energy. Solar energy, nuclear energy, Biogas, Tidal energy, Ocean thermal energy, wave energy etc. are non-conventional energy sources.
The various methods to harness non-conventional energy sources are:
  • Solar cookers, solar water heaters, solar cells and solar panels harness the solar energy.
  • Nuclear reactors convert nuclear power to electric energy.
  • The sun heats the water at the surface of the sea while the water in lower sections is relatively cold. This temperature difference is exploited to obtain energy in ocean-thermal-energy conversion plants.
  • The kinetic energy possessed by huge waves near the seashore can be trapped to generate electricity.

26. Why is there a need for harnessing non-conventional sources of energy? How can energy be harnessed from the sea in different ways?
Solution
Following are the reason for the need to harness non-conventional sources of energy:
  • In the near future, the conventional sources of energy will get exhausted in the near future.
  • Non-conventional source of energy do not cause environmental pollution. 
Energy can be harnessed from the sea in following ways:
  • Tidal energy: Tidal energy is harnessed by construction of a dam. A turbine fixed at the opening of the dam converts tidal energy to electricity.
  • Wave energy: Near the sea shore, the kinetic energy possessed by huge waves can be harnessed to generate electricity. Strong winds blowing across the sea, generate the waves.
  • Ocean Thermal Energy: The sun heats the water at the surface of the sea or ocean while the water in deeper layers is relatively cold. This temperature difference can be used to obtain energy in ocean-thermal-energy conversion plants.

27. What are the environmental consequences of using fossil fuels? Suggest the steps to minimise the pollution caused by various sources of energy including non-conventional sources of energy.

Solution
The various environmental consequences of using fossil fuels are:
  • Increased levels of carbon dioxide in the atmosphere lead to global warming.
  • Sulphur oxides causes acid rain.
  • Accumulation of smog in big cities which in turn reduces visibility
  • Air pollution which might cause health hazards to humans and animals.
The steps to minimize the pollution caused by various sources of energy including non-conventional sources of energy are as follows:
  • Use of public transport
  • Use of vehicles like electric bikes and bi-cycle
  • Adopting the 3R’s policy-reduce, reuse and recycle
  • Planting of trees

28. Energy from various sources is considered to have been derived from the sun. Do you agree? Justify your answer.
Solution
Energy from various sources is considered to have been derived from the sun. This statement can be justified by the following points: 
  • Plants trap sunlight and produce food which ultimately forms the biomass. Biomass gets converted to fossil fuels. So, in general the energy derived from the fossil fuels is indirectly harnessed from sunlight.
  • Biomass helps in obtaining wood and gobar gas.
  • The kinetic energy of the sun circulates the water on the earth.
  • The heating of water in the oceans causes waves and tides. Therefore, sun is indirectly responsible for harnessing tidal, wave and geothermal energy.

29. What is biomass? Explain the principle and working of a biogas plant using a labelled schematic diagram.
Solution
The fuel obtained from organic material is known as biomass. The anaerobic decomposition of cow-dung, various plant materials, vegetable waste and sewage gives bio-gas.
The biogas plant has a dome-shaped structure built with bricks. A slurry of cow-dung and water is made in the mixing tank which is then fed to the digester. The digester does not any supply of oxygen. Anaerobic micro-organisms decompose the complex compounds of the slurry. The decomposition process is completed in a few days and gases like methane, carbon dioxide, hydrogen and hydrogen sulphide are generated. This gas is stored in the gas tank above the digester from which it can be drawn through pipes for use.
Bio-gas contains up to 75% methane, hence it is an excellent fuel. It burns without smoke, leaves no residue like ash in wood, charcoal and coal burning. The heating capacity of bio-gas is also high.
The slurry left behind can be used as a manure as it is rich in nitrogen and phosphorous.

NCERT Exemplar Chapter 15 Our Environment Class 10 Science Solutions

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NCERT Exemplar Class 10 Science Chapter 15 Our Environment Solutions

NCERT Exemplar Solutions for Class 10 Science Chapter 15 Our Environment covers all the important questions and answers as well as advanced level questions. It helps in learning about the Ecosystem and its Components, biotic Components include living organisms like plants, animals, microbes, etc., Abiotic components include non-living organisms like soil, water, etc., food chain and food web.

The NCERT Exemplar solutions for class 10 science is very important for board exams. NCERT Exemplar Solutions for Class 10 Science Chapter 15 Our Environment is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like Trophic levels, producers, primary consumers, secondary consumers, tertiary consumers, herbivores, carnivores, decomposers, Pyramid of Trophic Levels, Ozone Layer and its Depletion, Garbage Management, Biodegradable Waste, Biodegradation and Non-biodegradable Waste.


Chapter Name

Chapter 15 Our Environment

Book Title

NCERT Exemplar for Class 10 Science

Related Study

  • NCERT Solutions for Class 10 Science Chapter 15 Our Environment
  • Revision Notes for Class 10 Science Chapter 15 Our Environment
  • MCQ for Class 10 Science Chapter 15 Our Environment
  • Important Questions for Class 10 Science Chapter 15 Our Environment

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 15 Our Environment Class 10 Science

Multiple Choice Questions

1. Which one of the following is an artificial ecosystem?
(a) Pond
(b) Crop field
(c) Lake
(d) Forest

Solution
(b) Crop field

Crop field is a man made ecosystem.


2. In a food chain, the third trophic level is always occupied by
(a) carnivores
(b) herbivores
(c) decomposers
(d) producers

Solution
(a) carnivores

First trophic level is producers, second tropic level is herbivores and the third trophic level is occupied by carnivores.


3. An ecosystem includes
(a) all living organisms
(b) non-living objects
(c) both living organisms and non-living objects
(d) sometimes living organisms and sometimes non-living objects

Solution
(c) both living organisms and non-living objects

An ecosystem includes living and nonliving organisms along with their interactions.


4. In the given food chain, suppose the amount of energy at fourth trophic level is 5 kJ, what will be the energy available at the producer level?
Grass → Grasshopper → Frog → Snake → Hawk
(a) 5 k J
(b) 50 k J
(c) 500 k J
(d) 5000 k J

Solution
(d) 5000 k J

The available energy at a particular trophic level is 10 times that of the energy available at the next tropic level. Hence, energy at third level trophic level is 50kJ that at the second level trophic is500 kJ and at first level trophic level (Producer) is5000 kJ.


5. Accumulation of non-biodegradable pesticides in the food chain in increasing amount at each higher trophic level is known as
(a) eutrophication
(b) pollution
(c) biomagnification
(d) accumulation

Solution
(c) biomagnification

  • Eutrophication is richness of nutrient beyond optimum level. Eutrophication occurs due to run off from land.
  • Introduction of an undesired substance into environment. Pollution leads to harmful effects on living organisms.

6. Depletion of ozone is mainly due to
(a) chlorofluorocarbon compounds
(b) carbon monoxide
(c) methane
(d) pesticides
Solution
(a) chlorofluorocarbon compounds

7.  Organisms which synthesise carbohydrates from inorganic compounds using radiant energy are called
(a) decomposers
(b) producers
(c) herbivores
(d) carnivores
Solution
(b) producers
Producers synthesize food from water and carbon dioxide in the presence of sunlight. Plants and few micro-organisms are the producers. Decomposers decompose the organic materials. Herbivores are the organisms that feed on plant and its products. Carnivores feed on other organisms.

8. In an ecosystem, the 10% of energy available for transfer from one trophic level to the next is in the form of
(a) heat energy
(b) light energy
(c) chemical energy
(d) mechanical energy
Solution
(c) chemical energy
Energy is transferred from one trophic level to the next in the form of food. Food is a chemical form of energy.

9. Organisms of a higher trophic level which feed on several types of organisms belonging to a lower trophic level constitute the
(a) food web
(b) ecological pyramid
(c) ecosystem
(d) food chain
Solution
(a) Food Web
In an ecological pyramid, number of organisms at the higher tropic level is always less than the number of organisms at the lower tropic level. An organism at the higher tropic level usually feeds on more than one type of organism from the lower tropic level.

10. Flow of energy in an ecosystem is always
(a) unidirectional
(b) bidirectional
(c) multi directional
(d) no specific direction
Solution
(a) unidirectional
Flow of energy takes place from the prey to the predator and it cannot be in reversed. Hence flow of energy is a unidirectional process.

11. Excessive exposure of humans to U V-rays results in
(i) damage to immune system
(ii) damage to lungs
(iii) skin cancer
(iv) peptic ulcers 
(a) (i) and (ii)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (iii) and (iv)
Solution
(c) (i) and (iii)
The upper surface of the body is affected by the UV rays. UV rays lead to skin cancer affect the immune system. This is because skin is considered as primary level of immune barrier.

12. In the following groups of materials, which group (s) contains only non-biodegradable items?
(i) Wood, paper, leather
(ii) Polythene, detergent, PVC
(iii) Plastic, detergent, grass
(iv) Plastic, bakelite, DDT 
(a) (iii)
(b) (iv)
(c) (i) and (iii)
(d) (ii) and (iv)
Solution
(d) (ii) and (iv)
Wood and leather are biodegradable. Grass is biodegradable too.

13. Which of the following limits the number of trophic levels in a food chain?
(a) Decrease in energy at higher trophic levels
(b) Sufficient food supply
(c) Polluted air
(d) Water
Solution
(a) Decrease in energy at higher trophic levels
At a particular tropic level, the available energy is 10 times the energy level available at the next tropic level. Therefore, a minute portion of energy from the producer is available when we reach the fourth trophic level.

14. Which of the statement is incorrect?
(a) All green plants and blue green algae are producers
(b) Green plants get their food from organic compounds
(c) Producers prepare their own food from inorganic compounds
(d) Plants convert solar energy into chemical energy
Solution
(b) Green plants get their food from organic compounds 
Green plants harness the solar energy and produce their food. Solar energy is used to prepare carbohydrate with the help of CO2 and water which are inorganic substances.

15. Which group of organisms are not constituents of a food chain?
(i) Grass, lion, rabbit, wolf
(ii) Plankton, man, fish, grasshopper
(iii) Wolf, grass, snake, tiger
(iv) Frog, snake, eagle, grass, grasshopper 
(a) (i) and (iii)
(b) (iii) and (iv)
(c) (ii) and (iii)
(d) (i) and (iv)
Solution
(c) (ii) and (iii)
In option (i) plankton does not eat grass. In option iii) none of them eat grass hence option c) is the right answer.

16. The percentage of solar radiation absorbed by all the green plants for the process of photosynthesis is about
(a) 1 %
(b) 5 %
(c) 8 %
(d) 10 %
Solution
(a) 1%
Green plants utilize 1% of radiation absorbed by leaf and use it for photosynthesis.

17. In the given Figure 15.1 the various trophic levels are shown in a pyramid. At which trophic level is maximum energy available?

(a) T4
(b) T2
(c) T1
(d) T3
Solution
(c) T1

Maximum energy is available for the producers. Only 10% of the energy is consumed by an organism at next level. Hence T1>T2>T3>T4.


18. What will happen if deer is missing in the food chain given below? Grass → Deer → Tiger

(a) The population of tiger increases
(b) The population of grass decreases
(c) Tiger will start eating grass
(d) The population of tiger decreases and the population of grass increases

Solution
(d) The population of tiger decreases and the population of grass increases

Lack of predators will increase population of grass. Due to lack of food, the population of tigerswill reduce.


19. The decomposers in an ecosystem
(a) convert inorganic material, to simpler forms
(b) convert organic material to inorganic forms
(c) convert inorganic materials into organic compounds
(d) do not breakdown organic compounds

Solution
(b) convert organic material to inorganic forms


20. If a grass hopper is eaten by a frog, then the energy transfer will be from
(a) producer to decomposer
(b) producer to primary consumer
(c) primary consumer to secondary consumer
(d) secondary consumer to primary consumer

Solution
(c) primary consumer to secondary consumer

As the grass hopper feeds on grass, it is a primary consumer. If the grasshopper is being eaten by a frog then the frog will be a secondary consumer.


21. Disposable plastic plates should not be used because
(a) they are made of materials with light weight
(b) they are made of toxic materials
(c) they are made of biodegradable materials
(d) they are made of non-biodegradable materials

Solution
(d) they are made of non-biodegradable materials

Plastics start accumulating in nature, thereby harming living organisms as they are non-biodegradable.


Short Answer Questions 

22. Why is improper disposal of waste a curse to environment? 

Solution
Wastes pollute the environment and cause harmful effects on all living organisms.


23. Write the common food chain of a pond ecosystem.

Solution

Phyto planktons and aquatic plants → small aquatic animals larvae, insects etc. → fish → bird


24. What are the advantages of cloth bags over plastic bags during shopping?

Solution
Advantages of cloth bags are:

  • capable of carrying more things
  • made of biodegradable material
  • do not pollute our environment
  • can be reused


25. Why are crop fields known as artificial ecosystems?

Solution
Crop fields are man-made. Therefore, the biotic and abiotic components are manipulated by humans.


26. Differentiate between biodegradable and non-biodegradable substances. Cite examples. 

Solution

Biodegradable substances

Non-biodegradable substances

Substances which can be broken down into simpler substances by biological processes are known as biodegradable substances.

Substances that cannot be broken down into simpler substances by biological processes are said to be non-biodegradable.

Example –wood, paper, etc.

Example –plastic, DDT, etc.


27. Suggest one word for each of the following statements/ definitions
(a) The physical and biological world where we live in
(b) Each level of food chain where transfer of energy takes place
(c) The physical factors like temperature, rainfall, wind and soil of an ecosystem
(d) Organisms which depend on the producers either directly or indirectly for food

Solution

(a) Environment 
(b) Trophic level 
(c) Abiotic factors 
(d) Consumers or heterotrophs 


28. Explain the role of decomposers in the environment?

Solution

Decomposers break down the dead and decaying organic matter and return the nutrients to the soil. Thus, they play a very important role in the nutrient recycling in the environment.


29. Select the mis-matched pair in the following and correct it.
(a) Biomagnification — Accumulation of chemicals at the successive trophic levels of a food chain (b) Ecosystem — Biotic components of environment
(c) Aquarium — A man-made ecosystem
(d) Parasites — Organisms which obtain food from other living organisms

Solution
(b) Ecosystem — Biotic components of environment 

Both biotic and abiotic components of environment constitute an ecosystem.


30. We do not clean ponds or lakes, but an aquarium needs to be cleaned. Why?

Solution
An aquarium is an artificial ecosystem which is incomplete ecosystem when compared to pond or lake which is natural and complete ecosystem. The abiotic components are not supplied naturally to the aquarium.


Long Answer Questions 

31. Indicate the flow of energy in an ecosystem. Why is it unidirectional? Justify.

Solution

  1. In a terrestrial ecosystem, the green plants capture about 1% of the energy of sunlight and convert it into food energy.
  2. When primary consumers eat these green plants, a great deal of energy is lost as heat to the environment. But some amount of energy goes into digestion and in doing work. The rest goes towards growth and reproduction of the primary consumer.
  3. Approximately 10% of the food eaten by the primary consumer is turned into its own body and made available for the next level of consumers.
  4. Therefore, 10% can be taken as the average value for the amount of organic matter that is present at each step and reaches the next level of consumers.
  5. As very little energy is available for the next level of consumers, the food chains generally consist of only three or four steps.
  6. Generally, there are a greater number of individuals at the lower trophic levels.
  7. The length and complexity of food chains vary greatly.
  8. Each organism is generally eaten by two or more other kinds of organisms which in turn are eaten by several other organisms.
  9. So instead of a straight line food chain, the relationship can be shown as a series of branching lines called a food web.

The energy captured by the autotrophs does not revert back to the sun hencethe flow of energy is unidirectional. The energy is no longer available to the previous level as it moves progressively through the various trophic levels. Secondly, due to loss of energy at each level the energy available at each trophic level gets diminished progressively.


32. What are decomposers? What will be the consequence of their absence in an ecosystem?

Solution
Decomposers are microorganisms like bacteria and fungi which break-down the dead remains and waste products of organisms. They break-down the complex organic substances into simple inorganic substances that go into the soil and are used up once more by the plants.

Recycling of material in the biosphere will not take place if the decomposers are absent in the ecosystem. This in turn will lead to accumulation of dead plants and animals in the environment. Moreover, the environment will be completely devoid of all its resources needed to maintain and sustain life.


33. Suggest any four activities in daily life which are eco-friendly

Solution
The following activities in daily life are eco-friendly:

  • Use of bicycles and electrical bikes instead of vehicles running on fossil fuels.
  • Avoid use plastic bags.
  • Plant trees in our surroundings.
  • Recycle old items and use them.


34. Give two differences between food chain and food web.
Solution 

Food Chain

Food web

The food chain is the straight and single pathway for the flow of energy in an ecosystem, through different species of organisms.

Food web is defined as the convoluted or complicated pathway of an ecosystem consist of numerous food chains of the different trophic level, through which the energy flow.

Food chain is hypothetical situation

Food web is a real life situation.

Members of higher trophic level feed on single type of organism of lower trophic level

Members of higher trophic level can feed upon organisms of the lower trophic levels of other food chain.


35. Name the wastes which are generated in your house daily. What measures would you take for their disposal?

Solution
The following wastes are generated in our house daily:

  • Kitchen wastes
  • Paper wastes like newspapers, bags, envelopes
  • Plastic bags

Measures to be taken in order to dispose house waste areas follows:

  • Segregation of biodegradable and non-biodegradable wastes.
  • Safe disposal of plastic bags.
  • Vegetable/fruit peels can be placed near trees/plants, which on decomposition will enrich the soil with nutrients.
  • Give paper wastes for recycling.
  • Prepare a compost pit for kitchen wastes.


36. Suggest suitable mechanism (s) for waste management in fertiliser industries.

Solution
The following steps must be taken in order to manage the waste in fertiliser industry:

  • Use of catalytic converters in chimneys: Catalytic converters help in minimizing the release of harmful gases and suspended particulate matter.
  • Use of electrostatic precipitators in chimneys: These help in minimizing the release of SPMs, thereby reducing air pollution.
  • Effluent treatment: An effluent treatment plant should be installed in the factory. The plant should have a proper facility for physical, chemical and biological treatment of effluent.


37. What are the by-products of fertiliser industries? How do they affect the environment?

Solution
The major by-products of fertilizer industry are ammonia, nitrates, urea, acids and alkalis, oil, arsenic, MEA, cyanides, sulphides, phosphates, etc.

The effects of these by-products on the environment are as follows:

  • Suspended solid and ash increase the air pollution.
  • Arsenic and cyanides may enter the food chain and cause bio magnification.
  • Fluorides in drinking water may be harmful for the teeth.
  • The pH of water is altered if the by-products reach the water bodies. This harms the aquatic ecosystem.


38. Explain some harmful effects of agricultural practices on the environment.

Solution
Some harmful effects of agricultural practices on the environment are as follows:

  • The chemistry of the soil changes because of excessive use of chemical fertilizers which kills useful microbes.
  • Bio-magnification is caused due to excessive use of non-biodegradable chemical pesticides.
  • Extensive cropping causes loss of soil fertility.
  • The ground water table lowers because of excessive use of ground water for agriculture.
  • Natural ecosystem/habitat is damaged.

NCERT Exemplar Chapter 16 Sustainable Management of Natural Resources Class 10 Science Solutions

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NCERT Exemplar Class 10 Science Chapter 16 Sustainable Management of Natural Resources Solutions

NCERT Exemplar Solutions for Class 10 Science Chapter 16 Sustainable Management of Natural Resources covers all the important questions and answers as well as advanced level questions. It helps in learning about the Ecosystem and its Components, biotic Components include living organisms like Pollution in Ganga, GAP (Ganga Action Plan) 1985, 5 R’s to save the environment, Forests and Wildlife, biodiversity hotspots, conservation of forests, Monoculture and People intervention in forests.

The NCERT Exemplar solutions for class 10 science is very important for board exams. NCERT Exemplar Solutions for Class 10 Science Chapter 16 Sustainable Management of Natural Resources is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like Chipko Movement, Water Management, dams, water harvesting systems, Groundwater, Coal, Petroleum and fossil fuels.


Chapter Name

Chapter 16 Sustainable Management of Natural Resources

Book Title

NCERT Exemplar for Class 10 Science

Related Study

  • NCERT Solutions for Class 10 Science Chapter 16 Sustainable Management of Natural Resources
  • Revision Notes for Class 10 Science Chapter 16 Sustainable Management of Natural Resources
  • MCQ for Class 10 Science Chapter 16 Sustainable Management of Natural Resources
  • Important Questions for Class 10 Science Chapter 16 Sustainable Management of Natural Resources

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 16 Sustainable Management of Natural Resources Class 10 Science

Multiple Choice Questions

1. From the list given below pick the item that is not a natural resource
(a) Soil
(b) Water
(c) Electricity
(d) Air

Solution
(c) Electricity

Soil, water and air are natural resources whereas electricity is produced by humans.


2. The most rapidly dwindling natural resource in the world is
(a) water
(b) forests
(c) wind
(d) sunlight

Solution
(b) forests

Water, wind and sunlight are non-exhaustible resources. But due to human activities, forests are depleting.


3. The most appropriate definition of a natural resource is that it is a substance/commodity that is
(a) present only on land
(b) a gift of nature which is very useful to mankind
(c) a man-made substance placed in nature
(d) available only in the forest

Solution
(b) a gift of nature which is very useful to mankind

Natural resources are not man made and they are present everywhere on earth.


4. The main cause for abundant coliform bacteria in the river Ganga is
(a) disposal of unburnt corpses into water
(b) discharge of effluents from electroplating industries
(c) washing of clothes
(d) immersion of ashes

Solution
(a) disposal of unburnt corpses into water

Human excreta mainly contain coliforms. Therefore, unburnt corpses are the sources of coliform which pollute Ganga river water.


5. The pH of water sample collected from a river was found to be acidic in the range of 3.5 – 4.5, on the banks of the river were several factories that were discharging effluents into the river. The effluents of which one of the following factories is the most likely cause for lowering the pH of river water? 

(a) Soap and detergent factory
(b) Lead battery manufacturing factory
(c) Plastic cup manufacturing factory
(d) Alcohol distillery

Solution
(d) Alcohol distillery

Acid spillage from lead battery manufacturing factory can lower down the pH.


6. The pH range most conducive for life of fresh water plants and animals is
(a) 6.5 – 7.5
(b) 2.0 – 3.5
(c) 3.5 – 5.0
(d) 9.0 – 10.5

Solution
(a) 6.5 – 7.5

Neutral pH is the most conducive for life of fresh water plants and animals.


7. The three R’s that will help us to conserve natural resources for long term use are
(a) recycle, regenerate, reuse
(b) reduce, regenerate, reuse
(c) reduce, reuse, redistribute
(d) reduce, recycle, reuse

Solution
(d) reduce, recycle, reuse

Reduce, recycle and reuse can conserve the natural resources.


8. Given below are a few statements related to biodiversity. Pick those that correctly describe the concept of biodiversity 

(i) Biodiversity refers to the different species of flora and fauna present in an area
(ii) Biodiversity refers to only the flora of a given area
(iii) Biodiversity is greater in a forest
(iv) Biodiversity refers to the total number of individuals of a particular species living in an area 

(a) (i) and (ii)
(b) (ii) and (iv)
(c) (i) and (iii)
(d) (ii) and (iii)

Solution
(c) (i) and (iii)

Biodiversity includes flora and fauna and it does not describe just one species.


9. Among the statements given below select the ones that correctly describe the concept of sustainable development
(i) Planned growth with minimum damage to the environment
(ii) Growth irrespective of the extent of damage caused to the environment
(iii) Stopping all developmental work to conserve the environment
(iv) Growth that is acceptable to all the stakeholders 

(a) (i) and (iv)
(b) (ii) and (iii)
(c) (ii) and (iv)
(d) (iii) only

Solution
(a) (i) and (iv)

Sustainable development refers to planned growth with minimum damage to the environment while keeping the future of coming generation in mind.


10. In our country, vast tracts of forests are cleared and a single species of plant is cultivated. This practice promotes
(a) biodiversity in the area
(b) monoculture in the area
(c) growth of natural forest
(d) preserves the natural ecosystem in the area

Solution
(b) monoculture in the area

Monoculture is cultivation of a single species of plant.


11. A successful forest conservation strategy should involve
(a) protection of animals at the highest trophic level
(b) protection of only consumers
(c) protection of only herbivores
(d) comprehensive programme to protect all the physical and biological components

Solution
(d) comprehensive programme to protect all the physical and biological components

Balance in the biotic and abiotic community is very essential in the conservation of forests.


12. The important message conveyed by the ‘Chipko Movement’ is
(a) to involve the community in forest conservation efforts
(b) to ignore the community in forest conservation efforts
(c) to cut down forest trees for developmental activities
(d) government agencies have the unquestionable right to order destruction of trees in forests

Solution
(a) to involve the community in forest conservation efforts

'Chipko Movement' represents sacred feelings of community towards the local forest areas.


13. In our country, there are attempts to increase the height of several existing dams like Tehri and Almati, dams across Narmada. Choose the correct statements among the following that are a consequence of raising the height of dams
(i) Terrestrial flora and fauna of the area is destroyed completely
(ii) Dislocation of people and domestic animals living in the area
(iii) Valuable agricultural land may be permanently lost
(iv) It will generate permanent employment for people 

(a) (i) and (ii)
(b) (i), (ii) and (iii)
(c) (ii) and (iv)
(d) (i), (iii) and (iv)

Solution
(d) (i), (iii) and (iv)

Dam construction causes destruction of downstream agricultural land and adversely affects the living organisms present there.


14. Expand the abbreviation GAP
(a) Governmental Agency for Pollution Control
(b) Gross Assimilation by Photosynthesis
(c) Ganga Action Plan
(d) Governmental Agency for Animal Protection

Solution
(c) Ganga Action Plan


15. Select the incorrect statement
(a) Economic development is linked to environmental conservation
(b) Sustainable development encourages development for current generation and conservation of resources for future generations
(c) Sustainable development does not consider the view points of stakeholders
(d) Sustainable development is a long planned and persistent development

Solution
(c) Sustainable development does not consider the view points of stakeholders

Sustainable developments take into consideration both economic growth and ecological conservation simultaneously.


16. Which of the following is not a natural resource?
(a) Mango tree
(b) Snake
(c) Wind
(d) Wooden house

Solution
(d) Wooden house

Resources that occur naturally within environment are called as natural resources.


17. Select the wrong statement
(a) Forests provide variety of products
(b) Forests have greater plant diversity
(c) Forests do not conserve soil
(d) Forests conserve water

Solution
(c) Forests do not conserve soil

Forests prevent soil erosion, add fertility to the soil by adding humus, and thus they conserve the soil and its fertility.


18. Arabari forests of Bengal is dominated by
(a) Teak
(b) Sal
(c) Bamboo
(d) Mangroove

Solution
(b) Sal

Arabari forests of Bengal are dominated by sal trees.


19. Ground water will not be depleted due to
(a) afforestation
(b) thermal power plants
(c) loss of forest, and decreased rainfall
(d) cropping of high water demanding crops

Solution
(b) thermal power plants

The roots of the plants make the soil pervious. Thus, afforestation helps in recharging the ground water.


20. Opposition to the construction of large dams is due to
(a) social reasons
(b) economic reasons
(c) environmental reasons
(d) all the above

Solution
(d) all the above

The ecosystem near the dam is destroyed. This is an environmental reason. Relocation of people is a social reason. Agricultural land and employment are economic reasons.


21. Khadins, Bundhis, Ahars and Kattas are ancient structures that are examples for
(a) grain storage
(b) wood storage
(c) water harvesting
(d) soil conservation

Solution
(c) water harvesting


22. Pick the right combination of terms which has no fossil fuel.
(a) Wind, ocean and coal
(b) Kerosene, wind and tide
(c) Wind, wood, sun
(d) Petroleum, wood, sun

Solution
(c) Wind, wood, sun

Coal, kerosene and petroleum are fossil fuels.


23. Select the eco-friendly activity among the following
(a) Using car for transportation
(b) Using polybags for shopping
(c) Using dyes for colouring clothes
(d) Using windmills to generate power for irrigation

Solution
(d) Using windmills to generate power for irrigation

Using car for transportation, using polybags and dyes for coloring clothes increases the environmental pollution.


24. It is important to make small check dams across the flooded gullies because they
(i) hold water for irrigation
(ii) hold water and prevent soil erosion
(iii) recharge ground water
(iv) hold water permanently 

(a) (i) and (iv)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iv)

Solution
(b) (ii) and (iii)

Check dams prevent soil erosion. Check dams also help in improvement of groundwater level by holding the water.


Short Answer Questions 

25. Prepare a list of five items that you use daily in the school. Identify from the list such items that can be recycled.

Solution
The following items are used in our daily life in schools: paper, rexin bag, blade, pen, plastic box, scale, eraser, compass and dividers (metallic), steel lunch box, steel spoon.

Paper, blade, plastic box, eraser, compass, steel lunch box and steel spoon can be recycled.


26. List two advantages associated with water harvesting at the community level.

Solution
Two advantages associated with water harvesting at the community level are:

  • Due to recharging of wells the ground water level increases.
  • Ground water keeps the layers of soil above it moist and prevents loss of water by evaporation.


27. In a village in Karnataka, people started cultivating crops all around a lake which was always filled with water. They added fertilisers to their field in order to enhance the yield. Soon they discovered that the waterbody was completely covered with green floating plants and fishes started dying in large numbers.
Analyse the situation and give reasons for excessive growth of plants and death of fish in the lake.

Solution
Excessive use of fertilisers in the fields carried the chemicals down to the lake during rains. Many fertilisers contain phosphates and nitrates. The surface of water gets completely covered with plants as these chemicals promote excessive growth of aquatic plants. This process is known as eutrophication. This in turn depletes the availability of light, dissolved oxygen and nutrients in water, thereby resulting in the death of fishes.


28. What measures would you take to conserve electricity in your house?

Solution

The measures taken to conserve electricity in our houses are as follows:

  • Put off the fans and lights when they are not in use.
  • Maximum use of solar radiation.
  • During winters, use solar water heating system.
  • Use of fluorescent tubes or CFL.


29. Although coal and petroleum are produced by degradatioin of bio- mass, yet we need to conserve them. Why?

Solution
Both the energy sources take millions of years for their formation. These resources are being utilised at a much faster rate than their formation rate, hence they will be exhausted in the near future. This is the reason why they need to be conserved.


30. Suggest a few measures for controlling carbon dioxide levels in the atmosphere.

Solution

A few measures for controlling carbon dioxide levels in the atmosphere are as follows:
  • Reduce the consumption of petrol in the automobiles.
  • Use CNG or a clean fuel.
  • Instead of burning litter prepare manure out of it.
  • Plant more trees.


31. (a) Locate and name the water reservoirs in Figures16.1 (a) and (b).
(b) Which has an advantage over the other and why?

Solution

(a) The water reservoir is a pond in Figure 16.1 (a) and underground water body in the Figure 16.1 (b).

(b) Figure 16.1 (b) has more advantage than Figure 16.1 (a), because of the following reasons:

  • Water does not evaporate.
  • Water is spread out to recharge wells.
  • Water provides moisture for vegetation over a wide area.
  • Water is protected from contamination by animal and human wastes.
  • Water prevents breeding of insects


Long Answer Questions 

32. In the context of conservation of natural resources, explain the terms reduce, recycle and reuse. From among the materials that we use in daily life, identify two materials for each category. 

Solution

Reduce: Reduce means to use a material/commodity in lesser quantity. For example: electricity and water.

Recycle: Recycle means a used material is collected and sent back to a manufacturer so that the manufacturer can make some other useful material from it. For example: plastic cups and buckets, glass tumbler, paper, metal objects.

Reuse: Reuse means using a thing over and over again instead of throwing it away. For example: used envelopes, plastic carry bags, bottles of jam.


33. Prepare a list of five activities that you perform daily in which natural resources can be conserved or energy utilisation can be minimised.

Solution
A list of activities which we can perform daily to conserve the natural resources or minimize the energy utilisation is:

  1. Unused water in the water bottle can be used to water the plants.
  2. Avoid using a hose pipe for watering plants.
  3. Wash vehicles only when they are dirty.
  4. Use fan and light only when required.
  5. Use CFL in place of conventional bulbs/tubes.


34. Is water conservation necessary? Give reasons. 

Solution
The total amount of fresh water is more than enough to meet the needs of human beings. But because of its uneven distribution, wide seasonal as well as yearly fluctuation in rainfalls and wastage water shortage, there is a need to conserve water.


35. Suggest a few useful ways of utilising waste water.

Solution
Waste water can be used for:

  • Replenishing the ground water.
  • Irrigation
  • Washing cars and watering the gardens.


36. What is the importance of forest as a resource?

Solution
Forests are renewable resources which provide:

  • Habitat, food, protection to wild life.
  • Help in balancing CO2 and O2 of atmosphere.
  • Improve the water holding capacity of soil.
  • Regulate water cycle.
  • Provide useful products like fruits, resins, gums, essential oils, bidi wrapper etc.


37. Why are the Arabari forests of Bengal known to be a good example of conserved forest?

Solution
The forest department developed a strategy in which the villagers were involved in the protection of the forest. The villagers were paid in turn for the labour and they also had some benefit in harvesting operations. The villagers were allowed to collect wood and fodder on nominal fee. Therefore, in this way, by the active and willing participation of the local people, the sal forests of Arabari were conserved.

NCERT Exemplar Chapter 1 Matter in Our Surroundings Class 9 Science Solutions

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NCERT Exemplar Class 9 Science Chapter 1 Matter in Our Surroundings Solutions

NCERT Exemplar Solutions for Class 9 Science Chapter 1 Matter in Our Surroundings covers all the important questions and answers as well as advanced level questions. It helps in learning about the Matter like air we breathe, the cloud, the plants, the stars, the books, the food we eat, the animals and even a small water droplet are made of matter, classification of matter, Physical Nature of Matter and size of the particles that make matter.

The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 1 Matter in Our Surroundings is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like Particles of Matter, States of Matter which are solid, liquid, gasses, conversion of states of matter, force of attraction, Evaporation, boiling and latent heat of vapourization.


Chapter Name

Chapter 1 Matter in Our Surroundings

Book Title

NCERT Exemplar for Class 9 Science

Related Study

  • NCERT Solutions for Class 9 Science Chapter 1 Matter in Our Surroundings
  • Revision Notes for Class 9 Science Chapter 1 Matter in Our Surroundings
  • MCQ for Class 9 Science Chapter 1 Matter in Our Surroundings
  • Important Questions for Class 9 Science Chapter 1 Matter in Our Surroundings

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 1 Matter in Our Surroundings Class 9 Science

Multiple Choice Questions

1. Which one of the following sets of phenomena would increase on raising the temperature?
(a) Diffusion, evaporation, compression of gases
(b) Evaporation, compression of gases, solubility
(c) Evaporation, diffusion, expansion of gas
(d) Evaporation, solubility, diffusion, compression of gases

Solution

(c) Evaporation, diffusion, expansion of gas
The particles of matter possess kinetic energy as they are continuously moving. Increase in kinetic energy increases the rate of diffusion. Evaporation is the conversion of liquid state into vapour state. As the temperature is increased the molecules move and vibrate quickly, this enables them to escape into the atmosphere in the form of vapours. Increase in temperature also causes the molecules move faster, this results in expansion of gases.


2. Seema visited a Natural Gas Compressing Unit and found that the gas can be liquefied under specific conditions of temperature and pressure. While sharing her experience with friends she got confused. Help her to identify the correct set of conditions :
(a) Low temperature, low pressure
(b) High temperature, low pressure
(c) Low temperature, high pressure
(d) High temperature, high pressure

Solution

(c) Low temperature, high pressure

Low temperature and high pressure are required to compress a gas into a liquid. On applying pressure, the particles of the gas get closer and start attracting each other thereby forming a liquid. This generates a lot of heat. Hence it is necessary to cool the liquid. Cooling lowers the temperature of compressed gas and helps in liquefying it.

3. The property to flow is unique to fluids. Which one of the following statements is correct?
(a) Only gases behave like fluids
(b) Gases and solids behave like fluids
(c) Gases and liquids behave like fluids
(d) Only liquids are fluids

Solution

(c) Gases and liquids behave like fluids

The intermolecular force of attraction between the particles is less in case of gases and liquids. This facilitates the flow of gases and liquids.


4. During summer, water kept in an earthen pot becomes cool because of the phenomenon of :
(a) diffusion
(b) transpiration
(c) osmosis
(d) evaporation

Solution

(d) evaporation
Evaporation of water through pores of earthen pot reduces the temperature of immediate surroundings. This cools the water after some time. Transpiration is the movement of water through plants. The movement of a substance from a region of higher concentration to a region of lower concentration is known as diffusion. If the molecules of a solvent pass through a semipermeable membrane from a less concentrated solution into a more concentrated one then the process is known as osmosis.


5. A few substances are arranged in the increasing order of ‘forces of attraction’ between their particles. Which one of the following represents a correct arrangement ?
(a) Water, air, wind
(b) Air, sugar, oil
(c) Oxygen, water, sugar
(d) Salt, juice, air

Solution

(c) Oxygen, water, sugar
Intermolecular forces of attraction are least in gases than liquids and most in solids. Oxygen is a gas, water is a liquid and sugar is a crystalline solid.


6. On converting 25°C, 38°C and 66°C to Kelvin scale, the correct sequence of temperature will be :
(a) 298 K, 311 K and 339 K
(b) 298 K, 300 K and 338 K
(c) 273 K, 278 K and 543 K
(d) 298 K, 310 K and 338 K

Solution

(a) 298 K, 311 K and 339 K
Add 273 to respective temperature to get temperature in kelvin.
25° + 273 = 298K
38° + 273 = 311K
66° + 273 = 339K


7. Choose the correct statement of the following :
(a) Conversion of solid into vapours without passing through the liquid state is called vaporization.
(b) Conversion of vapours into solid without passing through the liquid state is called sublimation.
(c) Conversion of vapours into solid without passing through the liquid state is called freezing.
(d) Conversion of solid into liquid is called sublimation.

Solution

(b) Conversion of vapours into solid without passing through the liquid state is called sublimation.
Sublimation is the process in which a solid changes into vapours without passing through the liquid state. Vaporisation is a phase transition from liquid to vapour phase. Conversion of liquid to solid at the freezing temperature is called freezing.


8. The boiling points of diethyl ether, acetone and n-butyl alcohol are 35°C, 56°C and 118°C respectively. Which one of the following correctly represents their boiling points in Kelvin scale?
(a) 306 K, 329 K, 391 K
(b) 308 K, 329 K, 392 K
(c) 308 K, 329 K, 391 K
(d) 329 K, 392 K, 308 K

Solution

(c) 308 K, 329 K, 391 K
Temperature expressed in Celsius can be converted into kelvin by adding 273. Boiling points of diethyl ether, acetone and n-butyl alcohol are 308 K, 329 K, and 391 K respectively.


9. Which condition out of the following will increase the evaporation of water?
(a) Increase in temperature of water
(b) Decrease in temperature of water
(c) Less exposed surface area of water
(d) Adding common salt to water

Solution

(a) Increase in temperature of water
Increasing the temperature of water increases the kinetic energy of the water. Because of this more particles attain energy which is sufficient to convert into vapour state. This is the reason why an increase in temperature increases the rate of evaporation.


10. In which of the following conditions, the distance between the molecules of hydrogen gas would increase?
(i) Increasing pressure on hydrogen contained in a closed container.
(ii) Some hydrogen gas leaking out of the container.
(iii) Increasing the volume of the container of hydrogen gas.
(iv) Adding more hydrogen gas to the container without increasing the volume of the container.
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and ( (iv)

Solution

(c) (ii) and (iii)
Either the volume of hydrogen gas should be reduced or the volume of the container should be increased in order to increase the intermolecular interaction. The inter-molecular interaction would also decrease if the pressure is increased or hydrogen is added without increasing container volume.


Short Answer Questions

11. A sample of water under study was found to boil at 102°C at normal temperature and pressure. Is the water pure? Will this water freeze at 0°C? Comment.

Solution

The boiling point and melting point of pure water at 1 atmospheric pressure are100 °C is 0°C respectively. In the question, it is given that the water boils at 102°C, this means that the water is not pure.
The water will freeze at temperature below 0°C.


12. A student heats a beaker containing ice and water. He measures the temperature of the content of the beaker as a function of time. Which of the following figure would correctly represent the result? Justify your choice.

Solution

Graph (d) correctly represents the result. This is because the temperature of the mixture will be 0° at the start of the experiment as water and ice are in equilibrium. Now, as the students start heating, initially heat will be utilised to melt the ice. Therefore, there will be no change in temperature for certain amount of time. When the students heat the mixture further the temperature of the water increases.


13. Fill in the blanks :
(a) Evaporation of a liquid at room temperature leads to a _____ effect.
(b) At room temperature the forces of attraction between the particles of solid substances are_____ than those which exist in the gaseous state.
(c) The arrangement of particles is less ordered in the_____ state. However, there is no order in the _____ state.
(d) _____ is the change of gaseous state directly to solid state without going through the _____ state.
(e) The phenomenon of change of a liquid into the gaseous state at any temperature below its boiling point is called _____.

Solution

(a) Evaporation of a liquid at room temperature leads to a cooling effect.

(b) At room temperature the forces of attraction between the particles of solid substances are stronger than those which exist in the gaseous state.

(c) The arrangement of particles is less ordered in the liquid state. However, there is no order in the gaseous state.

(d) Sublimation is the change of solid state directly to vapour state without going through the liquid state.

(e) The phenomenon of change of a liquid into the gaseous state at any temperature below its boiling point is called evaporation.


14. Match the physical quantities given in column A to their SI units given in column B :

A

B

(a) Pressure

(i) cubic metre

(b) Temperature

(ii) kilogram

(c) Density

(iii) pascal

(d) Mass

(iv) kelvin

(e) Volume

(v) kilogram per cubic metre

Solution

A

B

(a) Pressure

(iii) pascal

(b) Temperature

(iv) kelvin

(c) Density

(v) kilogram per cubic metre

(d) Mass

(ii) kilogram

(e) Volume

(i) cubic metre


15. The non-SI and SI units of some physical quantities are given in column A and column B respectively. Match the units belonging to the same physical quantity :

A

B

(a) degree Celsius

(i) kilogram
(b) centimeter(ii) pascal

(c) gram per centimeter cube

(iii) metre

(d) bar

(iv) kelvin

(e) miligram

(v) kilogram per metre cube

Solution

A

B

(a) degree Celsius

(iv) kelvin
(b) centimeter(iii) metre

(c) gram per centimeter cube

(v) kilogram per metre cube

(d) bar

(ii) pascal

(e) miligram

(i) kilogram


16. ‘Osmosis is a special kind of diffusion.’ Comment.

Solution

Yes, osmosis is a special kind of diffusion. This is because there is movement of particles from regions of higher concentration to regions of lower concentration. But in osmosis the movement of solvent is through a semi permeable membrane which is permeable only to water molecules.


17. Classify the following into osmosis/diffusion :
(i) Swelling up of a raisin on keeping in water.
(ii) Spreading of virus on sneezing.
(iii) Earthworm dying on coming in contact with common salt.
(iv) Shrinking of grapes kept in thick sugar syrup.
(v) Preserving pickles in salt.
(vi) Spreading of smell of cake being baked throughout the house.
(vii) Aquatic animals using oxygen dissolved in water during respiration.

Solution

(i) Osmosis
(ii) Diffusion
(iii) Osmosis
(iv) Osmosis
(v) Osmosis
(vi) Diffusion
(vii) Diffusion


18. Water as ice has a cooling effect, whereas water as steam may cause severe burns. Explain these observations.

Solution

The water molecules have low energy incase of ice whereas in the case of steam the water molecules have high energy. This high energy of water molecules in case of steam is transformed and may cause burns. On the other hand, the water molecules in case of ice take heat energy from the body thereby producing a cooling effect.


19. Alka was making tea in a kettle. Suddenly she felt intense heat from the puff of steam gushing out of the spout of the kettle. She wondered whether the temperature of the steam was higher than that of the water boiling in the kettle. Comment.

Solution

The temperature of both boiling water and steam is 100°C, but steam has more energy because of latent heat of vapourisation.


20. A glass tumbler containing hot water is kept in the freezer compartment of a refrigerator (temperature < 0°C). If you could measure the temperature of the content of the tumbler, which of the following graphs (fig.) would correctly represent the change in its temperature as a function of time.

Solution

Graph (a) correctly represents the change in the temperature of the content of the glass tumbler as a function of time. This can be explained as the water cools initially, it reaches 0°C, i.e. the freezing point. At this stage the temperature will remain constant till all the water freezes. The temperature falls again after this temperature.


21. Look at fig. and suggest in which of the vessels A, B, C or D the rate of evaporation will be the highest. Explain.

Solution

The rate of evaporation will be maximum in case of vessel c.

This is because evaporation is a surface phenomenon. Hence the rate of evaporation increases with an increase of surface area. Also, the particles of water vapour will move away with the air when there is an increase in air speed, this also increases the rate of evaporation.


22. (a) Conversion of solid to vapour is called sublimation. Name the term used to denote the conversion of vapour to solid.
(b) Conversion of solid state to liquid state is called fusion; what is meant by latent heat of fusion?

Solution

(a) Conversion of vapour to solid is known as deposition.

(b) Latent heat of fusion is the amount of heat required to melt1 kg of solid into liquid at one atmosphere pressure at its melting point.\


Long Answer Questions

23. You are provided with a mixture of naphthalene and ammonium chloride by your teacher. Suggest an activity to separate them with well labelled diagram.

Solution

Naphthalene being a non-polar compound is insoluble in water but soluble in ether an organic solvent. Also, it is volatile at room temperature. Similarly, ammonium chloride is a polar compound and thus soluble in water. It is also volatile at higher temperature. By decantation of aqueous mixture, the mixture of naphthalene and ammonium chloride can be separated. Naphthalene remains undissolved in water and can be separated out in the funnel. The filtrate of ammonium chloride can be evaporated to obtain dry ammonium chloride.


24. It is a hot summer day, Priyanshi and Ali are wearing cotton and nylon clothes respectively. Who do you think and would be more comfortable and why?

Solution

Cotton absorbs the sweat as it is a better absorber of water than nylon. Further evaporation of sweat form the cotton shirt will lead to cooling. Hence Priyanshi is more comfortable.


25. Cotton being a better absorber of water than nylon and helps in absorption of sweat followed by evaporation which leads to cooling. So, Priyanshi is more comfortable, whereas Ali is not so comfortable.

Solution

The following conditions can increase the rate of evaporation of water:

  • Increase the surface area by spreading the shirt.
  • Increase the temperature by putting the shirt under the sun.
  • Increase the wind speed by spreading it under the fan.


26. Comment on the following statements :
(a) Evaporation produces cooling.
(b) Rate of evaporation of an aqueous solution decreases with increase in humidity.
(c) Sponge though compressible is a solid.

Solution

(a) The particles at the surface of the liquid gain energy from the surroundings and change into vapour. This is the reason why evaporation produces a cooling effect.

(b) Humidity is the amount of water vapour content in the air around us. Air has a definite capacity to hold water vapour at a given temperature. So, if the air already has water vapour, it cannot take up more water. This is the reason why the rate of evaporation decreases in humidity.

(c) Sponge has minute holes with trapped air. The material of the sponge is not rigid. The air present in the holes gets expelled when a sponge is pressed, this is the reason why it can be compressed.


27. Why does the temperature of a substance remain constant during its melting point or boiling point?

Solution

At the melting and boiling point, the temperature of a substance remains constant until all the substance melts or boils. This happens because the heat supplied is used in overcoming the forces of attraction between the particles and thereby changing the state of the substance. Latent heat of fusion or the latent heat of vaporisation is this heat energy absorbed without showing any rise in temperature.

NCERT Exemplar Chapter 2 Is Matter around Us Pure Class 9 Science Solutions

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NCERT Exemplar Class 9 Science Chapter 2 Is Matter around Us Pure Solutions

NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter around Us Pure covers all the important questions and answers as well as advanced level questions. It helps in learning about the types of mixtures, homogeneous mixtures, heterogeneous mixtures, solution, suspension, concentration of solution, types of solution, Separating the Components of Mixtures, separation methods, miscible liquids, immiscible liquids and centrifugation.

The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 2 Is Matter around Us Pure is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like physical and chemical changes, difference between the pure substances and a mixture, types of Pure substances, elements and compounds.


Chapter Name

Chapter 2 Is Matter around Us Pure

Book Title

NCERT Exemplar for Class 9 Science

Related Study

  • NCERT Solutions for Class 9 Science Chapter 2 Is Matter around Us Pure
  • Revision Notes for Class 9 Science Chapter 2 Is Matter around Us Pure
  • MCQ for Class 9 Science Chapter 2 Is Matter around Us Pure
  • Important Questions for Class 9 Science Chapter 2 Is Matter around Us Pure

Topics Covered

  • MCQ
  • Short Answers Questions
  • Long Answers Questions

NCERT Exemplar Solutions for Chapter 2 Is Matter around Us Pure Class 9 Science

Multiple Choice Questions

1. Which of the following statements are true for pure substances?
(i) Pure substances contains only one kind of particles.
(ii) Pure substances may be compounds or mixtures.
(iii) Pure substances have the same composition throughout.
(iv) Pure substances can be exemplified by all elements other than nickel.
(a) (i) and (ii)
(b) (i) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iii)

Solution

(b) (i) and (iii)


2. Rusting of an article made up of iron is called :
(a) corrosion and it is a physical as well as chemical change.
(b) dissolution and it is a physical change.
(c) corrosion and it is a chemical change.
(d) dissolution and it is a chemical change.

Solution

(c) corrosion and it is a chemical change.
Rusting of iron is also known as corrosion and it is a chemical change because rust is a chemical compound called hydrated iron oxide Fe2O3. nH2O, iron(III) which is different from elemental iron.


3. A mixture of sulphur and carbon disulphide is :
(a) heterogeneous and shows Tyndall effect.
(b) homogeneous and shows Tyndall effect.
(c) heterogeneous and does not show Tyndall effect.
(d) homogeneous and does not show Tyndall effect.

Solution

(d) homogeneous and does not show Tyndall effect.
A mixture of sulphur and carbon disulphide is heterogeneous in nature and it shows Tyndall effect. This happens because the particles are big enough to scatter light.


4. Tincture of iodine has antiseptic properties. This solution is made by dissolving :
(a) Iodine in potassium iodide
(b) Iodine in vaseline
(c) Iodine in water
(d) Iodine in alcohol

Solution

(d) Iodine in alcohol.
Tincture is prepared by using2-7% elemental iodine, either potassium iodide or sodium dissolved in alcohol. Alcohol is a good solvent and iodine does not dissolve in water, so alcohol should be used.


5. Which of the following are homogeneous in nature?
(i) Ice
(ii) Wood
(iii) Soil
(iv) Air
(a) (i) and (iii)
(b) (ii) and (iv)
(c) (i) and (iv)
(d) (iii) and (iv)

Solution

(c) (i) and (iv)
Air and ice are homogeneous mixture as their constituents are not visible and cannot be distinguished from one another.

6. Which of the following are physical changes?
(i) Melting of iron metal
(ii) Rusting of iron
(iii) Bending of an iron rod
(iv) Drawing a wire of iron metal
(a) (i), (ii) and (iii)
(b) (i), (ii) and (iv)
(c) (i), (iii) and (iv)
(d) (ii), (iii) and (iv)

Solution

(c) (i), (iii) and (iv)
During the process of rusting of iron, iron reacts with water and oxygen to produce iron oxide, hence it is a chemical change.


7. Which of the following are chemical changes?
(i) Decaying of wood
(ii) Burning of wood
(iii) Sawing of wood
(iv) Hammering of a nail into a piece of wood
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (i) and (iv)

Solution

(a) (i) and (ii)
Decaying of wood and burning of wood are chemical changes as there will be a change of chemical composition and wood cannot be restored to its original form.


8. Two substances, A and B were made to react to form a third substance, A2B according to the following reaction
2A + B 
→ A2B
Which of the following statements concerning this reaction are incorrect?
(i) The product A2B shows the properties of substances A and B.
(ii) The product will always have a fixed composition.
(iii) The product so formed cannot be classified as a compound.
(iv) The product so formed is an element.
(a) (i), (ii) and (iii),
(b) (ii), (iii) and (iv)
(c) (i), (iii) and (iv)
(d) (ii), (iii) and (iv)

Solution

(c) (i), (iii) and (iv)

A2B is a compound made up of two elements A and B in a fixed ratio. The properties of a compound (e.g., A2B) are entirely different from those of its constituent elements (i.e. A and B). The composition of a compound is fixed.


9. Two chemical species X and Y combine together to form a product P which contains both X and Y
X + Y 
→ P
X and Y cannot be broken down into simpler substances by simple chemical reactions. Which of the following concerning the species X, Y and P are correct?
(i) P is a compound.
(ii) X and Y are compounds.
(iii) X and Y are elements.
(iv) P has a fixed composition.
(a) (i), (ii) and (iii)
(b) (i), (ii) and (iv)
(c) (ii), (iii) and (iv)
(d) (i), (iii) and (iv)

Solution

(d) (i), (iii) and (iv)
As X and Y cannot be further broken down into simpler substances, they are elements. P can be broken down to simpler substances hence it is a compound with a fixed composition.


Short Answer Questions

10. Suggest separation technique(s) one would need to employ to separate the following mixtures :

(i) Mercury and water
(ii) Potassium chloride and ammonium chloride
(iii) Common salt, water and sand
(iv) Kerosene oil, water and salt

Solution

(i) Separation by using separating funnel (used for separating two immiscible liquids). The principle is that immiscible liquids separate out in layers depending on their densities.
(ii) Sublimation : It is a process by which solid changes directly into gas and vice versa without passing
through the liquid state. Ammonium chloride is sublime.
(iii) Filtration followed by evaporation or centrifugation followed by evaporation/distillation.
(iv) Separation by using separating funnel to separate kerosene oil followed by evaporation or distillation.


11. Salt can be recovered from its solution by evaporation. Suggest some other technique for the same.

Solution

Salt can be recovered from its solution by crystallization. Crystallization removes soluble impurities which are not removed in case of evaporation. Hence it is a more efficient process.


12. Which of the tubes in figure given here (a) and (b) will be more effective as a condenser in the distillation apparatus?

Solution

Beads in tube A increase the surface area for the cooling of vapours that come in contact with them. Hence, tube A is a good condenser.


13. The ‘sea water’ can be classified as a homogeneous as well as heterogeneous mixture. Comment.

Solution

The sea water on the surface is a homogenous mixture as it comprises of water and salts. Whereas the sea water from deep sea consists of salts, water, mud, decayed plants etc. hence it will be a heterogeneous mixture.


14. While diluting a solution of salt in water, a student by mistake added acetone (boiling point 56°C). What technique can be employed to get back the acetone? Justify your choice.

Solution

Fractional distillation can be used to separate acetone from the mixture of salt and water. There is a considerable difference in the boiling points of acetone (56°C) and water (100°C). Acetone evaporates first when the solution is heated. The water is collected in the distillation flask. The vapours of acetone are then condensed to obtain acetone.


15. What would you observe when :
(i) A saturated solution of potassium chloride prepared at 60°C is allowed to cool at room temperature?
(ii) An aqueous sugar solution is heated to dryness?
(iii) A mixture of iron filings and sulphur powder is heated strongly?

Solution 

(a) Solid potassium chloride will separate out.

(b) The water will evaporate initially and then sugar will get charred.

(c) Iron sulphide will be formed.


16. Explain why particles of a colloidal solution do not settle down when left undisturbed, while in the case of a suspension they do.

Solution

In a suspension, the particle size is larger than those in a colloidal solution. Also in a suspension, the molecular interaction is not strong enough to keep the particles suspended and hence they settle down.


17. Smoke and fog both are aerosols. In what way are they different?

Solution

Gas forms the dispersion medium for both fog and smoke. The only difference is that the dispersed phase in fog is liquid while in case of smoke it is a solid.


18. Classify the following as physical or chemical properties :
(i) The composition of a sample of steel is : 98% iron, 1.5% carbon and 0.5% other elements.
(ii) Zinc dissolves in hydrochloric acid with the evolution of hydrogen gas.
(iii) Metallic sodium is soft enough to be cut with a knife.
(iv) Most metal oxides form alkalis on interacting with water.

Solution

(i) Composition of a sample of steel is 98% iron, 1.5% carbon and 0.5% other elements. It is a chemical property as no new compound is formed. Steel is an alloy.

(ii) Zinc reacts with HCl to give out zinc chloride and hydrogen gas hence it is chemical property.

(iii) Cutting metallic sodium with knife will not involve any chemical reaction. Also, no new compounds are formed hence it is a physical property.

(iv) A new compound is form by the interaction of metal oxides with alkalis hence it is chemical property.


19. The teacher instructed three students ‘A’, ‘B’ and ‘C’ respectively to prepare a 50%. (mass by volume)
solution of sodium hydroxide (NaOH). ‘A’ dissolved 50 g of NaOH in 100 ml of water, ‘B’ dissolved 50 g
of NaOH in 100 g of water while ‘C’ dissolved 50 g of NaOH in water to make 100 ml of solution. Which one of them has made the desired solution and why?

Solution

Student ‘C’ has made the desired solution.


20. Name the process associated with the following:
(i) Dry ice is kept at room temperature and at one atmospheric pressure.
(ii) A drop of ink placed on the surface of water contained in a glass spreads throughout the water.
(iii) A potassium permanganate crystal is in a beaker and water is poured into the beaker with stirring.
(iv) A acetone bottle is left open and the bottle becomes empty.
(v) Milk is churned to separate cream from it.
(vi) Settling of sand when a mixture of sand and water is left undisturbed for some time.
(vii) Fine beam of light entering through a small hole in a dark room, illuminates the particles in its paths.

Solution

(i) Sublimation
(ii) Diffusion
(iii) Dissolution/diffusion
(iv) Evaporation, diffusion
(v) Centrifugation
(vi) Sedimentation
(vii) Tyndall effect (Scattering of light)


21. You are given two samples of water, labelled as ‘A’ and ‘B’. Sample ‘A’ boils at 100°C and sample ‘B’ boils at 102°C. Which sample of water will not freeze at 0°C? Comment.

Solution

Sample ‘B’ is not pure water and hence it will not freeze at 0°C. At a pressure of 1 atm, the boiling point of pure water is 100°C and the freezing point of pure water is 0°C.


22. What are the favourable qualities given to gold when it is alloyed with copper or silver for the purpose of making ornaments?

Solution

Pure gold (24 karats) is very soft and it does not have enough strength. Silver and copper are alloyed with gold, to give it strength. 24 karat gold is an alloy that has 20 parts of gold and 4 parts of silver.


23. An element is sonorous and highly ductile. Under which category would you classify this element? What other characteristics do you expect the element to possess?

Solution

The element is a metal. Other characteristics possessed by the element are:

  • Lustre
  • Malleability
  • Heat
  • Electrical conductivity


24. Give an example each for the mixture having the following characteristics. Suggest a suitable method to separate the components of these mixtures.
(i) A volatile and a non-volatile component.
(ii) Two volatile components with appreciable difference in boiling points.
(iii) Two immiscible liquids.
(iv) One of the components changes directly from solid to gaseous state.
(v) Two or more coloured constituents soluble in some solvent.

Solution

(i) Mixture of acetone and water can be separated by distillation.

(ii) Mixture of petrol and kerosene can be separated by distillation.

(iii) Mixture of oil and water can be separated by fractional distillation.

(i)v Separating naphthalene by filtration and then separation of ammonium chloride from water by evaporation.

(v) Mixture of pigments from a flower petal extract can be separated by chromatography.


25. Fill in the blanks
(i) A colloid is a _____ mixture and its components can be separated by the technique known as_____
(ii) Ice, water and water vapour look different and display different _____ properties but they are_____ the same.
(iii) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of _____ and the lower layer will be that of _____
(iv) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25K can be separated by the process called _____
(v) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the _____ of light by milk and the phenomenon is called _____ This indicates that milk is a _____ solution.

Solution

(i) A colloid is a heterogeneous mixture and its components can be separated by the technique known as centrifugation.

(ii) Ice, water and water vapour look different and display different physical properties but they are chemically the same. 

(iii) A mixture of chloroform and water taken in a separating funnel is mixed and left undisturbed for some time. The upper layer in the separating funnel will be of water and the lower layer will be that of chloroform.

(iv) A mixture of two or more miscible liquids, for which the difference in the boiling points is less than 25 K can be separated by the process called fractional distillation.

(v) When light is passed through water containing a few drops of milk, it shows a bluish tinge. This is due to the scattering of light by milk and the phenomenonis called Tyndall effect. This indicates that milk is a colloidal solution.


26. Sucrose (sugar) crystals obtained from sugarcane and beetroot are mixed together. Will it be a pure substance or a mixture? Give reasons for the same.

Solution

As the chemical composition of sugar crystals remains the same irrespective of its source, the mixture is a pure substance.


27. Give some examples of Tyndall effect observed in your surroundings.

Solution

When light passes through a heterogeneous mixture, Tyndall effect can be seen. Example, when sunlight passes through the canopy of a dense forest.


28. Can we separate alcohol dissolved in water by using a separating funnel? If yes, then describe the procedure. If not, explain.

Solution

Alcohol dissolved in water cannot be separated by separating funnel because both are miscible solvents.


29. On heating calcium carbonate gets converted into calcium oxide and carbon dioxide.
(i) Is this a physical or a chemical change?
(ii) Can you prepare one acidic and one basic solution by using the products formed in the above process?
If so, write the chemical equation involved.

Solution

(i) It is a chemical change Calcium carbonate → Calcium oxide + Carbon dioxide
CaCO3 → CaO + CO2
(ii) Acidic and basic solutions can be prepared by dissolving the products of the above process in water
CaO + H2O → Ca(OH)2 (basic solution)
CO2 + H2O → H2CO3(acidic solution)


30. Non-metals are usually poor conductors of heat and electricity. They are non-lustrous, non-sonorous, non-malleable and are coloured.
(a) Name a lustrous non-metal.
(b) Name a non-metal which exists as a liquid at room temperature.
(c) The allotropic form of a non-metal is a good conductor of electricity. Name the allotrope.
(d) Name a non-metal which is known to form the largest number of compounds.
(e) Name non-metals other than carbon which show allotropy.
(f) Name a non-metal which is required for combustion.

Solution

(a) Iodine
(b) Bromine
(c) Graphite
(d) Carbon
(e) Sulphur, Phosphorus
(f) Oxygen


31. Classify the substances given in figure into elements and compounds :

Cu

Sand

H2O

CaCO3

O3

Zn

NaCl(aq)

F2

Hg

Diamond(C)

Wood

Solution 

Elements

Compounds

Cu

CaCO3

Zn

H2O

F2

O2

Diamond (c)

Hg


32. Which of the following are not compound?
(a) Chlorine gas
(b) Potassium chloride
(c) Iron
(d) Iron sulphate
(e) Aluminium
(f) Iodine
(g) Carbon
(h) Carbon monoxide
(i) Sulphur powder

Solution

Chlorine gas, iron, aluminium, iodine, carbon and sulphur powder are not compounds.


Long Answer Questions

33. Fractional distillation is suitable for separation of miscible liquids with a boiling point difference of about 25 K or less. What part of fractional distillation apparatus makes it efficient and possess an advantage over a simple distillation process. Explain using a diagram.

    Solution

    The most important part of fractional distillation apparatus is the fractionating column. It has glass beads in it. This prevents the upward movement of the vapours of the two liquids. The vapours of the liquid having high boiling point get condensed earlier at the lower levels of the column. The latent heat released helps to take the vapours of low boiling liquid to a greater height in the fractionating column.

    The advantages of fractional distillation are as follows:
    1. The liquids with a boiling point difference of about or less than 25 K can be separated by this method.
    2. Both evaporation and condensation take place simultaneously during the process.
    3. A mixture (like petroleum) which contains several components can also be separated by this process.

    34. (i) Under which category of mixtures will you classify alloys and why?
    (ii) A solution is always a liquid. Comment.
    (iii) Can a solution be heterogeneous?

    Solution

    (i) An alloy is a homogenous mixture of two or more elements. The constituent elements of an alloy can be two metals or a metal with a non-metal. An alloy is classified as a homogenous mixture as it exhibits the properties of its constituent elements.

    Examples: Brass is an alloy which shows characteristics of copper and Zinc and their composition vary from 20 to 35 %.

    (ii) The solutions are usually liquids, but not always. Solutions can also include a mixture of two solids like in alloys, and a mixture of gases, such as air.

    (iii) Yes, solutions can be heterogeneous also.


    35. Iron filings and sulphur were mixed together and divided into two parts, ‘A’ and ‘B’. Part ‘A’ was heated strongly, while part ‘B’ was not heated. Dilute hydrochloric acid was added to both the parts and evolution of gas was seen in both the cases. How will you identify the gases evolved?

    Solution

    Part A
    When iron fillings and sulphur is heated it will give following reaction

    Part B

    Fe (s) + S (s)→ Mixture of iron filings and sulphur
    When dilute HCl is added to it
    Fe (s) + S (s) + 2 HCl (aq) → FeCl2+ H2 gas
    Sulphur remains unreacted.H2S gas formed has a foul smell and on passing through lead acetate solution, it turns the solution black. Hydrogen gas burns with a pop sound.


    36. A child wanted to separate the mixture of dyes constituting a sample of ink. He marked a line by the ink on the filter paper and placed the filter paper in a glass containing water as shown in fig. The filter paper was removed when the water moved near the top of the filter paper.

    (i) What would you expect to see, if the ink contains three different coloured components?
    (ii) Name the technique used by the child.
    (iii) Suggest one more application of this technique.

    Solution

    (i) Three different bands are observed on the paper if the ink contains three different coloured components.

    (ii) Child uses the technique of paper chromatography.

    (iii) The different pigments present in the chlorophyll can be separated by paper chromatography.


    37. A group of students took an old shoe box and covered it with a black paper from all sides. They fixed a source of light (a torch) at one end of the box by making a hole in it and made another hole on the other side to view the light. They placed a milk sample contained in a beaker/ tumbler in the box as shown in the fig. They were amazed to see that milk taken in the tumbler was illuminated. They tried the same activity by taking a salt solution but found that light simply passed through it.

    (a) Explain why the milk sample was illuminated. Name the phenomenon involved.
    (b) Same results were not observed with a salt solution. Explain.
    (c) Can you suggest two more solutions which would show the same effect as shown by the milk solution?

    Solution

    (a) Milk is a colloidal substance. The particulate matter present in the milk scatter the light scatter thereby resulting in Tyndall effect. Because of this, the milk gets illuminated.

    (b) Salt solution is a homogenous solution. Small particles present in salt solution do not scatter light. This is the reason why there is no Tyndall effect in salt solution. The salt solution does not get illuminated as it does not exhibit the Tyndall effect.

    (c) Detergent solution and sulphur solution will exhibit the Tyndall effect.


    38. Classify each of the following, as a physical or a chemical change. Give reasons.
    (i) Drying of a shirt in the Sun.
    (ii) Rising of hot air over a radiator.
    (iii) Burning of kerosene in a lantern.
    (iv) Change in the colour of black tea on adding lemon juice to it.
    (v) Churning of milk cream to get butter.

    Solution

    (i) There is no chemical reaction or any chemical changed involved in drying a shirt in the sun, hence it is a physical phenomenon.

    (ii) In a radiator water is converted to vapours. Hot air being lighter rises upwards. Thus, rising of hot air over radiator is a physical change.

    (iii) Kerosene burns by using the atmospheric oxygen and produces carbon dioxide, this is the reason why burning of kerosene in a lantern is a chemical change.

    (iv) Lemon juice is a source of citric acid, ascorbic acid and malic acid. These acids react with the flavin antioxidants present in black tea and change the colour of the tea. Thus, the change in the colour of black tea on adding lemon juice is a chemical change.

    (v) There is no chemical reaction taking place when milk is churned to make butter. This is the reason why churning of milk cream to get butter is a physical change.


    39. During an experiment, the students were asked to prepare a 10% (Mass/Mass) solution of sugar in water. Ramesh dissolved 10 g of sugar in 100 g of water while Sarika prepared it by dissolving 10 g of sugar in water to make 100 g of the solution.
    (i) Are the two solutions of the same concentration?
    (ii) Compare the mass % of the two solutions.

      Solution 

      (i) The two solutions are not of the same concentration.

      (b) Solution prepared by Ramesh:


      40. You are provided with a mixture containing sand, iron filings, ammonium chloride and sodium chloride.
      Describe the procedures you would use to separate these constituents from the mixture.

      Solution

      The procedures to be used to separate the constituents from the mixture are:

      • Step-1:With the help of a magnet separate iron filings.
      • Step-2:Sublimation of the remaining mixture separates ammonium chloride.
      • Step-3:Add water to the remaining mixture, stir and filter.
      • Step-4:The filtrate can be evaporated to get back sodium chloride.


      41. Arun has prepared 0.01% (by mass) solution of sodium chloride in water. Which of the following correctly represents the composition of the solutions?
      (i) 1.00 g of NaCl + 100 g of water
      (ii) 0.11 g of NaC1 + 100 g of water
      (iii) 0.01 g of NaC1 + 99.99 g of water
      (iv) 0.10 g of NaC1 + 99.90 g of water

      Solution

      Option (iii) correctly represents the composition of the solutions.


      42. Calculate the mass of sodium sulphate required to prepare its 20% (mass percent) solution in 100 g of water.

      Solution

      In a 20% solution containing 100 g water; the mass percentage of water = 100 — 20 = 80%
      80% of solution is 100 gram
      100% of solution is 100/80 gram
      20% of solution is 100/80 × 20 = 25 gram
      Hence, to prepare 20% (w/w) solution in 100 gram of water 25 gram of sodium sulphate is needed.

      NCERT Exemplar Chapter 3 Atoms and Molecules Class 9 Science Solutions

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      NCERT Exemplar Class 9 Science Chapter 3 Atoms and Molecules Solutions

      NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules covers all the important questions and answers as well as advanced level questions. It helps in learning about the Atom and its size, characteristics, modern symbols of atoms, Law of Conservation of mass, chemical symbols, valency, cation, anion, solubility of substances and molecules.

      The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 3 Atoms and Molecules is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like bonding, molecule of elements, molecules of compounds, ions and writing formula for the compound.


      Chapter Name

      Chapter 3 Atoms and Molecules

      Book Title

      NCERT Exemplar for Class 9 Science

      Related Study

      • NCERT Solutions for Class 9 Science Chapter 3 Atoms and Molecules
      • Revision Notes for Class 9 Science Chapter 3 Atoms and Molecules
      • MCQ for Class 9 Science Chapter 3 Atoms and Molecules
      • Important Questions for Class 9 Science Chapter 3 Atoms and Molecules

      Topics Covered

      • MCQ
      • Short Answers Questions
      • Long Answers Questions

      NCERT Exemplar Solutions for Chapter 3 Atoms and Molecules Class 9 Science

      Multiple Choice Questions

      1. Which of the following correctly represents 360 g of water?
      (i) 2 moles of H2O
      (ii) 20 moles of water
      (iii) 6.022 × 1023 molecules of water

      (iv) 1.2044 × 1025 molecules of water

      (a) (i)
      (b) (i) and (iv)
      (c) (ii) and (iii)
      (d) (ii) and (iv)

      Solution

      (d) (ii) and (iv)

      Number of moles = Mass of water/Molar mass of water

      Number of moles = 360/120 = 20

      Number of molecules = 20 ×6.022× 1023
      = 1.2044 ×1025 molecules of water


      2. Which of the following statements is not true about an atom?
      (a) Atoms are not able to exist independently.
      (b) Atoms are the basic units from which molecules and ions are formed.
      (c) Atoms are always neutral in nature.
      (d) Atoms aggregate in large numbers to form the matter that we can see, feel or touch.

      Solution

      (a) Atoms are not able to exist independently.


      3. The chemical symbol for nitrogen gas is :
      (a) Ni
      (b) N2
      (c) N+
      (d) N

      Solution

      (b) N2

      Chemical formula of Nitrogen is N but nitrogen exist as a molecule of two atoms. Therefore, the chemical symbol of nitrogen gas is N2.


      4. The chemical symbol for sodium is :
      (a) So
      (b) Sd
      (c) NA
      (d) Na

      Solution

      (d) Na

      Sodium word is derived from Latin word Natrium hence the chemical name of sodium is Na.


      5. Which of the following would weigh the highest?
      (a) 0.2 mole of sucrose (C12 H22 O11)
      (b) 2 moles of CO2
      (c) 2 moles of CaCO3
      (d) 10 moles of H2O

      Solution

      (c) 2 moles of CaCO3

      Weight of a sample in grant = Number of moles ×Molar mass

      (a) 0.2 moles of C12H22O11= 0.2 ×342 = 68.4 g

      (b) 2 moles of CO2is 2 × 44 = 88 g

      (c) 2 moles of CaCO3= 2 ×100=200 g

      (d) 10 moles of H2O= 10 ×18 = 180g


      6. Which of the following has maximum number of atoms ?
      (a) 18g of H2O
      (b) 18g of O2
      (c) 18g of CO2
      (d) 18 g of CH4

        Solution

        (d) 18 g of CH4


        7. Which of the following contains maximum number of molecules ?
        (a) 1 g CO2
        (b) 1 g N2
        (c) 1 g H2
        (d) 1 g CH4

          Solution

          (c) 1 g H2

          1 g of H2 = ½ ×NA
          = 0.5 NA
          = 0.5 × 6.022 × 1023
          = 3.011 × 1023


          8. Mass of one atom of oxygen is :
          (a) 16/(6.023 × 1023 g)
          (b) 32/(6.023 × 1023 g)
          (c) 1/(6.023 × 1023 g)
          (d) 8u

          Solution

          (a) 16/(6.023 ×1023 g)

          Mass of one atom of oxygen = Atomic mass/NA

          = 16/6.023× 1023 g


          9. 3.42 g of sucrose are dissolved in 18 g of water in a beaker. The number of oxygen atoms in the solution are :
          (a) 6.68 × 1023
          (b) 6.09 × 1022
          (c) 6.022 × 1023
          (d) 6.022 × 1021

            Solution

            (a) 6.68 × 1023

            1 mol of sucrose (C12H22O11) contains = 11× NA atoms of oxygen,

            Here, NA= 6.023×1023

            0.01 mol of sucrose (C12H22O11) contains = 0.01 × 11 × NA atoms of oxygen

            = 0.11× NA atoms of oxygen

            = 18 g/(1×2+ 16)gmol-1

            = 1mol

            1mol of water (H2O) contains 1× NA atom of oxygen

            Total number of oxygen atoms = Number of oxygen atoms from sucrose + Number of oxygen atoms from water

            = 0.11 NA + 1.0 NA = 1.11NA

            Number of oxygen atoms in solution = 1.11 × Avogadro’s number

            = 1.11 × 6.022 ×1023

            = 6.68 × 1023


            10. A change in the physical state can be brought about :
            (a) Only when energy is given to the system.
            (b) Only when energy is taken out from the system.
            (c) When energy is either given to, or taken out from the system.
            (d) Without any energy change.

            Solution

            (c) When energy is either given to, or taken out from the system.


            Short Answer Questions

            11. Which of the following represents a correct chemical formula? Name it.
            (a) CaCl
            (b) BiPO4
            (c) NaSO4
            (d) NaS

              Solution

              (a) CaCl = Wrong (valency of Ca = 2, Cl = 1)
              (b) BiPO4 = Correct (valency of Bi = 3, PO4 = 3)
              (c) NaSO4 = Wrong (valency of Na = 1, SO4 = 2)
              (d) NaS = Wrong (valency of Na = 1, Sulphide = 2)

              Bismuth phosphate is right because both ions are trivalent.


              12. Write the molecular formulae for the following compounds.

              Solution

              (a) Copper (II) bromide = CuBr2

              (b) Aluminium (III) nitrate = Al(NO3)3

              (c) Calcium (II) phosphate = Ca3(PO4)2

              (d) Iron (III) sulphide = Fe2S3

              (e) Mercury (II) chloride = HgCl2

              (f) Magnesium (II) acetate = Mg(CH3COO)2


              13. Write the molecular formulae of all the compounds that can be formed by the combination of given ions : Cu3+, Na+, Fe3+, Cl, SO34 PO43 .

              Solution

              CuCl2/CuSO4/Cu3(PO4)

              NaCl/Na2SO4/Na3PO4

              FeCl3/Fe(SO4)3/FePO4


              14. Write the cations and anions present (if any) in the following compounds :
              (a) CH3COONa
              (b) NaCl
              (c) H2
              (d) NH4NO3

              Solution

              Compounds

              Cations

              Anions

              CH3COONa

              Na+

              CH3COO

              NaCl

              Na+

              Cl

              H2

              Nil

              Nil

              NH4NO3

              NH4+

              NO3



              15. Give the formulae of the compounds formed from the following sets of elements.

                Solution

                (a) Calcium and fluoride -Calcium Fluoride (CaF2)

                (b) Hydrogen and sulphur-Hydrogen Sulphide(H2S)

                (c) Nitrogen and hydrogen-Ammonia(NH3)

                (d) Carbon and chlorine -Carbon Tetrachloride(CCl4)

                (e) Sodium and oxygen -Sodium Oxide(Na2O)

                (f) Carbon and oxygen-Carbon dioxide(CO2); Carbon Monoxide(CO)


                16. Which of the following symbols of elements are incorrect? Give their correct symbols

                (a) Cobalt CO
                (b) Carbon c
                (c) Aluminium AL
                (d) Helium He
                (e) Sodium So

                Solution

                (a) Incorrect, the correct symbol of cobalt is Co

                (b) Incorrect, the correct symbol of carbon is C

                (c) Incorrect, the correct symbol of aluminium is Al

                (d) Correct (He)

                (e) Incorrect, the correct symbol of sodium is Na


                17. Give the chemical formulae for the following compounds and compute the ratio by mass of the combining elements in each one of them.
                (a) Ammonia
                (b) Carbon monoxide
                (c) Hydrogen chloride
                (d) Aluminium fluoride
                (e) Magnesium sulphide

                Solution

                (a) NH3

                (b) CO

                (c) HCl

                (d) AlF3

                (e) MgS

                N : H × 3

                C : O

                H : Cl

                Al : F × 3

                Mg : S

                14 : 1 × 3

                12 : 16

                1 : 35.5

                27 : 19 × 3

                24 : 32

                14 : 3

                3 : 4

                1 : 35.5

                9 : 19

                3 : 4


                18. State the number of atoms present in each of the following chemical species.
                (a) CO2-3
                (b) PO3-4
                (c) P2O5
                (d) CO

                Solution

                (a)

                CO2-3

                1 + 3 = 4

                (b)

                PO3-4 

                1 + 4 = 5

                (c)

                P2O5

                2 + 5 = 7

                (d)

                CO

                1 + 1 = 2


                19. What is the fraction of the mass of water due to neutrons?

                Solution

                • Mass of 1 molecule of water = 18 amu
                • No. of proton in 2 atoms of H = 2 and no. of neutron = 0
                • No. of proton in 1 atom of O = 8 and no. of neutron = 8
                • Fraction of mass due to neutron in water 8/18 = 4/9


                20. Does the solubility of a substance change with temperature? Explain with the help of an example.

                Solution

                Yes, the solubility of a substance depends on its temperature. The solubility generally, increases with an increase in temperature.

                For example: one can dissolve more sugar in hot water than in cold water.


                21. Classify each of the following on the basis of their atomicity.

                (a) F2
                (b) NO2
                (c) N2O
                (d) C2H6
                (e) P4
                (f) H2O2
                (g) P4O10
                (h) O3
                (i) HCl
                (j) CH4
                (k) He
                (l) Ag
                Solution

                (a) Diatomic (2 atoms)

                (b) Triatomic (3 atoms)

                (c) Triatomic (3 atoms)

                (d) Polyatomic (8 atoms)

                (e) Polyatomic (4 atoms)

                (f) Polyatomic (4 atoms)

                (g) Polyatomic (14 atoms)

                (h) Triatomic (3 atoms)

                (i) Diatomic (2 atoms)

                (j) Polyatomic (5 atoms)

                (k) Monoatomic (1 atom)

                (l) Monoatomic(1 atom)


                22. You are provided with a fine white coloured powder which is either sugar or salt. How would you identify it without tasting?

                Solution

                We can identify the powder by heating. The powder will char if it is a sugar. Alternatively, the powder can be dissolved in water and checked for electrical conductivity. If it conducts, then the powder is salt.


                23. Calculate the number of moles of magnesium present in a magnesium ribbon weighing 12 g. Molar atomic mass of magnesium is 24gmol–1.

                Solution


                Long Answer Questions

                24. Verify by calculating that :
                (a) 5 moles of CO2 and 5 moles of H2O do not have the same mass.
                (b) 240 g of calcium and 240 g of magnesium elements have a mole ratio of 3 : 5.

                  Solution

                  (a) CO2 has molar mass = 44 g mol–1
                  5 moles of CO2 have molar mass = 44 × 5 = 220 g
                  H2O has molar mass = 18 g mol-1
                  5 moles of H2O have mass = 18 × 5g= 90 g

                  (b) Number of moles in 240 g Ca metal 240/40 = 6
                  Number of moles in 240 g of Mg metal 24 = 240/24 = 10
                  Ratio, 6 : 10 = 3 : 5


                  25. Find the ratio by mass of the combining elements in the following compounds :
                  (a) CaCO3
                  (b) MgCl2
                  (c) H2SO4
                  (d) C2H3OH
                  (e) NH3
                  (f) Ca(OH)2

                  Solution

                  (a) CaCO3
                  Ca : C : O × 3
                  40 : 12 : 16 × 3
                  40 : 12 : 48
                  10 : 3 : 12

                  (b) MgCl2
                  Mg : Cl × 2
                  24 : 35.5 × 2
                  24 : 71

                  (c) H2SO4
                  H × 2 : S : O × 4
                  1 × 2 : 32 : 16 × 4
                  2 : 32 : 64
                  1 : 16 : 32

                  (d) C2H5OH
                  C × 2 : H × 6 : O
                  12 × 2 : 1 × 6 : 16
                  24 : 6 : 16
                  12 : 3 : 8

                  (e) NH3
                  N : H × 3
                  14 : 1 × 3
                  14 : 3

                  (f) Ca(OH)2
                  Ca : O × 2 : H × 2
                  40: 16 × 2 : 1 × 2
                  40 : 32 : 2


                  26. Calcium chloride when dissolved in according to the following equation :
                  CaCl2(aq) 
                   Ca2+(aq) + 2Cl(aq)
                  Calculate the number of ions obtained from CaCl2 when 222 g of it is dissolved in water.

                  Solution

                  1 mole of calcium chloride = 111 g
                  ∴ 222 g of CaCl2 is equivalent to 2 moles of CaCl2

                  Since 1 formula unit CaCl3 gives 3 ions, therefore, 1 mol of CaCl2 will give 3 moles of ions.
                  2 moles of CaCl2 would give 3 × 2 = 6 moles of ions
                  No. of ions = No. of moles of ions × Avogadro number
                  = 6 × 6.022 × 1023
                  = 36.132 × 1023 = 3.6132 × 1024 ions


                  27. The difference in the mass of 100 moles each of sodium atoms and sodium ions is 5.48002 g. Compute the mass of an electron.

                  Solution

                  A sodium atom and ion, differ by one electron. For 100 moles each of sodium atoms and ions there would be a difference of 100 moles of electrons.
                  Mass of 100 moles of electrons= 5.48002 g


                  28. Cinnabar (HgS) is a prominent ore of mercury. How many grams of mercury are present in 225 g of pure HgS ? Molar mass of Hg and S are 200.6 g mol-1 and 32 g mol-1 respectively.

                    Solution

                    Molar mass of HgS = 200.6 + 32 = 323.6 g mol-1
                    Mass of Hg in 232.6 g of HgS 200.6 g


                    29. The mass of one steel screw is 4.11g. Find the mass of one mole of these steel screws. Compare this value with the mass of the Earth (5.98 × 1024kg). Which one of the two is heavier and by how many times?

                    Solution

                    One mole of screws weigh = 2.475 ×1024g
                    = 2.475×1021 kg


                    Mass of earth is 2.4×103 times the mass of screws.
                    The earth is 2400 times heavier than one mole of screws.


                    30. A sample of vitamin C is known to contain 2.58 × 1024 oxygen atoms. How many moles of oxygen atoms are present in the sample?

                    Solution

                    1 mole of oxygen atoms = 6.022 × 1023 atoms
                    Number of moles of oxygen atoms

                    = 4.28 moles of oxygen atoms.


                    31. Raunak took 5 moles of carbon atoms in a container arid Krish also took 5 moles of a sodium atoms in another container of same weight.
                    (a) Whose container is heavier ?
                    (b) Whose countainer has more number of atoms ?

                    Solution

                    (a) Mass of sodium atoms carried by Krish
                    = (5 × 23) g = 115 g
                    While mass of carbon atom carried by Raunak
                    = (5 × 12) g = 60 g
                    Thus, Krish’s container is heavy.
                    (b) As the two bags have the same number of moles of atoms, both the bags will have the same number of atoms.


                    32. Fill in the missing data in the table :

                    Species property

                    Water

                    CO2

                    Na – Atom

                    MgCl2

                    No. of moles

                    2

                    ………

                    ………

                    0.5

                    No. of particles

                    …….

                    3.011 × 1023

                    ……..

                    ……..

                    Mass

                    36 g

                    ……..

                    115 g

                    ……

                    Solution

                    Species property

                    Water

                    CO2

                    Na – Atom

                    MgCl2

                    No. of moles

                    2

                    0.5 mole

                    5 moles

                    0.5

                    No. of particles

                    2×6.022×1023

                    3.011×1023

                    5×6.022×1023

                    0.5×6.022×1023×3

                     Mass

                    36 g

                    22 g

                    115 g

                    47.5 g


                    33. The visible universe is estimated to contain 1022 stars. How many moles of stars are present in the visible universe?

                    Solution

                    Number of moles of stars = 0.0166


                    34. What is the SI prefix for each of the following multiples and submultiples of a unit?

                    (a)

                    103

                    (b)

                    10-3

                    (c)

                    10-2

                    (d)

                    10-6

                    (e)

                    10-9

                    (f)

                    10-19

                    Solution

                    (a)

                    kilo

                    (b)

                    deci

                    (c)

                    centi

                    (d)

                    micro

                    (e)

                    nano

                    (f)

                    pico


                    35. Express each of the following in kilograms :
                    (a) 5.84 × 10–3 mg
                    (b) 58.34 g
                    (c) 0.584 g
                    (d) 5.873 × 10–21 g

                    Solution

                    (a) 5.84 × 10–9 kg
                    (b) 5.834 × 10–2 kg
                    (c) 5.84 × 10–4 kg
                    (d) 5.873 × 10–24 kg


                    36. Compute the difference in masses of 103moles each of magnesium atoms and magnesium ions. (Mass of an electron = 9.1×10–31 kg).

                    Solution

                    A Mg2+ion and Mg atom differ by two electrons.

                    103moles of Mg2+and Mg atoms would differ by 103× 2 moles of electrons.

                    Mass of 2 ×103moles of electrons = 2×103× 6.023×1023×9.1×10–31kg

                    =2×6.022 × 9.1×10–5kg
                    =109.6004 ×10–5kg
                    =1.096 × 10–3kg


                    37. Which has more number of atoms?
                    100 g of N2 or 100 g of NH3

                    Solution

                    (i) 100 g of N2 = 100/28 moles
                    Number of molecules = 100/28 × 6.022 × 1023
                    Number of atoms = 2 × 100/28 × 6.022 × 1023
                    = 43.01 × 1023
                    (ii) 100 g of NH3 = 100/17 moles
                    = 100/17 × 6.022 × 1023 molecules
                    = 100/17 × 6.022 × 1023 × 4 atoms
                    = 141.69 × 1023
                    ∴ NH3 would have more atoms.


                    38. Compute the number of ions present in 5.85 g of sodium chloride.

                    Solution

                    5.85 g of NaCl = 5.85/58.5 = 0.1 mole
                    or 0.1 mole of NaCl particle.
                    Each NaCl particle is equivalent to one Na+ one Cl- = 2 ions
                    Total moles of ions = 0.1 × 2 = 0.2 mole
                    No. of ions = 0.2 × 6.022 × 1023
                    = 1.2042 × 1023 ions


                    39. A gold sample contains 90% of gold and the rest copper. How many atoms of gold are present in one gram of this sample of gold?

                    Solution

                    One gram of gold sample will contain = 90/100 = 0.9 g of gold
                    Number of moles of gold = Mass of gold/Atomic mass of gld
                    = 0.9/197 = 0.0046
                    One mole of gold contians Na atoms = 6.022 × 1023
                    ∴ 0.0046 mole of gold will contain = 0.0046 × 6.022 × 1023
                    = 2.77 × 1021 


                    40. What are ionic and molecular compounds? Give examples.

                    Solution

                    While forming some compounds, atoms may gain or lose electrons, thereby forming electrically charged particles called ions. Compounds that are formed by the attraction of cations and anions are known as ionic compounds.
                    Example: 2Na + Cl2→ 2Na+ Cl-→ 2NaCl (sodium chloride-common salt.)

                    Compounds formed by the bonding of uncharged species are known as molecular compounds. The bonding is called covalent bonding. Molecular compounds are formed by sharing of electrons between the two atoms.
                    Example: 2C + O2→ 2CO (Carbon monoxide)


                    41. Compute the difference in masses of one mole each of aluminium atoms and one mole of its ions (mass of an electron is 9.1 × 10–28 g). Which one is heavier?

                    Solution

                    Mass of 1 mole of aluminium atom = the molar mass of aluminium = 27 g mol–1
                    An aluminium atom needs to lose three electrons to become an ion, Al3+
                    For one mole of Al3+ ion, three moles of electrons are to be lost.
                    The mass of three moles of electrons
                    = 3 × (9.1 × 10-28) × 6.022 × 1023 g
                    = 27.3 × 6.022 × 10-5 g
                    = 164.400 × 10-5 g = 0.00164 g
                    Molar mass of Al3+ = (27 – 0.00164) g mol-1
                    = 26.9984 g mol-1
                    Difference = 27 – 26.9984 = 0.0016 g


                    42. A silver ornament of mass ‘m’ gram is polished with gold equivalent to 1% of the mass of silver. Compute the ratio of the number of atoms of gold and silver in the ornament.

                    Solution

                    Mass of silver = mg
                    Mass of gold = m/100 g
                    Number of atoms of silver = [Mass/(Atomic mass)] × NA
                    = m/108 × NA
                    Number of atoms of gold = (m/100) × 197 × NA
                    Ratio of number of atoms of gold to silver = An : Ag

                    = 108 : 100 × 197
                    = 108 : 19700 = 1 : 182.41


                    43. A sample of ethane (C2H6) gas has the same mass as 1.5 ×1020 molecules of methane (CH4). How many C2Hmolecules does the sample of gas contain?

                    Solution


                    44. Fill in the blanks :
                    (a) In a chemical reaction, the sum of the masses of the reactants and products remains unchanged.
                    This is called _____.
                    (b) A group of atoms carrying a fixed charge on them is called _____.
                    (c) The formula unit mass of Ca3(PO4)2 is _____.
                    (d) Formula of sodium carbonate is _____ and that of ammonium sulphate is _____.

                    Solution

                    (a) Law of conservation of mass
                    (b) Polyatomic ion
                    (c) (3 × atomic mass of Ca) + (2 × atomic mass of phosphorus) + (8 × atomic mass of oxygen) = 310
                    (d) Na2CO3: (NH4)2SO4


                    45. Complete the following crossword puzzle (Fig. 3.1) by using the name of the chemical elements. Use the data given in Table 3.2.

                    Across

                    Down

                    The element used by Rutherford during his α–scattering experiment

                    A white lustrous metal used for making ornaments and which tends to get tarnished black in the presence of moist air.

                    An element which forms rust on exposure to moist air

                    Both brass and bronze are alloys of the element

                    A very reactive non–metal stored under water

                    The metal which exists in the liquid state at room temperature

                    Zinc metal when treated with dilute hydrochloric acid produces a gas of this element which when tested with burning splinter produces a pop sound.

                    An element with symbol Pb

                    Answer


                    46. Write the formulae for the following and calculate the molecular mass for each one of them :
                    (a) Caustic potash
                    (b) Baking powder
                    (c) Limestone
                    (d) Caustic soda
                    (e) Ethanol
                    (f) Common salt

                      Solution

                      (a) KOH = (39 + 16 + 1) = 56 g mol–1
                      (b) NaHCO3 = 23 + 1 + 12 + (3 × 16) = 84 g mol–1
                      (c) CaCO3 = 40 + 12 + (3 × 16) = 100 g mol–1
                      (d) NaOH = 23 + 16 + 1 = 40 g mol–1
                      (e) C2H5OH = C2H6O = 2 × 12 + (6 × 1) + 16 = 46 g mol–1
                      (f) NaCl = 23 + 35.5 = 58.5 g mol–1 


                      47. In photosynthesis, 6 molecules of carbon dioxide combine with an equal number of water molecules through a complex series of reactions to give a molecule of glucose having a molecular formula C6H12O6. How many grams of water would be required to produce 18 g of glucose? Compute the volume of water so consumed, assuming the density of water to be 1 g cm–3.

                      Solution

                      6CO2 + 6H2O → C6H12O6 + 6O2
                      1 mole of glucose needs 6 moles of water.
                      180 g of glucose needs (6×18) g of water.
                      1 g of glucose will need 108/180 g of water.
                      18 g of glucose would need 108/180 × 18 g of water = 10.8 g
                      Volume of water used = Mass/Density
                      = 10.8 g/1 g cm-3
                      = 10.8 cm3


                      NCERT Exemplar Chapter 4 Structure of the Atom Class 9 Science Solutions

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                      NCERT Exemplar Class 9 Science Chapter 4 Structure of the Atom Solutions

                      NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom covers all the important questions and answers as well as advanced level questions. It helps in learning about the Charged Particles of Matter, where does this charge come from, Structure of an Atom, different models of atoms proposed, Thomson’s Model of an Atom, Rutherford’s Models of an Atom, nucleus of an atom, orbits and Bohr’s Model of an Atom.

                      The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 4 Structure of the Atom is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like neutrons, proton, electron of an atom, isotopes, isobars, how are the electrons distributed in different shells, valence electrons, atomic number and mass number.


                      Chapter Name

                      Chapter 4 Structure of the Atom

                      Book Title

                      NCERT Exemplar for Class 9 Science

                      Related Study

                      • NCERT Solutions for Class 9 Science Chapter 4 Structure of the Atom
                      • Revision Notes for Class 9 Science Chapter 4 Structure of the Atom
                      • MCQ for Class 9 Science Chapter 4 Structure of the Atom
                      • Important Questions for Class 9 Science Chapter 4 Structure of the Atom

                      Topics Covered

                      • MCQ
                      • Short Answers Questions
                      • Long Answers Questions

                      NCERT Exemplar Solutions for Chapter 4 Structure of the Atom Class 9 Science

                      Multiple Choice Questions

                      1. Which of the following correctly represent the electronic distribution in the Mg atom?
                      (a) 3, 8, 1
                      (b) 2, 8, 2
                      (c) 1, 8, 3
                      (d) 8, 2, 2

                      Solution

                      (b) 2, 8, 2

                      Atomic number of Mg is 12.Therefore the electronic distribution will be 1s22s22p63s2.


                      2. Rutherford’s ‘alpha (a ) particles scattering experiment’ resulted in to discovery of :
                      (a) Electron
                      (b) Proton
                      (c) Nucleus in the atom
                      (d) Atomic mass

                      Solution

                      (c) Nucleus in the atom

                      Rutherford’s ‘alpha (α) particles scattering experiment’ showed that few alpha particles returned to their original path. This showed the presence of nucleus in the centre.


                      3. The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element?
                      (a) 15X31
                      (b) 16X31
                      (c) 15X16
                      (d) 16X15

                      Solution

                      (a) 15X31

                      Atomic number is the number of protons in an element. Number of protons and electrons are equal in an element. The atomic number is written as a subscript and the mass number is written as the superscript of the element.


                      4. Dalton’s atomic theory successfully explained :
                      (i) Law of conservation of mass
                      (ii) Law of constant composition
                      (iii) Law of radioactivity
                      (iv) Law of multiple proportions
                      (a) (i), (ii) and (iii)
                      (b) (i), (iii) and (iv)
                      (c) (ii), (iii) and (iv)
                      (d) (i), (ii) and (iv)

                      Solution

                      (d) (i), (ii) and (iv)

                      Dalton’s theory explains the Law of conservation of mass, Law of constant composition, Law of multiple proportion.


                      5. Which of the following statements about Rutherford’s model of atom are correct?
                      (i) considered the nucleus as positively charged
                      (ii) established that the a-particles are four times heavy as a hydrogen atom
                      (iii) can be compared to solar system
                      (iv) was in agreement with Thomson’s model
                      (a) (i) and (iii)
                      (b) (ii) and (iii)
                      (c) (i) and (iv)
                      (d) Only (i)

                      Solution

                      (a) (i) and (iii)

                      The positively charged alpha particles were deflected by the nucleus. This indicates that the nucleus is positively charged. Rutherford also postulated that electrons are arranged in an atom around the nucleus like planets arranged around sun.


                      6. Which of the following are true for an element?
                      (i) Atomic number = number of protons + number of electrons
                      (ii) Mass number = number of protons + number of neutrons
                      (iii) Atomic mass = number of protons = number of neutrons
                      (iv) Atomic number = number of protons = number of electrons
                      (a) (i) and (ii)
                      (b) (i) and (iii)
                      (c) and (iii)
                      (d) (ii) and (iv)

                      Solution

                      (d) (ii) and (iv)

                      Atomic number Z is the number of proton present in an atom. This is also equal to number of electrons in an atom. As the mass of neutron is negligible, number of protons and electron are added to obtain mass number of an element.


                      7. In the Thomson’s model of atom, which of the following statements are correct?
                      (i) The mass of the atom is assumed to be uniformly distributed over the atom.
                      (ii) The positive charge is assumed to be uniformly distributed over the atom.
                      (iii) The electrons are uniformly distributed in the positively charged sphere.
                      (iv) The electrons attract each other to stabilise the atom.
                      (a) (i), (ii) and (iii)
                      (b) (i) and (iii)
                      (c) (i) and (iv)
                      (d) (i), (iii) and (iv)

                      Solution

                      (a) (i), (ii) and (iii)

                      Thomson proposed that negatively charged electron are stabilised by positively charged protons in the nucleus.


                      8. Rutherford’s a -particle scattering experiment showed that :
                      (i) Electrons have negative charge.
                      (ii) The mass and positive charge of the atom is concentrated in the nucleus.
                      (iii) Neutron exists in the nucleus.
                      (iv) Most of the space in atom is empty. Which of the above statements are correct?
                      (a) (i) and (iii)
                      (b) (ii) and (iv)
                      (c) (i) and (iv)
                      (d) (iii) and (iv)

                      Solution

                      (b) (ii) and (iv)


                      9. The ion of an element has 3 positive charges. Mass number of the atom is 27 and the number of neutrons is 14. What is the number of electrons in the ion?
                      (a) 13
                      (b) 10
                      (c) 14
                      (d) 16

                      Solution

                      (b) 10

                      Mass number (A) of the atom = 27
                      Number of neutron in the atom =14
                      Number of electrons = Mass number - Number of neutrons
                      = 27-14 = 13
                      Since ions of the element has 3 positive charges number of electron in the ion is 13-3 which equal 10.


                      10. Identify the Mg2+ ion from the fig. where, n and p represent the number of neutrons and protons respectively :

                      Solution
                      Number of protons in Mg atom = 2+ 8 + 2 = 12
                      Number of neutrons in Mg atom = 24 -12 = 12
                      [As mass number of Mg atom = 24 and number of neutrons = mass number – number of protons]


                      11. In a sample of ethyl ethanoate (CH3COOC2H5), the two oxygen atoms have the same number of electrons but different number of neutrons. Which of the following is the correct reason for it?
                      (a) One of the oxygen atoms has gained electrons.
                      (b) One of the oxygen atoms has gained two neutrons.
                      (c) The two oxygen atoms are isotopes.
                      (d) The two oxygen atoms are isobars.

                      Solution

                      (c) The two oxygen atoms are isotopes.

                      Two oxygen atoms in CH3COOC2H5 can have different number of neutrons only if the two O-atoms are isotopes. Isotopes of an element have same number of protons (and electrons) but different number of neutrons.


                      12. Elements with valency 1 are :
                      (a) Always metals
                      (b) Always metalloids
                      (c) Either metals or non-metals
                      (d) Always non-metals

                      Solution

                      (c) Either metals or non-metals

                      If an element shows positive valency then it is a metal and if it shows negative valency then it is a non-metal.


                      13. The first model of an atom was given by :
                      (a) N. Bohr
                      (b) E. Goldstein
                      (c) Rutherford
                      (d) J.J. Thomson

                      Solution

                      (d) J.J. Thomson


                      14. An atom with 3 protons and 4 neutrons will have a valency of :
                      (a) 3
                      (b) 7
                      (c) 1
                      (d) 4

                      Solution

                      (c) 1


                      15. The electron distribution in an aluminium atom is :
                      (a) 2, 8, 3
                      (b) 2, 8, 2
                      (c) 8, 2, 3
                      (d) 2, 3, 8

                      Solution

                      (a) 2, 8, 3

                      Atomic number of Aluminium is 13. Therefore, the first shell can have maximum of 2 electrons and the second shell holds a maximum of 8 electrons.


                      16. Which of the following fig. do not represent Bohr’s model of an atom correctly?

                      (a) (i) and (ii)
                      (b) (ii) and (iii)
                      (c) (ii) and (iv)
                      (d) (i) and (iv)

                      Solution

                      (c) (ii) and (iv)

                      First shell can have a maximum of 2 electrons and the second shell can have a maximum of 8 electrons.


                      17. Which of the following statement is always correct?
                      (a) An atom has equal number of electrons and protons.
                      (b) An atom has equal number of electrons and neutrons.
                      (c) An atom has equal number of protons and neutrons.
                      (d) An atom has equal number of electrons, protons and neutrons.

                      Solution

                      (a) An atom has equal number of electrons and protons.

                      In an atom the number of protons is always equal to number of electrons.


                      18. Atomic models have been improved over the years. Arrange the following atomic models in the order of their chronological order :

                      (i) Rutherford’s atomic model,

                      (ii) Thomson’s atomic model,

                      (iii) Bohr’s atomic model.
                      (a) (i), (ii) and (iii)
                      (b) (ii), (iii) and (i)
                      (c) (ii), (i) and (iii)
                      (d) (iii), (ii) and (i)

                      Solution

                      (c) (ii), (i) and (iii)


                      Short Answer Questions

                      19. Is it possible for the atom of an element to have one electron, one proton and no neutron? If so, name the element.

                      Solution

                      Yes, Hydrogen is the element which has only 1 proton and 1 electron and no neutron. Therefore there is no repulsive force in the nucleus and hence it is stable.
                      1 proton means atomic no. (Z) = 1
                      1 neutron means mass no. (A) = p + n = 1 + 1 = 2
                      1 electron and 1 proton mean that atom is electrically neutral.
                      Hence, the element is 11H (An isotop of Hydrogen – Deuterium)


                      20. Write any two observations which support the fact that atoms are divisible.

                      Solution

                      1. Discovery of electrons and protons support the fact that atoms are divisible.
                      2. During a chemical reaction, there is either transfer or sharing of electrons which leads to the rearrangement of electrons in the atom.


                      21. Will 35Cl and 37Cl have different valencies? Justify your answer.

                      Solution

                      No, 35Cl and 37Cl both are the isotopes of the same element hence they cannot have different valencies.
                      Both atoms have valency 1.


                      22. Why did Rutherford select a gold foil in his a-ray scattering experiment?

                      Solution

                      Gold is the most malleable metal and Rutherford wanted the thinnest layer. Therefore, Rutherford used gold for his scattering experiment as it is a heavy metal with high atomic number and it is highly malleable.


                      23. Find out the valency of the atoms represented by the fig. (a) and (b).

                      Solution

                      As atom (a) has 8 electron in its valence shell, it has zero valency.

                      Atom (b) has a valency of +1 as it has 7 electrons in it outermost shell. To achieve the octet configuration, atom (b) will accept 1 electron.


                      24. One electron is present in the outermost shell of the atom of an element X. What would be the nature and value of charge on the ion formed if this electron is removed from the outermost shell?

                      Solution

                      A cation is formed if an electron is removed from the outermost shell and the charge of the element will be +1.


                      25. Write down the electron distribution of chlorine atom. How many electrons are there in the L shell? (Atomic number of chlorine is 17.)

                      Solution

                      Atomic number of chlorine atom = 17

                      So, its electronic configuration is
                      17Cl = K-2, L-8, M-7

                      26. In the atom of an element X, 6 electrons are present in the outermost shell. If it acquires noble gas configuration by accepting requisite number of electrons, then what would be the charge on the ion so formed?

                      Solution

                      6 electrons are already present in the outermost orbital. In order to attain noble gas configuration the element has to accept two electrons. Therefore, the charge on an atom of element X is -2.


                      27. What information do you get from the fig. about the atomic number, mass number and valency of atoms X, Y and Z? Give your answer in a tabular form.

                      Solution

                      Sl. No.

                      Atomic number

                      Mass number

                      Valency

                      X

                      5

                      11

                      3

                      Y

                      8

                      18

                      2

                      Z

                      15

                      31

                      3, 5


                      28. In response to a question, a student stated that in an atom, the number of protons is greater than the number of neutrons, which in turn is greater than the number of electrons. Do you agree with the statement? Justify your answer.

                      Solution

                      The statement is incorrect. In a neutral atom the number of protons and electrons is always equal. Number of neutron can be greater than the number of electron. Number of neutron can be equal to or greater than the number of protons because mass number is equal to double the atomic number.


                      29. Calculate the number of neutrons present in the nucleus of an element X which is represented as 1531X.

                      Solution

                      Mass number = No. of protons + No. of neutrons = 31
                      ∴ Number of neutrons = 31–number of protons
                      = 31–15
                      = 16


                      30. Match the names of the scientists given in column A with their contributions towards the understanding of the atomic structure as given in column B :

                       

                      Column A

                       

                      Column B

                      (a)

                      Ernest Rutherford

                      (i)

                      Indivisibility of atoms

                      (b)

                      J.J. Thomson

                      (ii)

                      Stationary orbits

                      (c)

                      Dalton

                      (iii)

                      Concept of nucleus

                      (d)

                      Neils bohr

                      (iv)

                      Discovery of electrons

                      (e)

                      James Chadwick

                      (v)

                      Atomic number

                      (f)

                      E. Goldstein

                      (vi)

                      Neutron

                      (g)

                      Mosley

                      (vii)

                      Canal rays

                      Solution

                      (a)

                      Ernest Rutherford

                      Concept of nucleus

                      (b)

                      J.J. Thomson

                      Discovery of electrons

                      (c)

                      Dalton

                      Indivisibility of atoms

                      (d)

                      Neils Bohr

                      Stationary orbits

                      (e)

                      James Chadwick

                      Neutron

                      (f)

                      E. Goldstein

                      Canal rays

                      (g)

                      Mosley

                      Atomic number


                      31. The atomic number of calcium and argon are 20 and 18 respectively, but the mass number of both these elements is 40. What is the name given to such a pair of elements?

                      Solution

                      Isobars are elements with different atomic numbers but same mass numbers. Calcium and argon are isobars.


                      32. Complete the table on the basis of information available in the symbols given below :
                      (a) 17Cl35 (b) 6C12 (c) 35Br85

                      Solution

                      Elements

                      No. of protons

                      No. of neutrons

                      17Cl35

                      17

                      35 – 17 = 18

                      6C12

                      6

                      12 – 6 = 6

                      35Br85

                      35

                      81 – 35 = 46


                      33. Helium atom has 2 electrons in its valence shell but its valency is not 2. Explain.

                      Solution

                      As Helium atom has 2 electrons in its outermost shell, its duplet is complete. Therefore the valency is zero.


                      34. Fill in the blanks in the following statements :
                      (a) Rutherford’s a -particle scattering experiment led to the discovery of the _____.
                      (b) Isotopes have same _____ but different _____.
                      (c) Neon and chlorine have atomic numbers 10 and 17 respectively. Their valencies will be _____ and _____ respectively.
                      (d) The electronic configuration of silicon is _____ and that of sulphur is _____.

                      Solution

                      (a) nucleus
                      (b) same atomic numbers, mass numbers
                      (c) 0 and –1
                      (d) Si = 2,8,4 and S = 2,8,6


                      35. An element X has a mass number 4 and atomic number 2. Write the valency of this element?

                      Solution

                      As the K shell is completely filled, valency is zero.


                      Long Answer Questions

                      36. Why do helium, neon and argon have a zero valency?

                      Solution

                      Helium has two electrons in its only energy shell, while Argon and Neon have 8 electrons in their valence shells. These elements do not have any tendency to combine with other elements as they have maximum number of electrons in their valence shells. Therefore, they have zero valency.


                      37. The ratio of the radii of hydrogen atom and its nucleus is ~ 105. Assuming the atom and the nucleus to be spherical,

                      (i) what will be the ratio of their sizes?

                      (ii) If atom is represented by planet earth ‘Re’ = 6.4 ×106m, estimate the size of the nucleus.

                      Solution

                      (i) Volume of the sphere = 4/3 πr3

                      Let R be the radius of the atom and r be that of the nucleus.

                      Therefore ,R = 105r

                      (ii) If the atom is represented by the planet earth (Re = 6.4×106m) then the radius of the nucleus would be


                      38. Enlist the conclusions drawn by Rutherford from his a-ray scattering experiment.

                      Solution

                      The following conclusions were made from the Rutherford α-particle scattering experiment:

                      1. As most of the α-particles passed through the gold foil undeflected, most of the space inside the atom is empty.
                      2. Very few particles were deflected from their path, indicating that the positive charge of the atom occupies very little space.
                      3. A very small fraction of α-particles were deflected by 180°,indicating that all the positive charges and mass of the gold atom were concentrated in a very small volume within the atom.

                      From the data he also calculated that the radius of the nucleus is about 105 times less than the radius of the atom.


                      39. In what way is the Rutherford’s atomic model different from that of Thomson’s atomic model?

                      Solution

                      Rutherford’s atomic model proposed that the electrons revolve around the nucleus in well-defined orbits. The center of the atom is positively charged and is called the nucleus. The model also proposed that the size of the nucleus is very small as compared to the size of the atom and nearly the entire mass of an atom is centred in the nucleus.

                      Thomson’s model of an atom was similar to that of a christmas pudding. The electrons are studded like currants in a positively charged sphere like christmas pudding and the mass of the atom was supposed to be uniformly distributed.


                      40. What were the drawbacks of Rutherford’s model of an atom?

                      Solution

                      Rutherford’s model failed to explain the stability of the atom. Any particle in a circular orbit would undergo an acceleration and the charged particles would radiate energy. Thus, the revolving electron would lose energy and finally fall into the nucleus. In other words, the atom should collapse.


                      41. What are the postulates of Bohr’s model of an atom?

                      Solution

                      The postulates of Neils Bohr’s about the model of an atom are as follows:

                      1. Only certain special orbits known as discrete orbits of electrons, are allowed inside the atom.
                      2. While revolving in discrete orbits the electrons do not radiate energy.

                      These orbits are called energy levels. Energy levels in an atom are shown by circles. These orbits are represented by the letters K,L,M,N,... or the numbers, n=1,2,3,4,....


                      42. Show diagrammatically the electron distributions in a sodium atom and a sodium ion and also give their atomic number.

                      Solution

                      As the atomic number of sodium atom is 11, it has 11 electrons. A positively charged sodium ion (Na+) is formed by the removal of one electron from a sodium atom. Thus, a sodium ion has 11–1 = 10 electrons. The electronic configuration of sodium ion will be 2, 8. The atomic number of an element is equal to the number of protons in its atom. As, sodium atom and sodium ion have the same number of protons, therefore, the atomic number of both is 11.


                      43. In the Gold foil experiment of Geiger and Marsden, that paved the way for Rutherford’s model of an atom, ~ 1.00% of the α-particles were found to deflect at angles > 50º. If one mole of α-particles were bombarded on the gold foil, compute the number of α-particles that would deflect at angles less than 500.

                      Solution

                      % of α-particles deflected more than 50°=1% of α-particles.

                      % of α-particles deflected less than 50° = 100–1% =  99%

                      Number of α-particles bombarded = 1 mole = 6.022×1023 particles

                      Number of particles that deflected at an angle less than 50°

                      NCERT Exemplar Chapter 5 The Fundamental Unit of Life Class 9 Science Solutions

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                      NCERT Exemplar Class 9 Science Chapter 5 The Fundamental Unit of Life Solutions

                      NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life covers all the important questions and answers as well as advanced level questions. It helps in learning about the structure of units that make living organisms, unicellular organisms, multicellular organisms, Structure of a cell, cell organelles, plasma membrane, nucleus and cytoplasm.

                      The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like rough Endoplasmic reticulum (RER), smooth Endoplasmic reticulum (SER), lysosomes, vacuoles, bacteria, fungi, plastids, chromoplasts, chloroplasts and leucoplast.


                      Chapter Name

                      Chapter 5 The Fundamental Unit of Life

                      Book Title

                      NCERT Exemplar for Class 9 Science

                      Related Study

                      • NCERT Solutions for Class 9 Science Chapter 5 The Fundamental Unit of Life
                      • Revision Notes for Class 9 Science Chapter 5 The Fundamental Unit of Life
                      • MCQ for Class 9 Science Chapter 5 The Fundamental Unit of Life
                      • Important Questions for Class 9 Science Chapter 5 The Fundamental Unit of Life

                      Topics Covered

                      • MCQ
                      • Short Answers Questions
                      • Long Answers Questions

                      NCERT Exemplar Solutions for Chapter 5 The Fundamental Unit of Life Class 9 Science

                      Multiple Choice Questions

                      1. Which of the following can be made into crystal?
                      (a) A bacterium
                      (b) An amoeba
                      (c) A virus
                      (d) A sperm

                      Solution

                      (c) A virus

                      Viruses are considered to be intermediates between living and non-living. This is because they cannot metabolize or reproduce on their own. Moreover, virus requires a host for all its processes. Viruses can be stored as crystal like chemicals. Virus crystals are collection of millions of virus cells.


                      2. A cell will swell up if :
                      (a) The concentration of water molecules in the cell is higher than the concentration of water molecules in surrounding medium.
                      (b) The concentration of water molecules in surrounding medium is higher than water molecules concentration in the cell.
                      (c) The concentration of water molecules is same in the cell and in the surrounding medium.
                      (d) Concentration of water molecules does not matter.

                      Solution

                      (b) The concentration of water molecules in surrounding medium is higher than water molecules concentration in the cell.
                      The water from the surrounding enter the cell through osmosis and the cell swells up when the concentration of water molecules in surrounding medium is higher.


                      3. Chromosomes are made up of :
                      (a) DNA
                      (b) protein
                      (c) DNA and protein
                      (d) RNA

                      Solution

                      (c) DNA and protein


                      4. Which of these options are not a function of ribosomes?
                      (i) It helps in manufacture of protein molecules
                      (ii) It helps in manufacture of enzymes
                      (iii) It helps in manufacture of hormones
                      (iv) It helps in manufacture of starch molecules
                      (a) (i) and (ii)
                      (b) (ii) and (iii)
                      (c) (iii) and (iv)
                      (d) (iv) and (i)

                      Solution

                      (c) (iii) and (iv)

                      Proteins are produced in ribosomes. Enzymes are proteins. Hormones and starch are not produced in ribosome.


                      5. Which of these is not related to endoplasmic reticulum?
                      (a) It behaves as transport channel for proteins between nucleus and cytoplasm.
                      (b) It transports materials between various regions in cytoplasm.
                      (c) It can be the site of energy generation.
                      (d) It can be the site for some biochemical activities of the cell.

                      Solution

                      (c) It can be the site of energy generation.

                      Energy is produced in the mitochondria.



                      6. Following are a few definitions of osmosis, read carefully and select the correct definition :
                      (a) Movement of water molecules from a region of higher concentration to a region of lower concentration through a semipermeable membrane.
                      (b) Movement of solvent molecules from its higher concentration to lower concentration.
                      (c) Movement of solvent molecules from higher concentration to lower concentration of solution through a permeable membrane.
                      (d) Movement of solute molecules from lower concentration to higher concentration of solution through a semipermeable membrane.

                      Solution

                      (a) Movement of water molecules from a region of higher concentration to a region of lower concentration through a semipermeable membrane.


                      7. Plasmolysis in a plant cell is defined as :
                      (a) Break down (lysis) of plasma membrane in hypotonic medium
                      (b) Shrinkage of cytoplasm in hypertonic medium
                      (c) Shrinkage of nucleoplasm
                      (d) None of the above

                      Solution

                      (b) Shrinkage of cytoplasm in hypertonic medium

                      Living plant cell loses water through the process of osmosis. This results in shrinking or contraction of the contents of the cell away from the cell wall. This is known as plasmolysis. When a plant cell has more water content than the surrounding, the plant cell tends to transfer water to its surroundings.


                      8. Which of the following are covered by a single membrane?
                      (a) Mitochondria
                      (b) Vacuole
                      (c) Lysosome
                      (d) Plastid

                      Solution

                      (b) Vacuole, (c) Lysosome

                      Both (b) and (c)
                      Both vacuole and lysosome are single membrane bound cell organelles. Mitochondria and plastids are double membrane bound cell organelles


                      9. Find out the false sentences :
                      (a) Golgi apparatus is involved with the formation of lysosomes.
                      (b) Nucleus, mitochondria and plastid have DNA; hence, they are able to make their own structural proteins.
                      (c) Mitochondria is said to be the powerhouse of the cell as ATP is generated in them.
                      (d) Cytoplasm is called as protoplasm.

                      Solution

                      (a) Golgi apparatus is involved with the formation of lysosomes.

                      Golgi apparatus has a role in the storage and synthesis of certain biomolecules. It does not have any role in the synthesis of lysosome.


                      10. Find out the correct sentence :
                      (a) Enzymes packed in lysosomes are made by RER (rough endoplasmic reticulum).
                      (b) Rough endoplasmic reticulum and smooth endoplasmic reticulum produce lipid and protein respectively.
                      (c) Endoplasmic reticulum is related with the destruction of plasma membrane.
                      (d) Nucleoid is present inside the nucleoplasm of eukaryotic nucleus.

                      Solution

                      (a) Enzymes packed in lysosomes are made by RER (rough endoplasmic reticulum).

                      RER has ribosomes present on its surface. These ribosomes are responsible for the synthesis of proteins and enzymes.


                      11. Which cell organelle plays a crucial role in detoxifying many poisons and drugs in a cell?
                      (a) Golgi apparatus
                      (b) Lysosomes
                      (c) Smooth endoplasmic reticulum
                      (d) Vacuoles

                      Solution

                      (c) Smooth endoplasmic reticulum

                      SER plays an important role in detoxifying many poisons and drugs in the liver cells of vertebrates.


                      12. The proteins and lipids, essential for building the cell membrane, are manufactured by :
                      (a) rough endoplasmic reticulum
                      (b) golgi apparatus
                      (c) plasma membrane
                      (d) mitochondria

                      Solution

                      (a) rough endoplasmic reticulum

                      Endoplasmic reticulum plays important role in the production of both lipids and proteins. SER produces proteins and RER produces lipids.


                      13. The undefined nuclear region of prokaryotes are also known as :
                      (a) nucleus
                      (b) nucleolus
                      (c) nucleic acid
                      (d) nucleoid

                      Solution

                      (d) nucleoid


                      14. The cell organelle involved in forming complex sugars from simple sugars are :
                      (a) endoplasmic reticulum
                      (b) ribosomes
                      (c) plastids
                      (d) Golgi apparatus

                      Solution

                      (d) Golgi apparatus

                      Golgi apparatus is involved in the packaging and transport of many biomolecules such as proteins, lipids and carbohydrates.


                      15. Which out of the following is not a function of vacuole?
                      (a) Storage
                      (b) Providing turgidity and rigidity to the cell
                      (c) Waste excretion
                      (d) Locomotion

                      Solution

                      (d) Locomotion

                      Vacuoles are responsible for storage, turgidity and rigidity of the cell and waste excretion. Locomotion is carried out by specialised organelles present outside cytoplasm. Vacuoles are present inside the cytoplasm.


                      16. Amoeba acquires its food through a process, termed :
                      (a) exocytosis
                      (b) endocytosis
                      (c) plasmolysis
                      (d) exocytosis and endocytosis both

                      Solution

                      (b) endocytosis

                      The process of taking in of matter by a living cell by invagination of its membrane to form a vacuole is known as endocytosis. In endocytosis, substances that are external to a cell are brought into the cell.


                      17. Cell wall of which one of these is not made up of cellulose?
                      (a) Bacteria
                      (b) Hydrilla
                      (c) Mango tree
                      (d) Cactus

                      Solution

                      (a) Bacteria

                      Hydrilla, mango tree and cactus are plants and thus their cell wall is made up of cellulose. The cell wall of bacteria is made of polysaccharide.


                      18. Silver nitrate solution is used to study :
                      (a) endoplasmic reticulum
                      (b) Golgi apparatus
                      (c) nucleus
                      (d) mitochondria

                      Solution

                      (b) Golgi apparatus


                      19. Organelle other than nucleus, containing DNA is :
                      (a) endoplasmic reticulum
                      (b) Golgi apparatus
                      (c) mitochondria
                      (d) lysosome

                      Solution

                      (c) mitochondria

                      Mitochondria and chloroplast are the organelles that have separate DNA called as mitochondrial DNA and chloroplast DNA.


                      20. Kitchen of the cell is :
                      (a) mitochondria
                      (b) endoplasmic reticulum
                      (c) chloroplast
                      (d) golgi apparatus

                      Solution

                      (c) chloroplast

                      Food in plants is produced inside chloroplast.


                      21. Lipid molecules in the cell are sythesized by :
                      (a) smooth endoplasmic reticulum
                      (b) rough endoplasmic reticulum
                      (c) golgi apparatus
                      (d) plastids

                      Solution

                      (a) smooth endoplasmic reticulum



                      22. Cell arises from pre-existing cell was stated by :
                      (a) Haeckel
                      (b) Virchow
                      (c) Hooke
                      (d) Schleiden

                      Solution

                      (b) Virchow


                      23. Cell theory was given by :
                      (a) Schleiden and Schwann
                      (b) Virchow
                      (c) Hooke
                      (d) Haeckel

                      Solution

                      (a) Schleiden and Schwann

                      The cell theory which stated that all plants and animals are made up of cell and cell is the basic unit of life was proposed by Schleiden and Schwann.


                      24. The only cell organelle seen in prokaryotic cell is :
                      (a) mitochondria
                      (b) ribosomes
                      (c) plastids
                      (d) lysosomes

                      Solution

                      (b) ribosomes


                      25. Organelle without a cell membrane is :
                      (a) ribosome
                      (b) Golgi apparatus
                      (c) chloroplast
                      (d) nucleus

                      Solution

                      (a) ribosome

                      Golgi bodies, chloroplast and nucleus are all membrane bound organelles but ribosomes are organelles without membrane.


                      26. 1μm is :
                      (a) 10–6 m

                      (b) 10–9 m
                      (c) 10–10 m

                      (d) 10–3 m

                      Solution

                      (a) 10–6 m

                      10–3 m–millimetre

                      10–6 m –Micrometer

                      10–9 m –nanometer


                      27. Lysosome arises from :
                      (a) endoplasmic reticulum
                      (b) Golgi apparatus
                      (c) nucleus
                      (d) mitochondria

                      Solution

                      (b) Golgi apparatus

                      The main functions of golgi apparatus are secretion, packaging and modification of proteins. It is also involved in the synthesis of new membranes and lysosomes.

                      28. Living cells were discovered by :
                      (a) Robert Hooke
                      (b) Purkinje
                      (c) Leeuwenhoek
                      (d) Robert Brown

                      Solution

                      (c) Leeuwenhoek

                      Robert hook observed cells first but he observed the dead cork cell. It was Leeuwenhoek who observed living cell from his microscope.


                      29. Select the odd one out
                      (a) The movement of water across a semipermeable membrane is affected by the amount of substances dissolved in it.
                      (b) Membranes are made of organic molecules like proteins and lipids.
                      (c) Molecules soluble in organic solvents can easily pass through the membrane.
                      (d) Plasma membranes contain chitin sugar in plants.

                      Solution

                      (d) Plasma membranes contain chitin sugar in plants.

                      Plant plasma membrane contains cellulose in it.


                      Short Answer Questions

                      30. Why are lysosomes known as ‘suicide-bags’ of a cell?

                      Solution

                      Lysosomes have digestive enzymes. When it bursts, the digestive enzymes are released and it starts digesting its own cells. This is the reason why lysosomes are known as suicidal bags of a cell.


                      31. Do you agree that “A cell is a building unit of an organism.” If yes, explain why?

                      Solution

                      Yes, the cells are the building units of an organism. Cells which perform similar functions join together to form tissues which further forms organs and organ systems.

                      This forms an organism. In a unicellular organism, a single cell performs all the life processes.


                      32. Why does the skin of your finger shrink when you wash clothes for a long time?

                      Solution

                      Soap solution is hypertonic in nature. This makes the water move out of the cells which in turn result in shrinking of fingers when you wash clothes for a long time.


                      33. Why is endocytosis found in animals only?

                      Solution

                      As cell wall is absent in animals, endocytosis is found only in animals. Because of this the movement of substances inside the cell is easier in animals than in plants.


                      34. A person takes concentrated solution of salt, after sometime, he starts vomiting. What is the phenomenon responsible for such situation? Explain.

                      Solution

                      Upon consumption of concentrated salt solution, the process of osmosis process takes place which results in dehydration. This is the reason why a person vomits.


                      35. Name any cell organelle which is non-membranous.

                      Solution

                      The only non-membranous cell organelle is the ribosome.


                      36. We eat food composed of all the nutrients like carbohydrates, proteins, fats, vitamins, minerals and water. After digestion, these are absorbed in the form of glucose, amino acids, fatty acids, glycerol, etc. What mechanisms are involved in absorption of digested food and water?

                      Solution

                      Diffusion and osmosis are involved in absorption and digestion respectively. Water is absorbed through the process of osmosis. Fatty acids and glycerol are diffused through submissive transport mechanism. Active transport helps in the absorption of glucose, amino acids and some ions.


                      37. If you are provided with some vegetables to cook, you generally add salt into the vegetables during cooking process. After adding salt, vegetables release water. What mechanism is responsible for this?

                      Solution

                      Vegetables release water on adding salt due to the process of osmosis. The addition of salt makes the external environment hypertonic. That is the concentration of water outside is lowered as compared to the concentration of water inside the cell. This results in the elimination of water from the vegetables due to exosmosis.


                      38. If cells of onion peel and RBC are separately kept in hypotonic solution, what among the following will take place? Explain the reason for your answer.
                      (i) Both the cells will swell.
                      (ii) RBCs will burst easily while cells of onion peel will resist the bursting to some extent.
                      (iii) (i) and (ii) both are correct.
                      (iv) RBC and onion peel cells will behave similarly.

                      Solution

                      The hypotonic water surrounding the cell moves inside the cell. This leads to swelling of the cells. As RBCs do not have plasma membrane, they swell and burst easily. Plant cells have cell wall which prevents them from bursting.


                      39. Bacteria do not have chloroplast, but some bacteria are photoautotrophic in nature and perform photosynthesis. Which part of bacterial cell performs this?

                      Solution

                      Small vesicles associated with plasma membrane are present in bacteria. These vesicles have pigment which can trap sunlight to carry photosynthesis.


                      40. Match the following A and B :

                       

                      A

                       

                      B

                      (1)

                      Smooth endoplasmic

                      (a)

                      Amoeba

                      (2)

                      Lysosome

                      (b)

                      Nucleus

                      (3)

                      Nucleoid

                      (c)

                      Bacteria

                      (4)

                      Food vacuoles

                      (d)

                      Detoxification

                      (5)

                      Chromatin material

                      (e)

                      Suicidal bags

                      Solution

                       

                      A

                       

                      B

                      (1)

                      Smooth endoplasmic

                      (a)

                      Amoeba

                      (2)

                      Lysosome

                      (b)

                      Nucleus

                      (3)

                      Nucleoid

                      (c)

                      Bacteria

                      (4)

                      Food vacuoles

                      (d)

                      Detoxification

                      (5)

                      Chromatin material

                      (e)

                      Suicidal bags


                      41. Write the name of different plant parts in which chromoplast, chloroplast and leucoplast are present.

                      Solution

                      • Chromoplast is present in flower and fruit.
                      • Chloroplast is found in the leaves.
                      • Leucoplast is present in the roots.


                      42. Name the organelles which show the analogy written as under :
                      (a) Transporting channels of the cell _____
                      (b) Powerhouse of the cell _____
                      (c) Packaging and dispatching unit of the cell _____
                      (d) Digestive bag of the cell _____
                      (e) Storage sacs of the cell _____
                      (f) Kitchen of the cell _____
                      (g) Control room of the cell _____

                      Solution

                      (a) endoplasmic reticulum
                      (b) mitochondria
                      (c) golgi body
                      (d) lysosome
                      (e) vacuole
                      (f) chloroplast
                      (g) nucleus


                      43. How is a bacterial cell different from an onion peel cell?

                      Solution

                      Bacterial cell

                      Onion peel cell

                      Prokaryotic cell

                      Eukaryotic cell

                      Lack of organized nucleus, the genetic material present in the form of nucleoid

                      Well-organised Nucleus

                      Nuclear membrane absent

                      Nuclear membrane present

                      Only a single chromosome is present

                      More than one chromosome are present

                      Membrane-bound organelles absent

                      Membrane-bound organelles present

                      Nucleolus is absent

                      Nucleolus is present

                      Cell division takes place by fission or budding

                      Cell division takes place by mitosis or meiosis


                      44. How do substances like carbon dioxide (CO2) and water (H2O) move in and out of the cell?

                      Solution

                      Carbon dioxide (CO2) moves in and out of the cell by the process of diffusion and water (H2O) move in and out of the cell through osmosis.


                      45. How does amoeba obtain its food?

                      Solution

                      Amoeba obtains its food through the process of endocytosis. Amoeba has finger-like projections which fuse with the food particles to forma food vacuole. Inside the vacuole the complex food particles are broken down into simpler ones which are then diffused into the cytoplasm.


                      46. Name the two organelles in a plant cell that contain their own genetic material and ribosomes.

                      Solution

                      The two organelles in a plant cell that contain their own genetic material and ribosomes are chloroplast and mitochondria.


                      47. Why are lysosomes also known as “scavengers of the cells”?

                      Solution

                      Lysosomes have lytic enzymes which are used to destroy the pathogens and worn-out cells. Lysosomes also destroy waste materials which are harmful for the cell. This is the reason why Lysosomes also are known as “scavengers of the cells”.


                      48. Which cell organelle controls most of the activities of the cell?

                      Solution

                      Most of the activities of the cell are controlled by the nucleus.


                      49. Which kind of plastid is more common in :
                      (i) Roots of the plant?
                      (ii) Leaves of the plant?
                      (iii) Flowers and fruits of the plant?

                      Solution

                      (i) Leucoplasts are more common in roots of the plant.

                      (ii) Chloroplasts are more common in leaves of the plant.

                      (iii) Chromoplasts are more common in flowers and fruits.


                      50. Why do plant cells possess large sized vacuole?

                      Solution

                      Vacuoles not only store important material but they also contain a sap that gives turgidity to the cell, this is the reason why plant cells possess large-sized vacuole.


                      51. How are chromatin, chromatid and chromosomes related to each other?

                      Solution

                      • Chromatin is a thread-like structure which forms chromosomes.
                      • A copy of duplicated chromosome which is generally joined to the other copy by a centromere is called Chromatid.
                      • When a cell starts to divide, the tangled mass of chromatin condenses into long threads and finally, rod-like bodies called chromosomes.


                      52. What are the consequences of the following conditions?
                      (i) A cell containing higher water concentration than the surrounding medium.
                      (ii) A cell having low water concentration than the surrounding medium.
                      (iii) A cell having equal water concentration to its surrounding medium.

                      Solution

                      (i) The cell loses water and shrinks if it contains higher water concentration than the surrounding medium. This is known as exosmosis.

                      (ii)The cell bursts if it has a lower water concentration than the surrounding medium. This process is known as endosmosis.

                      (iii) There is no change in the cell if a cell having equal water concentration to its surrounding medium.


                      Long Answer Questions

                      53. Draw a plant cell and label the parts which :
                      (a) Determines the function and development of the cell.
                      (b) Packages materials coming from the endoplasmic reticulum.
                      (c) Provides resistance to microbes to withstand hypotonic external media without bursting.
                      (d) Is site for many biochemical reactions necessary to sustain life.
                      (e) Is a fluid contained inside the nucleus.

                      Solution

                      Following is the schematic diagram of plant cell containing the specific cell organelles as asked in the question :

                      (a) Nucleus determines the function and development of the cell.

                      (b) Golgi apparatus is the package material coming from the endoplasmic reticulum.

                      (c) Cell wall provides the resistance to microbes to withstand hypotonic external media without bursting.

                      (d) Cytoplasm is a site for many biochemical reactions necessary to sustain life.

                      (e) Nucleoplasm is a fluid contained inside the nucleus.


                      54. Illustrate only a plant cell as seen under electron microscope. How is it different from animal cell?

                      Solution

                      Difference between Plant Cell and Animal Cell:

                      Plant Cell

                      Animal Cell

                      Cell wall present

                      Cell wall absent

                      It has a large vacuole

                      It has a small vacuole

                      Plastids are present

                      Plastids are absent

                      Centriole absent

                      Centriole present


                      55. Draw a neat labelled diagram of an animal cell.

                      Solution


                      56. Draw a well labelled diagram of a eukaryotic nucleus. How is it different from nucleoid?

                      Solution

                      Difference between Nucleus and Nucleiod:

                      Nucleus

                      Nucleiod

                      A membrane-bound organelle in the cytoplasm of eukaryotes.

                      A particular area in the cytoplasm of eukaryotes.

                      Contains more than one chromosome.

                      Contains a single chromosome.

                      Well-organized structure

                      Poorly organized structure

                      Spherical shaped

                      Irregular shaped

                      Nucleolus and nucleoplasm present

                      Nucleolus and nucleoplasm absent


                      57. Differentiate between rough and smooth endoplasmic reticulum. How is endoplasmic reticulum important for membrane biogenesis?

                      Solution

                      Differences between Rough Endoplasmic Reticulum (RER) and Smooth Endoplasmic Reticulum (SER):

                      Rough endoplasmic reticulum

                      Smooth endoplasmic reticulum

                      It possesses ribosomes attached to the membrane.

                      It does not possess ribosomes attached to its membrane.

                      Found deep inside the cytoplasm.

                      Found on the periphery.

                      Involved in protein synthesis.

                      Involved in lipid synthesis.

                      May develop from the nucleus.

                      May develop from the rough endoplasmic reticulum.


                      58. In brief state, what happens when :
                      (a) Dry apricots are left for sometime in pure water and later transferred to sugar solution?
                      (b) A Red Blood Cell is kept in concentrated saline solution?
                      (c) The Plasma-membrane of a cell breaks down?
                      (d) Rheo leaves are boiled in water first and then a drop of sugar syrup is put on it?
                      (e) Golgi apparatus is removed from the cell?

                      Solution

                      (a) When we put few dried raisins or apricots in plain water and leave them for some time, the cell gains water and swells. If we put the same seeds in a concentrated sugar solution, the seeds lose water, and consequently shrink.

                      (b) The red blood cells lose water when kept in concentrated saline solution and shrink consequently.

                      (c) The cell dies when the plasma membrane of a cell breaks down.

                      (d) The cells of rheo leaves are dead on boiling. If we put sugar solution on the dead leaves, there will be no intake of water due to lack of osmosis. Here the cell undergoes plasmolysis. This concludes that only living cells undergo osmosis.

                      (e) Removal of golgi apparatus stops the formation of vesicles and transport of proteins and lipids.


                      58. Draw a neat diagram of plant cell and label any three parts which differentiate it from animal cell.

                      Solution

                      NCERT Exemplar Chapter 6 Tissues Class 9 Science Solutions

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                      NCERT Exemplar Class 9 Science Chapter 6 Tissues Solutions

                      NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues covers all the important questions and answers as well as advanced level questions. It helps in learning about the functions of tissues, animal tissues like epithelial tissue, connective tissue, muscular and nervous tissue, meristematic plant tissues, role of tissues in plants, permanent tissues like simple tissues and complex tissues.

                      The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 6 Tissues is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like meristematic tissue is the only dividing tissue, simple tissues like  parenchyma, collenchyma and sclerenchyma, complex tissues like xylem and phloem, types of epithelial tissue, connective tissues like areolar tissue, adipose tissue, bone, tendon, cartilage and blood and muscular tissues like striated, striated and cardiac.


                      Chapter Name

                      Chapter 6 Tissues

                      Book Title

                      NCERT Exemplar for Class 9 Science

                      Related Study

                      • NCERT Solutions for Class 9 Science Chapter 6 Tissues
                      • Revision Notes for Class 9 Science Chapter 6 Tissues
                      • MCQ for Class 9 Science Chapter 6 Tissues
                      • Important Questions for Class 9 Science Chapter 6 Tissues

                      Topics Covered

                      • MCQ
                      • Short Answers Questions
                      • Long Answers Questions

                      NCERT Exemplar Solutions for Chapter 6 Tissues Class 9 Science

                      Multiple Choice Questions

                      1. Which of the following tissues has dead cells?
                      (a) Parenchyma
                      (b) Sclerenchyma
                      (c) Collenchyma
                      (d) Epithelial tissue

                      Solution

                      (b) Sclerenchyma

                      Sclerenchyma is composed of dead cells which have thick cell wall due to deposition of lignin. They are generally located in the leaf veins, hard coverings of the seeds and can also be found surrounding the vascular bundle.


                      2. Find out incorrect sentence :
                      (a) Parenchymatous tissues have intercellular spaces.
                      (b) Collenchymatous tissues are irregularly thickened at corners.
                      (c) Apical and intercalary meristems are permanent tissues.
                      (d) Meristematic tissues, in its early stage, lack vacuoles.

                      Solution

                      (c) Apical and intercalary meristems are permanent tissues.

                      Apical and intercalary meristems are not permanent tissues but they are meristematic tissues which remain in continuous state of division. These tissues differentiate to give rise to the permanent tissues.


                      3. Girth of stem increases due to :
                      (a) apical meristem
                      (b) lateral meristem
                      (c) intercalary meristem
                      (d) vertical meristem

                      Solution

                      (b) lateral meristem

                      Lateral meristem occurs on the sides almost parallel to the long axis of the root, stem and its branches. It is responsible for an increase in girth of the stem, i.e. secondary growth.


                      4. Which cell does not have perforated cell wall?
                      (a) Tracheids
                      (b) Companion cells
                      (c) Sieve tubes
                      (d) Vessels

                      Solution

                      (b) Companion cells


                      5. Intestine absorb the digested food materials. What type of epithelial cells are responsible for that?
                      (a) Stratified squamous epithelium
                      (b) Columnar epithelium
                      (c) Spindle fibres
                      (d) Cuboidal epithelium

                      Solution

                      (b) Columnar epithelium

                      Columnar epithelium is pillar-like cell which have nuclei towards its base. They form the lining of stomach, small intestine and colon, forming the mucous membrane. They also the facilitate movement across the epithelial barrier. Their main function is absorption and secretion.


                      6. A person met with an accident in which two long bones of hand were dislocated. Which among the following may be the possible reason?
                      (a) Tendon break
                      (b) Break of skeletal muscle
                      (c) Ligament break
                      (d) Areolar tissue break

                      Solution

                      (c) Ligament break

                      When joints held by ligaments get separated dislocation of bones occur. Tendons join the skeletal muscle and they cause inflammation upon break.


                      7. While doing work and running, you move your organs like hands, legs, etc. Which among the following is correct?
                      (a) Smooth muscles contract and pull the ligament to move the bones.
                      (b) Smooth muscles contract and pull the tendons to move the bones.
                      (c) Skeletal muscles contract and pull the ligament to move the bones.
                      (d) Skeletal muscles contract and pull the tendon to move the bones.

                      Solution

                      (d) Skeletal muscles contract and pull the tendon to move the bones.

                      Skeletal muscles contract and they pull the tendon, connecting muscles to bones while doing work and running. This moves the bones and they do not pull the ligaments as that will cause a sprain or stretch.


                      8. Which muscles act involuntarily?
                      (i) Striated muscles
                      (ii) Smooth muscles
                      (iii) Cardiac muscles
                      (iv) Skeletal muscles
                      (a) (i) and (ii)
                      (b) (ii) and (iii)
                      (c) (iii) and (iv)
                      (d) (i) and (iv)

                      Solution

                      (b) (ii) and (iii)

                      Skeletal muscles are voluntary muscles which are also striated muscles.


                      9. Meristematic tissues in plants are :
                      (a) localised and permanent
                      (b) not limited to certain regions
                      (c) localised and dividing cells
                      (d) growing in volume

                      Solution

                      (c) localised and dividing cells

                      Meristematic tissues in plants are the dividing cells. They are responsible for plant growth in certain specific region. Based on the location, meristematic tissues are classified as apical, lateral and intercalary.


                      10. Which is not a function of epidermis?
                      (a) Protection from adverse condition
                      (b) Gaseous exchange
                      (c) Conduction of water
                      (d) Transpiration

                      Solution

                      (c) Conduction of water

                      Conduction of water is a function of xylem tissue whereas protection, gaseous exchange and transpiration are the function of epidermis.


                      11. Select the incorrect sentence :

                      (a) Blood has matrix containing proteins, salts’ and hormones.
                      (b) Two bones are connected with ligament.
                      (c) Tendons are non-fibrous tissue and fragile.
                      (d) Cartilage is a form of connective tissue.

                      Solution

                      (c) Tendons are non-fibrous tissue and fragile.

                      Tendons are fibrous tissues which are highly elastic and strong.


                      12. Cartilage is not found in :
                      (a) nose
                      (b) ear
                      (c) kidney
                      (d) larynx

                      Solution

                      (c) kidney

                      Cartilage is a connective tissue which provides support and flexibility to various parts of our body. Cartilage is found in nose, ear, larynx but not in kidney.


                      13. Fats are stored in human body as :
                      (a) cuboidal epithelium
                      (b) adipose tissue
                      (c) bones
                      (d) cartilage

                      Solution

                      (b) adipose tissue

                      Fat is stored in the adipose tissue present below the skin and between internal organs. Cells in these tissues are filled with fat globules.


                      14. Bone matrix is rich in :
                      (a) fluoride and calcium
                      (b) calcium and phosphorus
                      (c) calcium and potassium
                      (d) phosphorus and potassium

                      Solution

                      (b) calcium and phosphorus


                      15. Contractile proteins are found in :
                      (a) bones
                      (b) blood
                      (c) muscles
                      (d) cartilage

                      Solution

                      (c) muscles

                      Contractile proteins are responsible for the contraction and relaxation of muscles.


                      16. Voluntary muscles are found in :
                      (a) alimentary canal
                      (b) limbs
                      (c) iris of the eye
                      (d) bronchi of lungs

                      Solution

                      (b) limbs

                      Some muscles can be moved by our own will. Muscles present in our limbs can move at our will, and stop when we decide to. Such muscles are called voluntary muscles.


                      17. Nervous tissue is not found in :
                      (a) brain
                      (b) spinal cord
                      (c) tendons
                      (d) nerves

                      Solution

                      (c) tendons

                      Tendons are the connective tissue which connects muscles to bones. They are fibrous in nature and gives strength and limited flexibility.


                      18. Nerve cell does not contain :
                      (a) axon
                      (b) nerve endings
                      (c) tendons
                      (d) dendrites

                      Solution

                      (c) tendons

                      Neurons are connective tissues that join skeletal muscles to bones, so they do not contain tendons.


                      19. Which of the following helps in repair of tissue and fills up the space inside the organ?
                      (a) Tendon
                      (b) Adipose tissue
                      (c) Areolar
                      (d) Cartilage

                      Solution

                      (c) Areolar

                      Areolar connective tissue is found between the skin and muscles, around blood vessels and nerves and in the bone marrow. It fills the space inside the organs, supports internal organs and helps in repair of tissues.


                      20. The muscular tissue which function throughout the life continuously without fatigue is :
                      (a) skeletal muscle
                      (b) cardiac muscle
                      (c) smooth muscle
                      (d) voluntary muscle

                      Solution

                      (b) cardiac muscle

                      Heart muscles are cylindrical, branched and uninucleated. They show rhythmic contraction and relaxation throughout life. Skeletal muscle, smooth muscle and voluntary muscles work as and when required.


                      21. Which of the following cells is found in the cartilaginous
                      tissue of the body?
                      (a) Mast cells
                      (b) Basophils
                      (c) Osteocytes
                      (d) Chondrocytes

                      Solution

                      (d) Chondrocytes

                      Mast cells are found in areolar tissue. Basophils are found in blood. Osteocytes are found in bone.


                      22. The dead element present in the phloem is :
                      (a) companion cells
                      (b) phloem fibres
                      (c) phloem parenchyma
                      (d) sieve tubes

                      Solution

                      (b) phloem fibres

                      Phloem fibres possess narrow lumen and thick-walls. They are elongated spindle-shaped dead cell. It provides mechanical support to the tissue. Phloem parenchyma is the thin-walled living cells of parenchyma.


                      23. Which of the following does not lose their nucleus at maturity?
                      (a) Companion cells
                      (b) Red blood cells
                      (c) Vessel
                      (d) Sieve tube cells

                      Solution

                      (a) Companion cells


                      24. In desert plants, rate of water loss gets reduced due to the presence of :
                      (a) cuticle
                      (b) stomata
                      (c) lignin
                      (d) suberin

                      Solution

                      (a) cuticle

                      Cuticle is a protective layer covering epidermis of the leaf, young shoots and other aerial parts of a plant. Cuticle minimizes the effect of heat and reduces the loss of water as it contains lipids and polymers impregnated with wax.


                      25. A long tree has several branches. The tissue that helps in the sideways conduction of water in the branches is :
                      (a) collenchyma
                      (b) xylem parenchyma
                      (c) parenchyma
                      (d) xylem vessels

                      Solution

                      (d) xylem vessels

                      Collenchyma is an active tissue and has no role in the conduction of water. Parenchyma is a supportive tissue. Xylem vessels also known as xylem trachea are responsible for conduction of water in plants.


                      26. If the tip of sugarcane plant is removed from the field, even then it keeps on growing in length. It is due to the presence of :
                      (a) cambium
                      (b) apical meristem
                      (c) lateral meristem
                      (d) intercalary meristem

                      Solution

                      (d) intercalary meristem

                      Intercalary meristem is responsible for the longitudinal growth of the plant even in the absence of apical meristem.


                      27. A nail is inserted in the trunk of a tree at a height of 1 metre from the ground level. After 3 years the nail will :
                      (a) move downwards
                      (b) move upwards
                      (c) remain at the same position
                      (d) move sideways

                      Solution

                      (c) remain at the same position

                      Longitudinal growth in the stem always takes place on the top. The apical meristem in the lower region remains constant. This is the reason why there will be no change in position of nail inserted in the trunk of the tree.


                      28. Parenchyma cells are :
                      (a) relatively unspecified and thin walled
                      (b) thick walled and specialised
                      (c) lignified
                      (d) None of these

                      Solution

                      (a) relatively unspecified and thin walled

                      The most common simple permanent tissue is the parenchyma. It consists of relatively unspecialized cells with thin cell walls. They are living cells. They are usually loosely arranged, thus large spaces between cells (intercellular spaces) are found in this tissue.


                      29. Flexibility in plants is due to :
                      (a) collenchyma
                      (b) sclerenchyma
                      (c) parenchyma
                      (d) chlorenchyma

                      Solution

                      (a) collenchyma


                      30. Cork cells are made impervious to water and gases by the presence of :
                      (a) cellulose
                      (b) lipids
                      (c) suberin
                      (d) lignin

                      Solution

                      (c) suberin


                      31. Survival of plants in terrestrial environment has been made possible by the presence of :
                      (a) intercalary meristem
                      (b) conducting tissue
                      (c) apical meristem
                      (d) parenchymatous tissue

                      Solution

                      (b) conducting tissue

                      Xylem and phloem which are the conducting tissues are responsible for the survival of plants in terrestrial environment. Xylem conduct water from roots to all the parts of the plants and phloem transports food and other nutrients from leaves to other parts of the plant.


                      32. Choose the wrong statement.
                      (a) The nature of matrix differs according to the function of the tissue.
                      (b) Fats are stored below the skin and in between the internal organ
                      (c) Epithelial tissues have intercellular spaces between them.
                      (d) Cells of striated muscles are multinucleate and unbranched.

                      Solution

                      (c) Epithelial tissues have intercellular spaces between them.

                      Epithelial tissues do not have intercellular spaces between them. They are tightly bound together to make continuous sheet.


                      33. The water conducting tissue generally present in gymnosperm is :
                      (a) vessel
                      (b) sieve tube
                      (c) tracheids
                      (d) xylem fibres

                      Solution

                      (c) tracheids

                      Vessels are generally absent in gymnosperms. Sieve tubes are present in phloem. Xylem fibres provide structural rigidity and have no role in conduction of water.


                      Short Answer Questions

                      34. Why animals of colder regions and fishes of cold water have thicker layer of subcutaneous fat?

                      Solution

                      Animals of colder regions and fishes of cold water have thicker layer of subcutaneous fat. This acts as an insulator and prevents the escape of heat from their body.


                      35. Match the column (A) with the column (B).

                       

                      Column (A)

                       

                      Column (B)

                      (1)

                      Parenchyma

                      (1)

                      Thin walled, packing cells

                      (2)

                      Photosynthesis

                      (2)

                      Carbon fixation

                      (3)

                      Aerenchyma

                      (3)

                      Localized thickenings

                      (4)

                      Collenchyma

                      (4)

                      Buoyancy

                      (5)

                      Permanent tissue

                      (5)

                      Sclerenchyma

                      Solution

                      (A)

                      (B)

                      (a) Parenchyma

                      (i) Thin-walled, packing cells

                      (b) Photosynthesis

                      (ii) Carbon fixation

                      (c) Aerenchyma

                      (iv) Buoyancy

                      (d) Collenchyma

                      (iii) Localized thickenings

                      (e) Permanent tissue

                      (v) Sclerenchyma


                      36. Match the column (A) with the column (B).

                      (A)

                      (B)

                      (a) Fluid connective tissue

                      (i) Subcutaneous layer

                      (b) Filling of space inside the organs

                      (ii) Cartilage

                      (c) Striated muscle

                      (iii) Skeletal muscle

                      (d) Adipose tissue

                      (iv) Areolar tissue

                      (e) Surface of joints

                      (v) Blood

                      (f) Stratified squamous epithelium

                      (vi) Ski

                      Solution

                      (A)

                      (B)

                      (a) Fluid connective tissue

                      (v) Blood

                      (b) Filling of space inside the organs

                      (iv) Areolar tissue

                      (c) Striated muscle

                      (iii) Skeletal muscle

                      (d) Adipose tissue

                      (i) Subcutaneous layer

                      (e) Surface of joints

                      (ii) Cartilage

                      (f) Stratified squamous epithelium

                      (vi) Ski


                      37. If a potted plant is covered with a glass jar, water vapours appear on the wall of glass jar. Explain, why?

                      Solution

                      Water is released from the plants in the form of vapours because of the process called transpiration.


                      38. Name the different components of xylem and draw a living component.

                      Solution

                      The different components of xylem are:

                      1. tracheids
                      2. vessels
                      3. xylem parenchyma
                      4. xylem fibres.

                      The only living component of xylem is xylem parenchyma.


                      39. Draw and identify different elements of phloem.

                      Solution


                      40. Differentiate between voluntary and involuntary muscles. Give one example of each type.

                      Solution

                      Voluntary muscles

                      Involuntary muscles

                      They are also called striated muscles since they show stripes or striations.

                      They are also called non-striated muscles since they lack striations.

                      Their cells are long and cylindrical.

                      Their cells are small and spindle shaped.

                      They have multinucleate cells.

                      They have uninucleate cells.

                      They are under our will or control.

                      They are not under our will or control.

                      They get tired and need rest at intervals.

                      They can work continuously without getting tired.

                      Example: Limb muscles

                      Example: Cardiac muscles


                      41. Write true (T) or false (F) :
                      (a) Epithelial tissue is protective tissue in animal body.
                      (b) The lining of blood vessels, lung alveoli and kidney tubules are all made up of epithelial tissue.
                      (c) Epithelial cells have a lot of intercellular spaces.
                      (d) Epithelial layer is permeable layer.
                      (e) Epithelial layer does not allow regulation of materials between body and external environment.

                      (a) True,

                      (b) True

                      (c) False

                      (d) True

                      (e) False


                      42. Differentiate the following activities on the basis of voluntary or involuntary muscles :
                      (a) Jumping of frog
                      (b) Pumping of the heart
                      (c) Writing with hand
                      (d) Movement of chocolate in your intestine

                      Solution

                      (a) Jumping of frog is an activity of voluntary muscles.

                      (b) Pumping of the heart is a function of involuntary muscles.

                      (c) Writing with hand is a function of voluntary muscles.

                      (d) Movement of chocolate in your intestine is a function of involuntary muscles.


                      43. Fill in the blanks :
                      (a) Lining of blood vessels is made up of _____
                      (b) Lining of small intestine is made up of _____
                      (c) Lining of kidney tubules is made up of _____
                      (d) Epithelial cells with cilia are found in _____ of our body.

                      Solution

                      (a) Lining of blood vessels is made up of squamous epithelium.

                      (b) Lining of small intestine is made up of columnar epithelium.

                      (c) Lining of kidney tubules is made up of cuboidal epithelium.

                      (d) Epithelial cells with cilia are found in respiratory tract of our body.


                      44. Explain, why water hyacinth float on water surface?

                      Solution

                      Water hyacinth float on water surface because of the presence of the air cavities present in the parenchyma tissue.


                      45. Which structure protects the plant body against the invasion of parasites?

                      Solution

                      Epidermis protects the plant body against the invasion of parasites. It has thick cuticle and dermal tissue which help in preventing attack from parasites.


                      46. Fill in the blanks :
                      (a) Cork cells possesses _____ on their walls that makes it impervious to gases and water.
                      (b) _____ have tubular cells with perforated walls and are living in nature.
                      (c) Bone possesses a hard matrix composed of _____ and _____.

                      Solution

                      (a) suberin
                      (b) Sieve tubes
                      (c) calcium and phosphate


                      47. Why is epidermis important for the plants?

                      Solution

                      The importance of epidermis is as follows :

                      1. Epidermis gives protection against water loss. 
                      2. Epidermal cell present on the aerial parts of the plant often secrete a waxy, water resistant layer on their outer surface which provides protection against loss of water, mechanical injury and invasion by parasitic fungi. 
                      3. Epidermal tissue forms a continuous layer and helps to avoid mechanical stress.


                      48. Fill in the blanks :
                      (a) _____ are forms of complex tissue.
                      (b) _____ have guard cells.
                      (c) Cells of cork contain a chemical called _____
                      (d) Husk of coconut is made of _____ tissue.
                      (e) _____ gives flexibility in plants.
                      (f) _____ and _____ are both conducting tissues.
                      (g) Xylem transports _____ and _____ from soil.
                      (h) Phloem transport _____ from _____ to other parts of the plant.

                      Solution

                      (a) Xylem and phloem
                      (b) Stomata
                      (c) suberin
                      (d) sclerenchyma
                      (e) Collenchyma
                      (f) Xylem and phloem
                      (g) water and minerals
                      (h) food and leaves


                      Long Answer Questions

                      49. Differentiate between sclerenchyma and parenchyma tissues. Draw well labelled diagram.

                      Solution

                      Differences between parenchyma and sclerenchyma :

                      Sclerenchyma tissues

                      Parenchyma tissues

                      Consists of living cells

                      Consists of dead cells

                      Consists of thin walled cell

                      Consists of uniformly thickened cell walls

                      Cellwall is made up of cellulose

                      Cellwall is made up of complex lignin

                      Serves as packing tissue

                      Serves as mechanical tissue

                      Stores food and perform photosynthesis

                      Gives strength, rigidity and protects from parasites


                      50. Describe the structure and functions of different types of epithelial tissues with diagram.

                      Solution

                      Epithelial cells are the thin protective coverings that line most of the organs and cavities within the body. It forms a barrier to keep the different body systems separate. The skin, the lining of the mouth, the lining of blood vessels, lung alveoli and kidney tubules are all made of epithelial tissue. Epithelial tissue cells are tightly packed and form a continuous sheet.

                      Anything entering or leaving the body must cross at least one layer of epithelium. As a result, the permeability of the cells of various epithelia plays an important role in regulating the exchange of materials between the body and the external environment and also between different parts of the body. Regardless of the type, all epithelium is usually separated from the underlying tissue by an extracellular fibrous basement membrane.

                      Depending on the shape and function, epithelial cells are classified into:

                      1. Squamous epithelial tissue
                      2. Stratified squamous epithelial tissue
                      3. Cuboidal epithelial tissue
                      4. Columnar( Ciliated) epithelial tissues)

                      1. Squamous epithelial tissue: In cell lining, blood vessels or lung alveoli transportation of substances occurs through a selectively permeable surface, this epithelium is a flat kind of epithelium. This is called the simple squamous epithelium. These cells are extremely thin and flat and form a delicate lining. The oesophagus and the lining of the mouth are also covered with squamous epithelium. The skin, which protects the body, is also made of squamous epithelium.

                      2. Stratified squamous epithelium: Skin epithelial cells are arranged in many layers to prevent wear and tear. Since, they are arranged in a pattern of layers, the epithelium is called stratified squamous epithelium.

                      3. Columnar epithelium: Tall epithelial cells are present where absorption and secretion occurs, as in the inner lining of the intestine. This columnar epithelial facilitates movement across the epithelial barrier. In the respiratory tract, the columnar epithelial tissue also has cilia, which are hair-like projections on the outer surfaces of epithelial cells.

                      4. Cuboidal epithelium: These form the lining of the kidney tubules and ducts of salivary glands where these provide mechanical support. Sometimes, a portion of the epithelial tissue folds inward and a multicellular gland is formed. This is glandular epithelium.


                      51. Draw well labelled diagrams of various types of muscles found in human body.

                      Solution


                      52. Give reasons for :
                      (i) Meristematic cells have a prominent nucleus and dense cytoplasm but they lack vacuole.
                      (ii) Intercellular spaces are absent in sclerenchymatous tissues.
                      (iii) We get a crunchy and granular feeling, when we chew pear fruit.
                      (iv) Branches of a tree move and bend freely in high wind velocity.
                      (v) It is difficult to pull out the husk of a coconut tree.

                      Solution

                      (i) Meristamatic tissue cells divide and they have prominent nucleus and a dese cytoplasm. As they are dividing rigorously they need not store food or waste products. This is the reason they lack vacuoles.

                      (ii) Cell wall of sclerenchyma are lignified and are packed densely. This in turn protects the plant and gives mechanical strength to the plant. This is the reason why intercellular spaces are absent in sclerenchymatous tissues.

                      (iii) Sclerenchyma cells are called stone cells in pear fruit. They are small, thick and hard. Because of this we get a crunchy and granular feeling when we chew a pear fruit.

                      (iv) Junction of a tree branch is composed of collenchyma cells which provide rigidity and flexibility to the branches. This is the reason why branches of a tree move and bend freely in high wind velocity.

                      (v) Husk of coconut tree is made up of sclerenchyma which is hard. This is the reason why it is difficult to pull out the husk of a coconut tree.


                      53. List the characteristics of cork. How are they formed? Mention their role.

                      Solution

                      Characteristics of cork are as follows:

                      Cork cells are dead at maturity.

                      Cork cells are compactly arranged.

                      Cells do not possess intercellular spaces.

                      Cells possess a chemical substances suberin in their walls

                      They have several layers.

                      With age, the outer protective tissue of the plant undergoes changes. Epidermis of the stem is replaced by a secondary meristem called the phellogen or cork cambium. It is a simple tissue consisting of rectangular cells whose protoplasts are vacuolated. Cork cells contain tanins and chloroplasts.

                      On the outer side the cork cambium forms cork and on the inner side a secondary cortex by giving off new cells. The layer of cells which is cut by cork cambium on the outer side ultimately becomes several layered thick cork (bark) of trees.

                      Cork cells are compactly arranged dead cells. Suberin which is fat thickens the walls of cork. Suberin makes the cork cells impermeable to water and gases. Cork provides protection to plant and it prevents the loss of water from plant body. It also protects the plants from infection and mechanical injury. Cork is light in weight and it cannot catch fire. Because of this property it is used as insulators. Cork is hard in nature hence it is used to make sports goods.


                      54. Why are xylem and phloem called complex tissues ? How are they different from one another ?

                      Solution

                      A tissue which has more than one type of cells having a common origin which coordinate to perform a common function is known as specific tissue. As xylem and phloem are made of different types of cells they are called complex tissues.

                      Difference between Phloem and Xylem

                      Phloem

                      Xylem

                      Transports minerals and water.

                      Transports food.

                      Consists of tracheids, vessels, xylem, parenchyma and xylem fibres.

                      Consist of sieve tubes, companion cell, phloem parenchyma and phloem fibres.

                      They transport water and minerals vertically from soil to aerial parts of the plant.

                      They transport food from leaves to other parts of the plant.

                      Most of the cells except xylem parenchyma are dead cells.

                      Most of the cells except phloem fibres are living cells.


                      54. (a) Differentiate between meristematic and permanent tissues in plants.
                      (b) Define the process of differentiation.
                      (c) Name any two simple and two complex permanent tissues in plants.

                      Solution

                      (a) Difference between meristematic and permanent tissues in plants:

                      Meristematic tissues

                      Permanent tissues

                      Its component cells are living, small, spherical or polygonal and un-differentiated

                      Its components cells may be living or dead. They are large, differentiated with different shapes.

                      Cytoplasm is dense and vacuoles are nearly absent as they are metabolically active.

                      Large central vacuole occurs in living permanent cells as, they are less metabolically active.

                      Intercellular spaces are absent.

                      Intercellular spaces are often present.

                      Cell wall of its cells is thin and elastic.

                      Cell wall of its cells may either thin or thick.

                      Nucleus of each cell of this tissue is large and prominent.

                      Nucleus is less conspicuous.

                      Its cells grow and divide regularly.

                      Its cells do not normally divide.

                      It is a simple tissue.

                      It can be simple, complex or specialized tissues.

                      Cell organelles of its cells are simple

                      Cell organelles of its cells are well developed.

                      Cell contain crystals and other inclusions.

                      Cells crystals and other inclusions

                      It provides growth to the plant.

                      It provides protection. Support, conduction photosynthesis, storage, etc.

                      (b) Cells which are derived from the division of meristematic tissues take up specific roles and gradually lose their ability to divide. In this way they form permanent tissues. The process by which the cells divide meristematically to take a permanent shape, size and function is known as differentiation.

                      (c) The two simple permanent tissues are parenchyma and collenchyma whereas xylem and phloem are two complex permanent tissues.

                      NCERT Exemplar Chapter 7 Diversity in Living Organisms Class 9 Science Solutions

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                      NCERT Exemplar Class 9 Science Chapter 7 Diversity in Living Organisms Solutions

                      NCERT Exemplar Solutions for Class 9 Science Chapter 7 Diversity in Living Organisms covers all the important questions and answers as well as advanced level questions. It helps in learning about the biodiversity, classification, living organisms, their types, the classification of organisms, Classification System i.e. two kingdom classification and five kingdom classification, levels or groups and types of cellular organization.

                      The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 7 Diversity in Living Organisms is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like body organization, mode of obtaining food, major identification patterns for classification, 5 kingdoms that are protista, monera, fungi, plantae and animalia, binomial nomenclature, five groups of plants which are thallophytes, bryophytes, pteridophytes, gymnosperms and angiosperms, Porifera, Coelenterata, Platyhelminthes, Nematoda, Annelida, Arthropoda, Mollusca, Echinodermata, Protochordates and Vertebrata


                      Chapter Name

                      Chapter 7 Diversity in Living Organisms

                      Book Title

                      NCERT Exemplar for Class 9 Science

                      Related Study

                      • NCERT Solutions for Class 9 Science Chapter 7 Diversity in Living Organisms
                      • Revision Notes for Class 9 Science Chapter 7 Diversity in Living Organisms
                      • MCQ for Class 9 Science Chapter 7 Diversity in Living Organisms
                      • Important Questions for Class 9 Science Chapter 7 Diversity in Living Organisms

                      Topics Covered

                      • MCQ
                      • Short Answers Questions
                      • Long Answers Questions

                      NCERT Exemplar Solutions for Chapter 7 Diversity in Living Organisms Class 9 Science

                      Multiple Choice Questions

                      1. Find out incorrect sentence :
                      (a) Protista includes unicellular eukaryotic organisms.
                      (b) Whittaker considered cell structure, mode and source of nutrition for classifying the organisms in five kingdoms.
                      (c) Both Monera and Protista may be autotrophic and heterotrophic.
                      (d) Monerans have well defined nucleus.

                      Solution

                      (d) Monerans have well defined nucleus.

                      Monerans include single-celled prokaryotes, actinomycetes and photosynthetic blue-green algae. They don’t have well-defined nucleus and cell organelles.


                      2. Which among the following has specialized tissue for conduction of water?
                      (i) Thallophyta
                      (ii) Bryophyta
                      (iii) Pteridophyta
                      (iv) Gymnosperms
                      (a) (i) and (ii)
                      (b) (ii) and (iii)
                      (c) (iii) and
                      (iv) (d) (i) and (iv)

                      Solution

                      (c) (iii) and (iv)

                      Thallophytes and Bryophytes don’t have specialized tissues for water conduction whereas Pteridophytes and Gymnosperms have specialized tissues for conduction of water.


                      3. Which among the following produce seeds?
                      (a) Thallophyta
                      (b) Bryophyta
                      (c) Pteridophyta
                      (d) Gymnosperms

                      Solution

                      (d) Gymnosperms

                      Gymnosperms are the flowerless plants that produce seeds.


                      4. Which one is a true fish?
                      (a) Jellyfish
                      (b) Starfish
                      (c) Dogfish
                      (d) Silverfish

                      Solution

                      (c) Dogfish

                      Jellyfish is a coelenterate, starfish belongs to Echinodermata and silverfish is an Arthropod.


                      5. Which among the following is exclusively marine?
                      (a) Porifera
                      (b) Echinodermata
                      (c) Mollusca
                      (d) Pisces

                      Solution

                      (b) Echinodermata

                      Echinodermata are exclusively found in marine environment whereas Porifera, Molluscs and Pisces can be found in both marine and freshwater.


                      6. Which among the following have open circulatory system?
                      (i) Arthropoda
                      (ii) Mollusca
                      (iii) Annelida
                      (iv) Coelenterata
                      (a) (i) and (ii)
                      (b) (iii) and (iv)
                      (c) (i) and (iii)
                      (d) (ii) and (iv)

                      Solution

                      (a) (i) and (ii)

                      Annelida and Coelenterata have closed circulatory system whereas Arthropods and Mollusca have open circulatory system.


                      7. In which group of animals, coelom is filled with blood?
                      (a) Arthropoda
                      (b) Annelida
                      (c) Nematoda
                      (d) Echinodermata

                      Solution

                      (a) Arthropoda

                      Annelida, Nematoda and Echinodermata don’t have blood whereas the coelom in Arthropods is filled with blood.


                      8. Elephantiasis is caused by :
                      (a) Wuchereria
                      (b) Pinworm
                      (c) Planarians
                      (d) Liver flukes

                      Solution

                      (a) Wuchereria

                      Wuchereria is a human parasite which causes Elephantiasis. Elephantiasis is spread through mosquitos.


                      9. Which one is the most striking or (common) character of the vertebrates?
                      (a) Presence of notochord
                      (b) Presence of triploblastic condition
                      (c) Presence of gill pouches
                      (d) Presence of coelom

                      Solution

                      (a) Presence of notochord

                      Vertebrates and invertebrates have triploblastic condition, gill pouches and coelom but notochord is exclusively present in vertebrates.


                      10. Which among the following have scales?
                      (i) Amphibians
                      (ii) Pisces
                      (iii) Reptiles
                      (iv) Mammals
                      (a) (i) and (iii)
                      (b) (iii) and (iv)
                      (c) (ii) and (iii)
                      (d) (i) and (ii)

                      Solution

                      (c) (ii) and (iii)

                      Amphibians and mammals do not have scales on their body whereas pisces and reptiles have scales on their body.


                      11. Find out the false statement :
                      (a) Ayes are warm blooded, egg laying and have four chambered heart.
                      (b) Ayes have feather covered body, forelimbs are modified as wing and breathe through lungs.
                      (c) Most of the mammals are viviparous.
                      (d) Fishes, amphibians and reptiles are oviparous.

                      Solution

                      (d) Fishes, amphibians and reptiles are oviparous.

                      Some fishes are viviparous but amphibians show external fertilization. Amphibians can neither be kept under oviparous nor be viviparous.


                      12. Pteridophyta do not have :
                      (a) Root
                      (b) Stem
                      (c) Flowers
                      (d) Leaves

                      Solution

                      (c) Flowers


                      13. Identify a member of porifera :
                      (a) Spongilla
                      (b) Euglena
                      (c) Penicillium
                      (d) Hydra

                      Solution

                      (a) Spongilla

                      Euglena is a protozoan. Pencillium is a fungi and hydra is a Coelenterata.


                      14. Which is not an aquatic animal?
                      (a) Hydra
                      (b) Jellyfish
                      (c) Corals
                      (d) Filaria

                      Solution

                      (d) Filaria

                      Filaria is a disease caused by Wuchereria.


                      15. Amphibians do not have the following :
                      (a) Three chambered heart
                      (b) Gills or lungs
                      (c) Scales
                      (d) Mucus glands

                      Solution

                      (c) Scales

                      Amphibians have a 3 chambered heart. Adults have lungs and tadpoles have gills. Mucus glands are present on the skin of amphibians.


                      16. Organisms without nucleus and cell organelles belong to :
                      (i) Fungi
                      (ii) Protista
                      (iii) Cyano bacteria
                      (iv) Archae bacteria
                      (a) (i) and (ii)
                      (b) (iii) and (iv)
                      (c) (i) and (iv)
                      (d) (ii) and (iii)

                      Solution

                      (b) (iii) and (iv)

                      Cyanobacteria and archaebacteria are prokaryotes and they do not have a well-defined nucleus and cell organelles. Fungi and Protista are Eukaryote which possess cell organelles and nucleus.


                      17. Which of the following is not a criterion for classification of living organisms?
                      (a) Body design of the organism
                      (b) Ability to produce one’s own food
                      (c) Membrane bound nucleus and cell organelles
                      (d) Height of the plant

                      Solution

                      (d) Height of the plant

                      Height of a plant is related to bushes and trees which belong of Kingdom Plantae.


                      18. The feature that is not a characteristic of protochordata?
                      (a) Presence of notochord
                      (b) Bilateral symmetry and coelom
                      (c) Jointed legs
                      (d) Presence of circulatory system

                      Solution

                      (c) Jointed legs

                      Protochordata are triploblastic with bilaterally symmetric body and coelom. At some stage of life they show notochord and they are marine living.


                      19. The locomotory organs of Echinodermata are :
                      (a) Tube feet
                      (b) Muscular feet
                      (c) Jointed legs
                      (d) Parapodia

                      Solution

                      (a) Tube feet

                      Tube feet in Echinodermata help in locomotion and respiration.


                      20. Corals are :
                      (a) Poriferans attached to some solid support
                      (b) Cnidarians that are solitary living
                      (c) Poriferans present at the sea bed
                      (d) Cnidarians that live in colonies

                      Solution

                      (d) Cnidarians that live in colonies


                      21. Who introduced the system of scientific nomenclature of organisms?
                      (a) Robert Whittaker
                      (b) Carolus Linnaeus
                      (c) Robert Hooke
                      (d) Ernst Haeckel

                      Solution

                      (b) Carolus Linnaeus

                      Carolus Linnaeus introduced binomial nomenclature which is simplified method of naming organisms. Binomial nomenclature gives each organism a scientific name that has two parts. First part is a Genus and second part is Species.


                      22. Two chambered heart occurs in :
                      (a) Crocodiles
                      (b) Fish
                      (c) Ayes
                      (d) Amphibians

                      Solution

                      (b) Fish

                      Amphibians have 3 chambered heart whereas aves and crocodiles have 4 chambered heart.


                      23. Skeleton is made entirely of cartilage in :
                      (a) Sharks
                      (b) Tuna
                      (c) Rohu
                      (d) None of these

                      Solution

                      (a) Sharks

                      Sharks are cartilaginous whereas tuna and rohu are bony fishes.


                      24. One of the following is not an Annelid :
                      (a) Nereis
                      (b) Earthworm
                      (c) Leech
                      (d) Urchins

                      Solution

                      (d) Urchins

                      Urchins are coelenterates.


                      25. The book Systema Naturae was written by :
                      (a) Linnaeus
                      (b) Haeckel
                      (c) Whittaker
                      (d) Robert Brown

                      Solution

                      (a) Linnaeus


                      26. Karl Von Linne was involved with which branch of science?
                      (a) Morphology
                      (b) Taxonomy
                      (c) Physiology
                      (d) Medicine

                      Solution

                      (b) Taxonomy


                      27. Real organs are absent in :
                      (a) Mollusca
                      (b) Coelenterata
                      (c) Arthropoda
                      (d) Echinodermata

                      Solution

                      (b) Coelenterata

                      Coelenterates are at the tissue level of organization and thus they lack real organs.


                      28. Hard calcium carbonate structures are used as skeleton by :
                      (a) Echinodermata
                      (b) Protochordata
                      (c) Arthropoda
                      (d) Nematoda

                      Solution

                      (a) Echinodermata

                      Echinodermata are exclusively free-living marine animals with spiny skin. They are triploblastic and have a coelomic cavity. They use a unique water-driven tube for moving and also contain calcium carbonate structures which are used as skeletons.


                      29. Differentiation in segmental fashion occurs in :
                      (a) Leech
                      (b) Starfish
                      (c) Snails
                      (d) Ascaris

                      Solution

                      (a) Leech

                      Leech belongs to kingdom Annelids and it shows metameric body segmentation.


                      30. In taxonomic hierarchy family comes between :
                      (a) Class and Order
                      (b) Order and Genus
                      (c) Genus and Species
                      (d) Division and Class

                      Solution

                      (b) Order and Genus


                      31. 5-Kingdom classification has given by :
                      (a) Morgan
                      (b) R. Whittaker
                      (c) Linnaeus
                      (d) Haeckel

                      Solution

                      (b) R. Whittaker

                      R. Whittaker proposed 5 kingdom classification which includes Monera, Protista, Fungi, Plantae and Animalia.


                      32. Well defined nucleus is absent in :
                      (a) blue green algae
                      (b) diatoms
                      (c) algae
                      (d) yeast

                      Solution

                      (a) blue green algae

                      Blue-green algae belong to kingdom prokaryotes which do not have well defined nucleus and cell organelles.


                      33. The ‘Origin of Species’ is written by :
                      (a) Linnaeus
                      (b) Darwin
                      (c) Haeckel
                      (d) Whittaker

                      Solution

                      (b) Darwin


                      34. Meena and Hari observed an animal in their garden. Hari called it an insect while Meena said it was an earthworm. Choose the character from the following which confirms that it is an insect.
                      (a) Bilateral symmetrical body
                      (b) Body with jointed legs
                      (c) Cylindrical body
                      (d) Body with little segmentation

                      Solution

                      (b) Body with jointed legs

                      Kingdom Arthropoda has the characteristic feature of body with joined legs and all the insects belong to this kingdom.


                      Short Answer Questions

                      35 Write true (T) or false (F) :
                      (a) Whittaker proposed five kingdom classifications.
                      (b) Monera is divided into Archaebacteria and Eubacteria.
                      (c) Starting from Class, Species comes before the Genus.
                      (d) Anabaena belongs to the kingdom Monera.
                      (e) Blue green algae belong to the kingdom Protista.
                      (f) All prokaryotes are classified under Monera.

                      Solution

                      (a) True

                      (b) True

                      (c) False

                      (d) True

                      (e) False

                      (f) True


                      36. Fill in the blanks :
                      (a) Fungi shows _____ mode of nutrition.
                      (b) Cell wall of fungi is made up of _____
                      (c) Association between blue green algae and fungi is called as _____
                      (d) Chemical nature of chitin is _____
                      (e) _____ has smallest number of organisms with maximum number of similar characters.
                      (f) Plants without well differentiated stem, root and leaf are kept in _____
                      (g) _____ are called as amphibians of the plant kingdom.

                      Solution

                      (a) saprophytic,
                      (b) Chitin,
                      (c) lichens,
                      (d) carbohydrates,
                      (e) Species,
                      (f) thallophyta,
                      (g) Bryophytes


                      37. You are provided with the seeds of gram, wheat, rice, pumpkin, maize and pea. Classify them whether they are monocot or dicot.

                      Solution

                      Gram – dicot

                      Wheat – monocot

                      Rice – monocot

                      Pumpkin – dicot

                      Maize – monocot

                      Pea – dicot


                      38. Match items of column A with items of column B.

                       

                      Column A

                       

                      Column B

                      1.

                      Naked eye

                      (a)

                      Angiosperms

                      2.

                      Covered seed

                      (b)

                      Gymnosperms

                      3.

                      Flagella

                      (c)

                      Bryophytes

                      4.

                      Marchantia

                      (d)

                      Euglena

                      5.

                      Marsilea

                      (e)

                      Thallophyta

                      6.

                      Cladophora

                      (f)

                      Pteridophyta

                      7.

                      Penicillium

                      (g)

                      Fungi

                      Solution

                       

                      Column A

                       

                      Column B

                      1.

                      Naked eye

                      (a)

                      Gymnosperms

                      2.

                      Covered seed

                      (b)

                      Angiosperms

                      3.

                      Flagella

                      (c)

                      Euglena

                      4.

                      Marchantia

                      (d)

                      Bryophytes

                      5.

                      Marsilea

                      (e)

                      Pteridophyta

                      6.

                      Cladophora

                      (f)

                      Thallophyta

                      7.

                      Penicillium

                      (g)

                      Fungi


                      39. Match items of column A with items of column B :

                       

                      Column A

                       

                      Column B

                      1.

                      Pore bearing animals

                      (a)

                      Arthropoda

                      2.

                      Diploblastic

                      (b)

                      Coelenterata

                      3.

                      Metameric segmentation

                      (c)

                      Porifera

                      4.

                      Jointed legs

                      (d)

                      Echinodermata

                      5.

                      Soft bodied animals

                      (e)

                      Mollusca

                      6.

                      Spiny skinned animals

                      (f)

                      Annelida

                      Solution

                       

                      Column A

                       

                      Column B

                      1.

                      Pore bearing animals

                      (a)

                      Porifera

                      2.

                      Diploblastic

                      (b)

                      Coelenterata

                      3.

                      Metameric segmentation

                      (c)

                      Annelida

                      4.

                      Jointed legs

                      (d)

                      Arthropoda

                      5.

                      Soft bodied animals

                      (e)

                      Mollusca

                      6.

                      Spiny skinned animals

                      (f)

                      Echinodermata


                      40. Classify the following organisms based on the absence/presence of true coelom (i.e., acoelomate, pseudocoelomate and coelomate) : Spongilla, Sea anemone, Planaria, Liver fluke, Wuchereria, Ascaris, Nereis, Earthworm, Scorpion, Birds, Fishes, Horse.

                      Solution

                      Spongilla – Acoelomate,

                      Sea anemone – Acoelomate

                      Planaria – Acoelomate,

                      Liver fluke – Acoelomate

                      Wuchereria – Pseudocoelomate,

                      Ascaris – Psudocoelomate

                      Nereis – Coelomate,

                      Scorpion – Coelomate

                      Earthworm – Coelomate,

                      Birds, Fishes and Horse – Coelomate


                      41. Endoskeleton of fishes are made up of cartilage and bone; classify the following fishes as cartilagenous or bony : Torpedo, Sting ray, Dog fish, Rohu, Angler fish, Exocoetus.

                      Solution

                      Torpedo – Cartilagenous,

                      Sting ray – Cartilagenous

                      Dog fish – cartilagenous,

                      Rohu – Bony

                      Angler fish – Cartilagenous,

                      Exocoetus – Bony


                      42. Classify the following based on number of chambers in their heart :
                      Rohu, Scoliodon, Frog, Salamander, Flying lizard, King Cobra, Crocodile, Ostrich, Pigeon, Bat, Whale.

                      Solution

                      • Rohu-2 chambered
                      • Scoliodon-2 chambered
                      • Frog-3 chambered
                      • Salamander-3 chambered
                      • Flying lizard-3 chambered
                      • King Cobra-3 chambered
                      • Crocodile-4 chambered
                      • Ostrich-4chambered
                      • Bat-4chambered
                      • Whale-4 chambered


                      43. Classify Rohu, Scoliodon, Flying lizard, King Cobra, Frog, Salamander, Ostrich, Pigeon, Bat, Crocodile and Whale into the cold blooded/warm blooded animals.

                      Solution

                      • Rohu-Cold Blooded
                      • Scolidon-Cold Blooded
                      • Flying lizard-Cold Blooded
                      • King Cobra-Cold Blooded
                      • Frog-Cold Blooded
                      • Salamander-Cold Blooded
                      • Ostrich-Warm Blooded
                      • Pigeon-Warm Blooded
                      • Bat-Warm Blooded
                      • Crocodile-Cold Blooded
                      • Whale-Warm Blooded


                      44. Name two egg laying mammals.

                      Solution

                      Billed platypus and the echidna are two egg-laying mammals


                      45. Fill in the blanks :
                      (a) Five kingdom classification of living organisms is given by _____
                      (b) Basic smallest unit of classification is _____
                      (c) Prokaryotes are grouped in Kingdom _____
                      (d) Paramecium is a protista because of its _____
                      (e) Fungi do not contain _____
                      (f) A fungus _____ can be seen without microscope.
                      (g) Common fungi used in preparing the bread is _____
                      (h) Algae and fungi form symbiotic association called_____

                      Solution

                      (a) Robert whittaker,
                      (b) species,
                      (c) monera,
                      (d) eukaryotic unicellular, mobile organisms,
                      (e) chlorophyll,
                      (f) mushroom,
                      (g) yeast,
                      (h) lichens.


                      46. Give True (T) and False (F) :
                      (a) Gymnosperms differ from Angiosperms in having covered seed.
                      (b) Non-flowering plants are called Cryptogamae.
                      (c) Bryophytes have conducting tissue.
                      (d) Funaria is a moss.
                      (e) Compound leaves are found in many ferns.
                      (f) Seeds contain embryo.

                      Solution

                      (a) False,
                      (b) True,
                      (c) False,
                      (d) True,
                      (e) True,
                      (f) True,


                      47. Fill in the blanks :
                      (a) Bilateral, dorsiventral symmetry is found in _____
                      (b) Worms causing disease elephantiasis is _____
                      (c) Open circulatory system is found in _____ where coelomic cavity is filled with blood.
                      (d) _____ are known to have pseudocoelom.

                      Solution

                      (a) liver fluke,
                      (b) filarial worm,
                      (c) anthropods,
                      (d) Nematodes.


                      48. Label a, b, c and d given in the figure. Give the function of b.

                      Solution

                      (a) Dorsal fin

                      (b) Caudal fin

                      (c) Pelvic fin

                      (d) Pectoral fin

                      Function of Caudal fin: Caudal fin helps in streamlined movement in water.


                      49. Fill in the boxes given in figure with appropriate characteristics/plant group (s) :

                      Solution
                      (a) Thallophyta,
                      (b) Without specialized vascular tissue,
                      (c) Pteridophyta,
                      (d) Phanerogams,
                      (e) Bear naked seeds,
                      (f) Angiosperms,
                      (g) Have seeds with two cotyledons,
                      (h) Monocots


                      Long Answer Questions

                      50. Write names of few Thallophytes. Draw a labelled diagram of Spirogyra.

                        Solution

                        Ulva, Spirogvra, Cladophora, Ulothrix, Chara are the some examples of Thallophyta.


                        51. Why are Thallophyta, bryophyta and pteridophyta called ‘Cryptogams’ while Gymnosperms and Angiosperms are called ‘phanerogams’? Draw one example of Gymnosperm.

                        Solution

                        Thallophyta, bryophyta and pteridophyta are called ‘Cryptogams’ because the reproductive organs of plants in all these three group is hidden and plants do not bear flowers or seeds. On the other hand ‘Phanerogams’ are flowering and bear seeds. It includes gymnosperms and angiosperms which have well-differentiated reproductive tissue that finally make seeds. Seeds are the result of the reproductive process. Seeds consist of the embryo along with stored food, which serves for the primary growth of the embryo during germination.


                        52. Define the terms and give one example of each :
                        (a) Bilateral symmetry
                        (b) Coelom
                        (c) Triploblastic

                        Solution

                        (a) Bilateral symmetry: When the body have similar parts fixed around a common central axis. Parts of these animals at any point of radius have similar structure to that of other. The left and the right halves of the body are same in bilateral symmetry. The organisms which possess this type of symmetry are called bilaterally symmetrical organisms, e.g. butterfly, human, crab.

                        (b) Coelom: A coelom is fluid-filled body cavity that lies in between the body wall and alimentary canal. It is typically found in multicellular organisms.

                        (c) Triploblastic: Triploblastic animals are the animals whose cells body develops from three layers : the outer ectoderm, inner endoderm and the third germinal layer mesoderm which is in between outer ectoderm and the endoderm.
                        Example : Tapeworm.


                        53. You are given Leech, Nereis, Scolopendra, Prawn and Scorpion; and all have segmented body organisation. Will you classify them in one group? If no, give the important characters based on which you will separate these organisms into different groups.

                        Solution

                        All the given organisms do not belong to same group. Leech and Nereis belong to phylum annelida because as their body is bilaterally symmetrical and triploblastic. They have elongated and segmented body and bear lateral appendages for locomotion. They have true body cavity. Colopendra, Prawn and Scorpion belong to phylum arthropoda as these are bilaterally symmetrical, segmented animals having jointed legs and an open circulatory system.


                        54. Which organism is more complex and evolved among Bacteria, Mushroom and Mango tree. Give reasons.

                        Solution

                        Among these mango tree is the most complex, second most complex is the mushroom, and the bacteria are the least complex. Bacteria are generally having single cell. They are prokaryotic and some of the simplest life forms on the planet (example : bacteria). But mushroom is more complex than the bacteria. Mushroom is heterotrophic, eukaryotic and simple thallophyte with no tissue systems. Mango tree is eukaryotic, autotrophic and multicellular plant. It is an angiosperm in which seeds are produced inside the fruit and flowers are formed. It is a complete plant.


                        55. Differentiate between flying lizard and bird. Draw the diagram.

                        Solution

                        Flying Lizards

                        Birds

                        It belongs to group reptiles.

                        It belongs to group ayes

                        It is cold blooded.

                        Birds are warm blooded

                        Its body covered with scales.

                        It persist on the feet but have outside covering of feathers.

                        Flying lizards have three chambered heart.

                        Birds have four-chambered heart.

                        Teeth are present in all reptiles except in tortoises and turtles.

                        Narrow jaws form a horny beak, which is modified for different purpose. Teeth are absent.


                        56. Write some common features in cat, rat and bat.

                        Solution

                        Bat, rat and cat belong to class Mammalia and have following common features :

                        1. All are multicellular animals.
                        2. All have notochord at some stage of life cycle.
                        3. Mammals are warm-blooded animals with four- chambered hearts.
                        4. They have mammary glands for the production of milk to nourish their young ones.
                        5. Their skin has hairs as well as sweat and oil glands.
                        6. All the three mammals given above produce live young ones.
                        7. All are heterotrophic in nature.
                        8. All have paired gill pouches.


                        57. Why do we keep both snake and turtle in the same class?

                        Solution

                        Both turtle and snake are kept in the same class because they have the following common features:

                        1. Both the animals are cold-blooded
                        2. They have scales and breathe through lungs
                        3. They belong to Reptilia group.
                        4. Most of the animals in this group have three- chambered heart except crocodiles which have four chambered heart.
                        5. They lay eggs with tough coverings.

                        NCERT Exemplar Chapter 9 Force and Laws of Motion Class 9 Science Solutions

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                        NCERT Exemplar Class 9 Science Chapter 9 Force and Laws of Motion Solutions

                        NCERT Exemplar Solutions for Class 9 Science Chapter 9 Force and Laws of Motion covers all the important questions and answers as well as advanced level questions. It helps in learning about the balanced and unbalanced forces, Newton's laws of motion i.e. first law of motion, second law of motion and third law of motion.

                        The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 9 Force and Laws of Motion is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like inertia, action and reaction forces, momentum and conversion of momentum.


                        Chapter Name

                        Chapter 9 Force and Laws of Motion

                        Book Title

                        NCERT Exemplar for Class 9 Science

                        Related Study

                        • NCERT Solutions for Class 9 Science Chapter 9 Force and Laws of Motion
                        • Revision Notes for Class 9 Science Chapter 9 Force and Laws of Motion
                        • MCQ for Class 9 Science Chapter 9 Force and Laws of Motion
                        • Important Questions for Class 9 Science Chapter 9 Force and Laws of Motion

                        Topics Covered

                        • MCQ
                        • Short Answers Questions
                        • Long Answers Questions

                        NCERT Exemplar Solutions for Chapter 9 Force and Laws of Motion Class 9 Science

                        Multiple Choice Questions

                        1. Which of the following statement is not correct for an object moving along a straight path in an accelerated motion?

                        (a) Its speed keeps changing.
                        (b) Its velocity always changes.
                        (c) It always goes away from the Earth.
                        (d) A force is always acting on it.

                        Solution

                        (c) It always goes away from the Earth.
                        Object’s acceleration should be more than the acceleration due to gravity in order to move away from the earth. Only moving along a straight path is not enough to escape from gravity.


                        2. According to the third law of motion, action and reaction :
                        (a) Always act on the same body.
                        (b) Always act on different bodies in opposite directions.
                        (c) Have same magnitude and directions.
                        (d) Act on either body at normal to each other.

                        Solution

                        (b) Always act on different bodies in opposite directions.

                        Newton’s third law of motion states that “For every action there is equal and opposite reaction”.


                        3. A goalkeeper in a game of football pulls his hands backwards after holding the ball shot at the goal. This enables the goalkeeper to :
                        (a) Exert larger force on the ball.
                        (b) Reduce the force exerted by the ball on hands.
                        (c) Increase the rate of change of momentum.
                        (d) Decrease the rate of change of momentum.

                        Solution

                        (b) Reduce the force exerted by the ball on hands.

                        Pulling hands backwards help the goalkeeper to reduce the momentum of the ball. This in turn reduces the force of the ball exerted on goalkeeper’s hands.


                        4. The inertia of an object tends to cause the object :
                        (a) To increase its speed
                        (b) To decrease its speed
                        (c) To resist any change in its state of motion
                        (d) To decelerate due to friction

                        Solution

                        (c) To resist any change in its state of motion

                        Inertia is the property which resists any change in the state of motion of the object. Object remains in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.


                        5. A passenger in a moving train tosses a coin which falls behind him. It means that motion of the train is.
                        (a) Accelerated
                        (b) Uniform
                        (c) Retarded
                        (d) Along circular tracks

                        Solution

                        (a) Accelerated

                        The coin would have fallen in the passenger’s hand if the motion of train was uniform. If the motion of the train would have been retarded then the coin would have fallen ahead of him. As the coin falls behind the passenger, the motion of the train is an accelerated one.


                        6. An object of mass 2 kg is sliding with a constant velocity of 4 ms-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is :
                        (a) 32 N
                        (b) 0 N
                        (c) 2 N
                        (d) 8 N

                        Solution

                        (b) 0 N

                        The friction is opposing the force. Thus no force is required to keep the object in uniform motion.


                        7. Rocket works on the principle of conservation of :
                        (a) Mass
                        (b) Energy
                        (c) Momentum
                        (d) Velocity

                        Solution

                        (c) Momentum

                        Velocity of hot gases coming out of rocket provided large momentum in opposite direction. This makes the rocket move upwards. Here the conservation of momentum takes place.


                        8. A water tanker filled up to 2/3 of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would :
                        (a) Move backward
                        (b) Move forward
                        (c) Be unaffected
                        (d) Rise upwards

                        Solution

                        (b) Move forward

                        The water tanker comes to rest on applying the brakes but water will be in motion. This makes the water to come forward.


                        Short Answer Questions

                        9. There are three solids made up of aluminum, steel and wood, of the same shape and same volume. Which of them would have highest inertia ?

                        Solution

                        As the mass is a measure of inertia, the ball of same shape and size, having more mass than other balls will have highest inertia. Since steel has greatest density and greatest mass, therefore, it has highest inertia.


                        10. Two balls of the same size but of different materials, rubber and iron are kept on the smooth floor of a moving train. The brakes are applied suddenly to stop the train. Will the balls start rolling? If so, in which direction will they move with the same speed? Give reasons for your answer.

                        Solution

                        Yes, both the balls will start rolling due to inertia of and they roll in the direction in which the train was moving. When the brakes are applied, train will come to rest and the balls try to attain rest. Now, because of inertia, the balls remain in motion and they continue to roll. The mass of two balls is not the same, thus the inertia of iron ball is greater than the inertia of the rubber ball. This is the reason why the rubber ball rolls faster than the iron ball.


                        11. Two identical bullets are fired one by a light rifle and another by a heavy rifle with the same force. Which rifle will hurt the shoulder more and why?

                        Solution

                        As both the bullets are said to be identical and are fired with the same force, therefore, as per Newton’s third law of motion, ‘Every action has an equal and opposite reaction’; same force will be applied on both the rifles. As the same amount force is applied on both the rifles, the lighter rifle will move more quickly in the backward direction causing serious injury to the shoulder.


                        12. A horse continues to apply a force in order to move a cart with a constant speed. Explain, why?

                        Solution

                        Once the cart starts moving, the force of friction comes into action. It starts working on the wheels of the cart in a direction opposite to the direction of motion of the cart, the horse should overcome the frictional force. Therefore it continues to apply the force.


                        13. Suppose a ball of mass m is thrown vertically upward with an initial speed v, its speed decreases continuously till it becomes zero. Thereafter, the ball begins to fall downward and attains the speed v again before striking the ground. It implies that the magnitude of initial and final momentums of the ball is same. Yet, it is not an example of conservation of momentum. Explain, why?

                        Solution

                        Law of conservation of momentum is applicable to isolated system. An isolated system is a system that is free from the influence of a net external force that alters the momentum of the system. In the given example, the change in velocity is due to the gravitational force of Earth. This is an external force. Therefore, it is not an example of conservation of momentum.


                        14. Velocity versus time graph of a ball of mass 50 g rolling on a concrete floor is shown in figure. Calculate the acceleration and frictional force of the floor on the ball.

                        Solution

                        Given, m = 50 g, F = ?
                        The velocity of the ball zero time is 80 ms-1 . It decelerates due to the friction of the floor with itself and comes to rest after 8 s

                        The negative sign indicates that the frictional force exerted opposes the motion of the ball. Now, using Newton’s relation
                        F = ma
                        ⇒ F = 0.5 N


                        15. A truck is then loaded with an object equal to the mass of the truck and the driving force is halved, then how does the acceleration change?

                        Solution

                        Force = mass ×acceleration

                        F = ma

                        a = F/m

                        When mass is doubled, force is halved, m becomes 2m and F becomes F/2.

                        Thus, the new acceleration will be 1/4 th of the old acceleration.


                        16. Two friends on roller-skates are standing 5 m apart facing each other. One of them throws a ball of 2 kg towards the other, who catches it, how will this activity affect the position of the two? Explain your answer.

                        Solution

                        Separation between them will increase. Initially the momentum of both of them are zero. In order to conserve the momentum the one who throws the ball would move backward. The second will experience a net force after catching the ball and therefore, will move backwards that is in the direction of the force.


                        17. Water sprinkler used for grass lawns begins to rotate as soon as the water is supplied. Explain the principle on which it works.

                        Solution

                        The working of sprinkler which rotates as soon as water is supplied to it is based on the thirdlaw of motion. As the water comes out of the nozzle of the sprinkler, an equal and opposite reaction force comes into play which rotates the sprinkler.


                        Long Answer Questions

                        18. Using second law of motion, derive the relation between force and acceleration. A bullet of 10 g strikes a sand-bag at a speed of 103 ms–1 and gets embedded after travelling 5 cm. Calculate :
                        (i) The resistive force exerted by the sand on the bullet.
                        (ii) The time taken by the bullet to come to rest.

                        Solution

                        (ii) v = u + at
                        ⇒ 0 = 103 - 107 t
                        ⇒ 107 t = 103


                        19. Derive the unit of force using the second law of motion. A force of 5 N produces an acceleration of 8 ms–2 on a mass m1 and an acceleration of 24 ms–2 on a mass m2. What acceleration would the same force provide if both the masses are tied together?

                        Solution

                        F = m a = kg m s–2

                        This unit is also called newton. Its symbol is N.


                        Acceleration produced in M,


                        20. What is momentum? Write its SI unit. Interpret force in terms of momentum. Represent the following graphically :
                        (a) Momentum versus velocity when mass is fixed.
                        (b) Momentum versus mass when velocity is constant.

                        Solution

                        Momentum is the quantity of motion of a moving body. It is measured as product of mass and velocity.

                        Momentum = mass × velocity

                        SI unit of momentum is kg m s–1

                        Force = Rate of change in momentum

                        (a) Mass is fixed

                        (b) Velocity is fixed

                        BSEB Bihar Board Solutions for Matric Class 10

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                        Class 10 (Matric) BSEB Bihar Board Solutions All Subjects

                        BSEB (बिहार विद्यालय परीक्षा समिति) Solutions are available for the students who study in Matric. Bihar Secondary School Examination of Matric Class 10 is conducted by Bihar School Examination Board. There are total 6 subject which need to be studied in matric examination of BSEB. We have provided solutions for all the subjects including Science, Maths, Social Science, Hindi, English and Sanskrit for the students of Class 10. These solutions of the Bihar Board will help the students in understanding the key concepts of the chapter. We have included both objective type questions and subjective questions for the matric examination of BSEB. Bihar Board solutions for Class 10 Matric is very useful in getting good marks in Board examination.


                        BSEB (बिहार विद्यालय परीक्षा समिति) मैट्रिक में पढ़ने वाले छात्रों के लिए यहाँ पठन सामग्री उपलब्ध हैं। बिहार स्कूल परीक्षा बोर्ड द्वारा मैट्रिक कक्षा 10 की बिहार माध्यमिक विद्यालय परीक्षा आयोजित की जाती है। कुल 6 विषय हैं जिनका BSEB की मैट्रिक परीक्षा में अध्ययन करने की आवश्यकता है। हमने कक्षा 10 के छात्रों के लिए विज्ञान, गणित, सामाजिक विज्ञान, हिंदी, अंग्रेजी और संस्कृत सहित सभी विषयों के Solutions प्रदान किए हैं। बिहार बोर्ड के ये समाधान छात्रों को अध्याय की मुख्य अवधारणाओं को समझने में मदद करेंगे। हमने बीएसईबी की मैट्रिक परीक्षा के लिए वस्तुनिष्ठ प्रकार (Objective Type Questions) के प्रश्न और व्यक्तिपरक प्रश्न (Subjective Type Questions) दोनों को शामिल किया है। बोर्ड परीक्षा में अच्छे अंक प्राप्त करने के लिए कक्षा 10 मैट्रिक के लिए बिहार बोर्ड समाधान बहुत उपयोगी है।

                        List of Subjects for Bihar Board Matric Solutions

                        SubjectsBihar Board Solutions
                        Science (विज्ञान)
                        Social Science (सामाजिक विज्ञान)
                        • Hindi Medium Solutions
                        Hindi
                        • Textbook Solutions
                        English
                        • Textbook Solutions
                        Mathematics
                        Sanskrit
                        • Textbook Solutions

                        Solutions for all subjects of Bihar Board Matric Examination has been provided. Yo can also check the below links for more details of the examination like syllabus, Model Paper and Previous Year Question Papers.

                        • BSEB Board महत्वपूर्ण प्रश्न for Class 10 Matric (Important Questions)
                        • BSEB Board बहुवैकल्पिक  प्रश्न for Class 10 Matric (MCQ)
                        • Model Papers of Matric Examination Bihar Board Class 10
                        • Syllabus of Matric Examination Bihar Board Class 10
                        • Old Question Papers of Matric Examination Bihar Board Class 10

                        These will help you in understanding the pattern of the board examination. It will act as a guide in getting good marks.

                        NCERT Exemplar Chapter 8 Motion Class 9 Science Solutions

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                        NCERT Exemplar Class 9 Science Chapter 8 Motion Solutions

                        NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion covers all the important questions and answers as well as advanced level questions. It helps in learning about the concepts of motion, Straight-line motion, Measuring the rate of motion, Uniform motion, Non-uniform motion, distance, displacement, average speed and velocity.

                        The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 8 Motion is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like rate of change of velocity like acceleration and retardation, graphical representation of motion, distance time graphs, velocity time graphs, uniform circular motion and velocity-time graph.


                        Chapter Name

                        Chapter 8 Motion

                        Book Title

                        NCERT Exemplar for Class 9 Science

                        Related Study

                        • NCERT Solutions for Class 9 Science Chapter 8 Motion
                        • Revision Notes for Class 9 Science Chapter 8 Motion
                        • MCQ for Class 9 Science Chapter 8 Motion
                        • Important Questions for Class 9 Science Chapter 8 Motion

                        Topics Covered

                        • MCQ
                        • Short Answers Questions
                        • Long Answers Questions

                        NCERT Exemplar Solutions for Chapter 8 Motion Class 9 Science

                        Multiple Choice Questions

                        1. A particle is moving in a circular path of radius r. The displacement after half a circle would be :
                        (a) Zero
                        (b) 𝛑r
                        (c) 2r
                        (d) 2𝛑r

                        Solution

                        (c) 2r

                        After half revolution
                        Displacement = Final position-Initial Position
                        Diameter of the circle = 2r


                        2. A body is thrown vertically upward with velocity u, the greatest height h to which it will rise is :
                        (a) u/g
                        (b) u2/2g
                        (c) u2/g
                        (d) u/2g

                        Solution

                        (b) u2/2g

                        v2 = u2 + 2as
                        Given,
                        v = 0, a = -g, s = H
                        Therefore,
                        0 = u² -2gH
                        ⇒ H = u²/2g


                        3. The numerical ratio of displacement to distance for a moving object is :
                        (a) Always less than 1
                        (b) Always equal to 1
                        (c) Always more than 1
                        (d) Equal or less than 1

                        Solution

                        (d) Equal or less than 1
                        Shortest distance between initial and end point is called displacement. Distance is the total path length. Displacement is vector and it may be positive or negative.

                        Distance is scalar and it can never be negative. The distance travelled by a body can be equal to or greater than the displacement which means ratio of displacement to distance is always equal to or less than 1.


                        4. If the displacement of an object is proportional to square of time, then the object moves with :
                        (a) Uniform velocity
                        (b) Uniform acceleration
                        (c) Increasing acceleration
                        (d) Decreasing acceleration

                        Solution

                        (b) Uniform acceleration

                        Velocity is measured in distance/second and acceleration is measured in distance/second2.


                        5. From the given v – t graph figure, it can be inferred that the object is :

                        (a) In uniform motion
                        (b) At rest
                        (c) In non-uniform motion
                        (d) Moving with uniform acceleration

                        Solution

                        (a) In uniform motion

                        From the given graph it is clear that velocity of the object remains constant throughout the motion.


                        6. Suppose a boy is enjoying a ride on a merry-go-round which is moving with a constant speed of 10 m s–1. It implies that the boy is :
                        (a) At rest
                        (b) Moving with no acceleration
                        (c) In accelerated motion
                        (d) Moving with uniform velocity

                        Solution

                        (c) In accelerated motion

                        Boy is moving in a circular motion and circular motion is an accelerated motion.


                        7. Area under a v – t graph represents a physical quantity which has the unit

                        (a) m2

                        (b) m

                        (c) m3

                        (d) m s–1

                        Solution

                        (b) m

                        Area given in the graph represents displacement and its unit is meter.


                        8. Four cars A, B, C and D are moving on a levelled road. Their distance versus time graphs are shown in figure.

                        Choose the correct statement :
                        (a) Car A is faster than car D.
                        (b) Car B is the slowest.
                        (c) Car D is faster than car C.
                        (d) Car C is the slowest.

                        Solution

                        (b) Car B is the slowest.

                        Graph shows that Car B covers the least distance in a given time.


                        9. Which of the following figure represents uniform motion of a moving object correctly?

                        Solution
                        (a)
                        Distance in graph (a) is uniformly increasing with time.

                        10. Slope of a velocity-time graph gives :
                        (a) The distance
                        (b) The displacement
                        (c) The acceleration
                        (d) The speed

                        Solution

                        (c) The acceleration


                        11. In which of the following cases of motions, the distance moved and the magnitude of displacement are equal?
                        (a) If the car is moving on straight road.
                        (b) If the car is moving in circular path.
                        (c) The pendulum is moving to and fro.
                        (d) The Earth is revolving around the Sun.

                        Solution

                        (a) If the car is moving on straight road.

                        In all the other cases, displacement can be less than distance.


                        Short Answer Questions

                        12. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero ? Justify your answer.

                        Solution

                        When displacement is zero, it does not mean that the distance is zero. Distance can be zero when the object moves back to the place it started from. Displacement is either equal or less than distance. Distance is always greater than one and it cannot have a negative value.


                        13. How will the equations of motion for an object moving with a uniform velocity change?

                        Solution

                        In uniform motion, if acceleration (a) = 0
                        So in this case, 1st equation of motion becomes,
                        v = u + at
                        If a = 0 then v = u
                        If a = 0, 2nd equation of motion becomes,
                        s = ut + ½at2
                        ⇒ s = ut
                        If a = 0 , 3rd equation of motion becomes,
                        v2 – u2 = 2as
                        ⇒ v2 – u2 = 0


                        14. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in figure 8.4. Plot a velocity-time graph for the same.

                        Solution

                        From the graph, we get.
                        Initial velocity (u) = 0 (As time and displacement are zero.)
                        Velocity after 50 s, v = Displacement/Time
                        By substituting the values, we get
                        v = 100/50 = 2 ms-1
                        Velocity after 100 s, v = Displacement/Time
                        Displacement = 0; Time = 100 s
                        Therefore,
                        By substituting the values, we get
                        v = 0/100 = 0

                        V

                        0

                        2

                        0

                        T

                        0

                        50

                        100

                        Velocity – time graph for above data:


                        15. A car starts from rest and moves along the x-axis with constant acceleration 5 ms–2 for 8 seconds. If it then continues with constant velocity, what distances will the car cover in 12 seconds since it started from the rest?

                        Solution

                        As the car moves from rest, therefore, u = 0, a = 5 ms2, t = 8 sec, distance (s) = ?
                        s’ = ut + ½at2
                        = 0 + ½ × 5 × (8)2
                        = 0 + 320/2
                        = 160 m
                        To calculate the velocity after 8 sec,
                        v = u + at
                        = 0 + 5 × 8 = 40 m/s
                        So, the distance travelled with this velocity for remaining 4 sec (12 s – 8s = 4s)
                        s” = 40 × 4 = 160 m
                        Therefore total distance travelled by the car
                        = s’ + s”
                        = 160 +160 = 320 m


                        16. A motorcyclist drives from A to B with a uniform speed of 30 km h-1 and returns back with a speed of 20 km h–1. Find its average speed.

                        Solution

                        Let AB = x,


                        17. The velocity-time graph figure shows the motion of a cyclist. Find
                        (i) its acceleration
                        (ii) its velocity and
                        (iii) the distance covered by the cyclist in 15 seconds.

                        Solution

                        (i) Since velocity is not changing, acceleration is equal to zero.
                        (ii) Through graph, velocity = 20 m s–1
                        (iii) Distance covered in 15’ seconds,
                        s = u × t
                        = 20 × 15 = 300 m


                        18. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.

                        Solution

                        The velocity versus time graph of a stone thrown upwards vertically is as given below:


                        Long Answer Questions

                        19. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height of 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?

                          Solution

                          Initial difference in height
                          h = (150 – 100) m = 50 m
                          Both the objects are dropped simultaneously from rest (zero velocity),
                          Distance travelled by first body in 2 s
                          d1 = ut + ½at2
                          Initial speed = 0 (as the object was at rest), t = 2s
                          Therefore, by substituting the values we get
                          = 0 + ½ g(2)2 = 2g
                          = 2 × 9.8 = 19.6 m
                          Where, g = 9.8
                          Distance travelled by another body in
                          2s = d2 = 0 + ½g(2)2
                          = 2g = 19.6 m
                          After 2s, height at which the first body will be
                          h1 = 150 – 19.6
                          After 2s, height at which the second body will be
                          h2 = 100 – 19.6
                          Thus, after 2s, difference in height
                          = h1 – h2
                          = 150 – 19.6 – (100 – 19.6)
                          = 50 m
                          After 2s, the difference in height will be 50 m = Initial difference in height (h)
                          Thus, difference in height does not vary with time.


                          20. An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start?

                          Solution

                          Let’s acceleration of the body be ‘a’.

                          Since acceleration is the same, we have v′ =0+ (10 × 7) = 70 m s–1


                          21. Using following data, draw time-displacement graph for a moving object-

                          Time (s)

                          0

                          2

                          4

                          6

                          8

                          10

                          12

                          14

                          16

                          Displacement(m)

                          0

                          2

                          3

                          4

                          4

                          6

                          4

                          2

                          0

                          Solution

                          Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.

                          Average velocity for first 4s :
                          Average velocity = Change in displacement/Total time taken


                          22. An electron moving with a velocity of 5 × 10m s–1 enters into a uniform electric field and acquires a uniform acceleration of 104 m s–2 in the direction of its initial motion.
                          (i) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
                          (ii) How much distance the electron would cover in this time?

                          Solution

                          Given,
                          Initial velocity, u = 5 × 10ms–1
                          and acceleration, a = 104 m s–2
                          (i) Final velocity = v = 2 u
                          v = 2 × (5 × 104 ms-1)
                          = 10 × 104 ms-1 ; t = ?
                          To find t, use v = u + at


                          23. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds. Using the equation of motion s = ut + ½ at2 .

                          Solution

                          Using the equation of motion,

                          S = ut + ½at2

                          The distance travelled in 5 s,


                          Similarly, the distance travelled in 4 s,
                          Therefore, the distance travelled in the interval between 4thand 5thsecond = (S –S’)


                          24. Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the heights reached by them would be in the ratio of (u1)2 : (u2)2 (Assume upward acceleration is – g and downward acceleration to be + g).

                          Solution 

                          For upward motion, v2 = u2– 2gh



                          NCERT Exemplar Chapter 10 Gravitation Class 9 Science Solutions

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                          NCERT Exemplar Class 9 Science Chapter 10 Gravitation Solutions

                          NCERT Exemplar Solutions for Class 9 Science Chapter 10 Gravitation covers all the important questions and answers as well as advanced level questions. It helps in learning about the Universal Law of Gravitation, freefall, mass, inertia, greater the mass the greater will be the inertia, weight, mass of an object is constant, buoyancy, buoyant law, thrust and pressure.

                          The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 10 Gravitation is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like Archimedes principle i.e. weight of the object immersed in water is equal to the weight of the fluid that is displaced by the object, law of basic fluid mechanics and relative density.


                          Chapter Name

                          Chapter 10 Gravitation

                          Book Title

                          NCERT Exemplar for Class 9 Science

                          Related Study

                          • NCERT Solutions for Class 9 Science Chapter 10 Gravitation
                          • Revision Notes for Class 9 Science Chapter 10 Gravitation
                          • MCQ for Class 9 Science Chapter 10 Gravitation
                          • Important Questions for Class 9 Science Chapter 10 Gravitation

                          Topics Covered

                          • MCQ
                          • Short Answers Questions
                          • Long Answers Questions

                          NCERT Exemplar Solutions for Chapter 10 Gravitation Class 9 Science

                          Multiple Choice Questions

                          1. Two objects of different masses falling freely near the surface of Moon would :
                          (a) Have same velocities at any instant
                          (b) Have different accelerations
                          (c) Experience forces of same magnitude
                          (d) Undergo a change in their inertia

                          Solution

                          (a) Have same velocities at any instant

                          Acceleration of an object depends on acceleration due to gravity irrespective of its mass.


                          2. The value of acceleration due to gravity :
                          (a) Is same on equator and poles
                          (b) Is least on poles
                          (c) Is least on equator
                          (d) Increases from pole to equator

                          Solution

                          (c) Is least on equator

                          As the distance between surface of the earth and its centre is more on equator than in poles, the acceleration due to gravity is least at the equator.


                          3. The gravitational force between two objects is F. If masses of both objects are halved without changing distance between them, then the gravitational force would become :
                          (a) F/4
                          (b) F/2
                          (c) F
                          (d) 2 F

                          Solution

                          (a) F/4


                          4. A boy is whirling a stone tied with a string in an horizontal circular path. If the string breaks, the stone :
                          (a) Will continue to move in the circular path
                          (b) Will move along a straight line towards the centre of the circular path
                          (c) Will move along a straight line tangential to the circular path
                          (d) Will move along a straight line perpendicular to the circular path away from the boy

                          Solution

                          (c) Will move along a straight line tangential to the circular path

                          At any instance of time an object in circular motion will tend to be in rectilinear motion. The object keeps on moving due to the centripetal force and it moves along a straight line tangential to the circular path when strings break.


                          5. An object is put one by one in three liquids having different densities. The object floats with 1/9, 2/11 and 3/7 parts of their volumes outside the liquid surface in liquids of densities d1, d2 and drespectively. Which of the following statement is correct?

                          (a) d1> d2> d3

                          (b) d1> d2< d3

                          (c) d1< d2> d3

                          (d) d1< d2< d3

                          Solution

                          (d) d1< d2< d3


                          6. In the relation F = G M m/d2, the quantity G :
                          (a) Depends on the value of g at the place of observation
                          (b) Is used only when the Earth is one of the two masses
                          (c) Is greatest at the surface of the Earth
                          (d) Is universal constant of nature

                          Solution

                          (d) Is universal constant of nature

                          G is called as Newton’s constant. It is the force of gravity on a body. Value of G is 6.66×10-11Nm2kg-2.


                          7. Law of gravitation gives the gravitational force between :
                          (a) The Earth and a point mass only
                          (b) The Earth and Sun only
                          (c) Any two bodies having some mass
                          (d) Two charged bodies only

                          Solution

                          (c) Any two bodies having some mass


                          8. The value of quantity G in the law of gravitation :
                          (a) Depends on mass of Earth only
                          (b) Depends on radius of Earth only
                          (c) Depends on both mass and radius of Earth
                          (d) Is independent of mass and radius of the Earth

                          Solution

                          (d) Is independent of mass and radius of the Earth

                          As G is the universal constant, it is independent of mass and radius of the earth.


                          9. Two particles are placed at some distance. If the of each of the two particles is doubled, keeping the distance between them unchanged, the value of gravitational force between them will be :
                          (a) 1/4 times
                          (b) 4 times
                          (c) 1/2 times
                          (d) Unchanged

                          Solution

                          (b) 4 times


                          10. The atmosphere is held to the Earth by :
                          (a) Gravity
                          (b) Wind
                          (c) Clouds
                          (d) Earth’s magnetic field

                          Solution

                          (a) Gravity


                          11. The force of attraction between two unit point masses separated by a unit distance is called :
                          (a) Gravitational potential
                          (b) Acceleration due to gravity
                          (c) Gravitational field
                          (d) Universal gravitational constant

                          Solution

                          (d) Universal gravitational constant

                          Here, point masses are separated by unit distance.

                          That is m1, m2 and r = 1

                          Therefore, F = G which is a universal constant.


                          12. The weight of an object at the centre of the Earth of radius R is :
                          (a) Zero
                          (b) Infinite
                          (c) R times the weight at the surface of the Earth
                          (d) 1/R2 times the weight at surface of the Earth

                          Solution

                          (a) Zero

                          The acceleration due to gravity is zero at the centre of the earth. As the weight is the product of mass and gravity. Weight of the object at the centre of the earth will be zero.


                          13. An object weighs 10 N in air. When immersed fully in water, it weighs only 8 N. The weight of the liquid displaced by the object will be :
                          (a) 2 N
                          (b) 8 N
                          (c) 10 N
                          (d) 12 N

                          Solution

                          (a) 2 N

                          Object of weight displace d by liquid = weight in air-weight in liquid

                          =10N - 8N

                          = 2N


                          14. A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be :
                          (a) Maximum when length and breadth form the base
                          (b) Maximum when breadth and width form the base
                          (c) Maximum when width and length form the base
                          (d) The same in all the above three cases

                          Solution

                          (b) Maximum when breadth and width form the base

                          Pressure is inversely proportional to the surface area. So, the pressure will be maximum when the surface area is minimum.


                          15. An apple falls from a tree because of gravitational attraction between the Earth and apple. If F1 is the magnitude of force exerted by the Earth on the apple and F2 is the magnitude of force exerted by apple on Earth, then :
                          (a) F1 is very much greater than F2
                          (b) F2 is very much greater than F1
                          (c) F1 is only a little greater than F2
                          (d) F1 and F2 are equal

                          Solution

                          (d) F1 and F2 are equal

                          Newton’s third law of motion states that for every action there is an equal and opposite reaction.


                          Short Answer Questions

                          16. What is the source of centripetal force that a planet requires to revolve around the Sun? On what factors does that force depend?

                            Solution

                            Gravitational force provides the necessarycentripetal force which makes the planetsrevolve around the sun.

                            1. Mass of the planet and the Sun, i.e., depends on the product of the masses of the planet and the Sun.
                            2. Distance between the planet and the Sun, i.e. depends on the square of distance between the planet and the Sun.


                            17. On the Earth, a stone is thrown from a height in a direction parallel to the Earth’s surface while another stone is simultaneously dropped from the same height. Which stone would reach the ground first and why?

                            Solution

                            Both the stones will reach the ground at the same time because the two stones fall from the same height.
                            As both the stones will have initial velocity (u) = 0,
                            a = g (acceleration due to gravity), distance (s) = s; t = ?
                            Using equation of motion,


                            18. Suppose gravity of Earth suddenly becomes zero, then in which direction will the Moon begin to move if no other celestial body affects it?

                            Solution

                            If gravity of Earth suddenly becomes zero then the moon will start moving in a straight line tangent to its circular path. This change happens because the circular motion of the moon is due to the centripetal force provided by the gravitational force of the Earth.


                            19. Identical packets are dropped from two aeroplanes, one above the equator and the other above the north pole, both at height h. Assuming, all conditions are identical, will those packets take same time to reach the surface of Earth. Justify your answer.

                            Solution

                            The value of acceleration due to gravity is constant. But it depends upon the surface of the earth. The surface of the earth varies from place to place as it is not completely spherical. The earth is flattened at the poles, hence the value of ‘g’ is maximum at the poles. There is bulge at the equator which has the minimum value of ‘g’. The value of ‘g’ value increases as we move towards the poles. Therefore, the packets fall gradually at the equator in comparison to the poles. This is the reason why the packets stay in air for longer when dropped at the equator.


                            20. The weight of any person on the Moon is about 1/6 times that on the Earth. He can lift a mass of 15 kg on the Earth. What will be the maximum mass, which can be lifted by the same force applied by the person on the Moon ?

                            Solution

                            Weight of person on moon = 1/6 thof weight on earth
                            Therefore, ‘g’ on moon = 1/6th ‘g’ on earth
                            The force that is applied by the man to lift mass ‘m’ is
                            F = mg = 15g (on earth)
                            If he can lift a certain mass ‘m’ by applying the same force on moon, then
                            m = F ÷ g
                            = 15× 6
                            =90kg
                            Therefore on moon a person can lift a mass 6 times heavier than that on earth.


                            21. Calculate the average density of the Earth in terms of g, G and R. ‘g’ is related to Earth’s mass “M” and radius “R”.

                            Solution


                            22. The Earth is acted upon by gravitation of Sun, even thought it does not fall into the Sum. Why ?

                            Solution

                            According to Newton’s first law of motion, an object in motion tries to move in straight-line at a constant speed unless external pressure is not applied. When the Earth comes close to the Sun that has a large gravitational force, the path of the Earth is altered due to the unbalanced force of gravity on it. The Sun exerts an attractive force on the Earth, accelerating the Earth directly towards the Sun. It moves toward the Sun as described by Newton’s second law. To avoid falling into the Sun, the Earth also counteracts the force that is pulling it towards the Sun. Thus, a force, which tends to make a body move in a curved path, is called a centripetal force is applied by the Earth. Therefore, we can say that the Sun provides the necessary centripetal force to keep the earth in its orbit and the earth provides centrifugal force due to its motion. These two forces counter balance each other and this prevents the earth from falling into the sun.


                            Long Answer Questions

                            23. How does the weight of an object vary with respect to mass and radius of the Earth? In a hypothetical case, if the diameter of the Earth becomes half of its present value and its mass becomes four times of its present value, then how would the weight of any object on the surface of the Earth be affected ?

                              Solution

                              Let R and M be the radius and mass of the earth.

                              Then,

                              Hypothetically m and R becomes 4m and R/2.


                              24. How does the force of attraction between the two bodies depend upon their masses and distance between them? A student thought that two bricks tied together would fall faster than a single one under the action of gravity. Do you agree with his hypothesis or not? Comment.

                              Solution

                              From Newton’s law of gravitation, we get Force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
                              F ∝ m1m2 …(1)
                              F ∝ 1/d2 …(2)
                              The two bricks, like a single body, fall with the same speed to reach the ground at the same time in case of free-fall. This is because acceleration due to gravity is independent of the mass of the falling body.


                              25. Two objects of masses m1 and m2 having the same sizes are dropped simultaneously from heights h1 and h2 respectively. Find out the ratio of time they would take in reaching the ground. Will this ratio remain the same if

                              (i) one of the objects is hollow and the other one is solid and

                              (ii) both of them are hollow, size remaining the same in each case? Give reason.

                              Solution

                              Two objects of masses m1 and m2 having the same sizes are dropped, therefore, initial velocity (u) = 0 and a = g (Acceleration due to gravity–falling object) and it is independent of mass.
                              By using Newton’s equation of motion namely.

                              For object 1 = h1

                              For object 2 = h2

                              Time taken by object 1 = t1

                              Time taken by object 2 = t2

                              (i) This ratio of time will be the same even if one of the objects is hollow and the other one is solid. This is because acceleration due to gravity is independent of the mass of the falling body.

                              (ii) This ratio of time will be the same even if both of them are hollow, size remaining the same in each case. This is because acceleration due to gravity is independent of the mass of the falling body.


                              26. (a) A cube of side 5 cm is immersed in water and then in saturated salt solution. In which case will it experience a greater buoyant force. If each side of cube is reduced to 4 cm and then immersed in water. What will be the effect on buoyant force experienced by the cube as compared to the first case for water? Give reason for each case.
                              (b) A ball weighing 4 kg of density 4000 kg m–3 is completely immersed in water of density 103 Kg m–3. Find the force of buoyancy on it. (Given g = 10 ms–2)

                              Solution

                              (a) (i) Buoyant force, F = Vρg

                              ρ= Density of water,

                              V = Volume of water displaced by the body

                              Volume and density of an object decides its Buoyancy.

                              As the density of the salt solution is greater than that of water the cube will experience a greater buoyant force in the saturated salt solution.

                              (ii) The smaller cube will experience lesser buoyant force as its volume is lesser than the initial cube.

                              (b) Buoyant force = weight of the liquid displaced

                              = density of water × volume of water displaced ×g

                              = 1000 × 4/4000 × 10

                              = 10 N

                              NCERT Exemplar Chapter 11 Work and Energy Class 9 Science Solutions

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                              NCERT Exemplar Class 9 Science Chapter 11 Work and Energy Solutions

                              NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work and Energy covers all the important questions and answers as well as advanced level questions. It helps in learning about the work, unit of work, Energy, forms of energy, kinetic energy, potential energy, heat energy, chemical energy, electrical energy, light energy and mechanical energy.

                              The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 11 Work and Energy is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like law of conservation of energy, power and commercial unit of energy.


                              Chapter Name

                              Chapter 11 Work and Energy

                              Book Title

                              NCERT Exemplar for Class 9 Science

                              Related Study

                              • NCERT Solutions for Class 9 Science Chapter 11 Work and Energy
                              • Revision Notes for Class 9 Science Chapter 11 Work and Energy
                              • MCQ for Class 9 Science Chapter 11 Work and Energy
                              • Important Questions for Class 9 Science Chapter 11 Work and Energy

                              Topics Covered

                              • MCQ
                              • Short Answers Questions
                              • Long Answers Questions

                              NCERT Exemplar Solutions for Chapter 11 Work and Energy Class 9 Science

                              Multiple Choice Questions

                              1. When a body falls freely towards the Earth, then its total energy :
                              (a) Increases
                              (b) Decreases
                              (c) Remains constant
                              (d) First increases and then decreases

                              Solution

                              (c) Remains constant

                              Body falling freely towards the earth possesses the same kinetic and potential energy. This is because the body obeys the Law of conservation of energy.


                              2. A car is accelerated on a levelled road and attains a velocity 4 times of its initial velocity. In this process the potential energy of the car :
                              (a) Does not change
                              (b) Becomes twice to that of initial
                              (c) Becomes 4 times that of initial
                              (d) Becomes 16 times that of initial

                              Solution

                              (a) Does not change

                              Potential energy is the product of height, mass and gravity. Therefore, height is a factor in determining potential energy.


                              3. In case of negative work the angle between the force and displacement is :
                              (a) 0°
                              (b) 45°
                              (c) 90°
                              (d) 180°

                              Solution

                              (d) 180°

                              Work done = FS cosθ

                              ⇒ cos 180°= -1


                              4. An iron sphere of mass 10 kg has the same diameter as an aluminium sphere of mass is 3.5 kg. Both spheres are dropped simultaneously from a tower. When they are 10 m above the ground, they have the same :
                              (a) Acceleration
                              (b) Momentum
                              (c) Potential energy
                              (d) Kinetic energy

                              Solution

                              (a) Acceleration

                              Momentum, potential energy and kinetic energy vary with weight. But here, the acceleration is due to acceleration due to gravity which is independent of mass.


                              5. A girl is carrying a school bag of 3 kg mass on her back and moves 200 m on a levelled road. The work done against the gravitational force will be : (g = 10 m s–2)
                              (a) 6 × 103 J
                              (b) 6 J
                              (c) 0.6 J
                              (d) zero

                              Solution

                              (d) zero

                              Here, the direction of displacement is perpendicular to the direction of gravitational force. Therefore, the work done against the gravity is zero.


                              6. Which one of the following is not the unit of energy?
                              (a) Joule
                              (b) Newton metre
                              (c) Kilowatt
                              (d) Kilowatt hour

                              Solution

                              (c) Kilowatt

                              Kilowatt is SI unit of electrical power.


                              7. The work done on an object does not depend upon the :
                              (a) Displacement
                              (b) Force applied
                              (c) Angle between force and displacement
                              (d) Initial velocity of the object

                              Solution

                              (d) Initial velocity of the object

                              Work done is the product of force and displacement. But work done is not dependent on initial velocity.


                              8. Water stored in a dam possesses :
                              (a) No energy
                              (b) Electrical energy
                              (c) Kinetic energy
                              (d) Potential energy

                              Solution

                              (d) Potential energy

                              The energy stored in an object because of the position is known as the potential energy.


                              9. A body is falling from a height h. After it has fallen a height 2 m, it will possess :
                              (a) Only potential energy
                              (b) Only kinetic energy
                              (c) Half potential and half kinetic energy
                              (d) More kinetic and less potential energy

                              Solution

                              (c) Half potential and half kinetic energy

                              The body will have only potential energy when it is at height h. Similarly, when the body reaches the ground its potential energy will be zero and kinetic energy will be maximum.

                              At height h/2, both potential energy and kinetic energy of the body will be half.


                              Short Answer Questions

                              10. A rocket is moving up with a velocity v. If the velocity of this rocket is suddenly tripled, what will be the ratio of two kinetic energies?

                              Solution

                              Initial velocity, u= v
                              New initial velocity, u' = 3v
                              Therefore, the ratio between initial and final kinetic energy is 1:9.


                              11. Avinash can run with a speed of 8 ms–1 against the frictional force of 10 N, and Kapil can move with a speed of 3 ms–1 against the frictional force of 25 N. Who is more powerful and why?

                              Solution

                              As work equals to product of force and distance, you can write the equation for power in the following way, assuming that the force acts along the direction of travel :
                              Power = Work done/Time
                              But, work done = Force × distance
                              Therefore, P = W/t = F×s/t = Fv
                              where, s is the distance travelled and v is the speed.
                              Power of Avinash = 10 × 8 = 80 W
                              Power of kapil = 25 × 3 = 75 W
                              Therefore, Avinash is more powerful than Kapil.


                              12. A boy is moving on a straight road against a frictional force of 5 N. After travelling a distance of 1.5 km he forgot the correct path at a roundabout figure of radius 100 m. However, he moves on the circular path for one and half cycle and then he moves forward upto 2.0 km. Calculate the work done by him.

                              Solution

                              F = 5 N
                              As work equals force times distance
                              W = f×s
                              ⇒ W = 5 × [1500 + 200 + 2000]
                              = 18500 J
                              Here, 1.5 km = 1500 m; 2 km = 2000 m


                              13. Can any object have mechanical energy even if its momentum is zero ? Explain.

                              Solution

                              Momentum is the product of mass and velocity. The velocity of a body is zero if the body is at rest. The body at rest at certain height possess gravitational potential energy. Therefore, an object can have mechanical energy even if its momentum is zero.


                              14. Can any object have momentum even if its mechanical energy is zero? Explain.

                              Solution

                              No, an object with zero mechanical energy cannot have momentum asits potential and kinetic energy are zero.


                              15. The power of a motor pump is 2 kW. How much water per minute the pump can raise to a height of 10 m? (Given, g = 10 m s–2)

                              Solution

                              Power of the pump = 2kW = 2000 W;
                              t = 1 min = 60 sec;
                              height = 10 m;
                              g = 10 ms-2
                              Power = work done/time
                              (The force on an object of mass m at the surface of the earth is mg, from F = ma, when acceleration is g, the acceleration at the surface of the Earth. If the object falls through a distance h, then the work done on the object by the force of gravity is mg times h, force times distance.)
                              Work done = mgh
                              = m ×10×10
                              =100m
                              Now,
                              2000 W= 100m/60
                              Or, m = 1200 kg
                              Therefore, the pump can raise 1200 kg of water in one minute.


                              16. The weight of a person on a planet A is about half that on the Earth. He can jump up to 0.4 m height on the surface of the Earth. How high he can jump on the planet A?

                              Solution

                              As per the definition of force, the equation of force due to gravity is given by, W = mg, i.e., weight is equal to mass times gravitational acceleration In this case, the force is better known as the weight of the object. Weight of a person on Earth = w = mg1 (given);
                              height the person can jump (h1) = 0.4 m
                              U = potential energy from height,
                              m = mass of the object,
                              g = gravity and height = h
                              U = mgh
                              Therefore,
                              potential energy = mg1h1 ...(1)
                              where(h1) = 0.4 m; g1 = g
                              Now,
                              Weight of the person on planet A = W/2 = mg2/2
                              Let the height the person can jump = h2 :
                              g2 = ½ , g1 = ½ g ….(2)
                              Therefore,
                              Potential energy of planet A = mg2h2
                              Potential energy of the person on the Earth = mg1h1 …(3)
                              h2 = 0.4 × 2 = 0.8 m
                              Hence, he can jump double the height with the same muscular force.


                              17. The velocity of a body moving in a straight line is increased by applying a constant force F, for some distance in the direction of the motion. Prove that the increase in the kinetic energy of the body is equal to the work done by the force on the body.

                              Solution

                              Consider that a force ‘F’ is applied on a body having mass ‘m’ and the distance travelled be ‘s’.
                              Work Done (joules) = Force (Newton) × Distance (meter)
                              W = F × s …(1)
                              As,
                              Force (F) = ma ...(2)
                              By substituting (2) in (1), we get
                              W = ma × s …(3)
                              Using the Newton’s third equation of motion,
                              v2 – u2 = 2as


                              Putting the value of s in (3),
                              W = final kinetic energy – initial kinetic energy

                              Work done = change in kinetic energy


                              18. Is it possible that an object is in the state of accelerated motion due to external force acting on it, but no work is being done by the force ? Explain it with an example.

                              Solution

                              Yes, Force always acts in perpendicular to the displacement. If the object is moving in circular path then the force acts perpendicular to the direction of the motion and therefor will be no work done despite the action of force.


                              19. A ball is dropped from a height of 10 m. If the energy of the ball reduces by 40% after striking the ground, how much high can the ball bounce back? (g = 10 m s–2)

                              Solution

                              Given, g = 10 ms-2 ; h = 10 m
                              energy possessed by the ball = mgh
                              = m × 10 × 10 = 100 m joules
                              Energy left in the ball after striking the ground
                              = (100 – 40)/100 = 60/100
                              ( As energy is reduced by 40% after striking the ground)
                              Therefore,
                              remaining energy = 60 m joules ….(1)
                              Let the height at which the ball bounces back = h1 ….(2)
                              Therefore,
                              Energy possessed by the ball = mgh
                              Using (1) and (2), we get
                              60 m = m × 10 × h1
                              h1 = 6 metres.
                              Thus, the height at which the ball bounces back = 6 m


                              20. If an electric iron of 1200 W is used for 30 minutes every day, find electric energy consumed in the month of April.

                              Solution

                              Power of electric iron = 1200W
                              Usage per day = 30min
                              = 0.5hrs
                              Number of days in the month of April =30 days
                              Electrical energy consumed, E = P× t
                              = 1200×0.5×30
                              = 18000 Wh
                              = 18 kWh
                              = 18 units
                              Therefore, the total electricity consumed in April is 18 units.


                              Long Answer Questions

                              21. A light and a heavy object have the same momentum. Find out the ratio of their kinetic energies. Which one has a larger kinetic energy?

                                Solution

                                Consider the mass and velocity of the light object to be m1 and v1 respectively. Similarly, let the mass and velocity of the heavy object to be m2 and v2 respectively
                                Momentum = mass × velocity
                                i.e., p = mv ...(1)
                                Momentum of light object = m1v1
                                Momentum of heavy object = m2v2
                                Given,
                                light and a heavy object have the same momentum
                                Therefore, m1v1 = m2v2
                                But, we know that
                                Kinetic energy = ½ mv2
                                Thus,
                                Kinetic energy of light object (KE1)

                                But, V1 > V2 [from (4)]
                                Therefore, (K.E.)l > (K.E.)h
                                ‘The lighter one will have more kinetic energy than the heavy one. Moreover, Kinetic energy is directly proportional to the mass of the object.


                                22. An automobile engine propels a 1000 kg car (A) along a levelled road at a speed of 36 km h–1. Find the power if the opposing frictional force is 100 N. Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now, car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.

                                Solution
                                Given, Final velocity of ear
                                A(vA) = 0 (since it comes to rest after colliding with car B);  (uA) = 36 km/h = 10 m/s (since, 1 km/hr = 518 m/s);
                                Frictional force = 100 N
                                Since, the car A moves with a uniform speed, it means that the engine of car applies a force equal to the frictional force

                                mA = mB = 1000 kg
                                Initial velocity of the car, v = 36 km/h = 10 m/s
                                Frictional force = 100 N
                                Car A moves with a uniform speed, which means engine of car A applies a force equal to the frictional force.
                                Power = (Force ×distance)/time
                                = F×v
                                = 100 N × 10 m/s
                                = 1000 W
                                After collision
                                mA uA + mB uB = mA vA + mB vB
                                1000 × 10 + 1000 × 0= 1000 × 0 + 1000 × vB
                                Therefore, vB = 10 m s–1


                                23. A girl having mass of 35 kg sits on a trolley of mass 5 kg. The trolley is given an initial velocity of 4 ms–1 by applying a force. The trolley comes to rest after traversing a distance of 16 m.
                                (a) How much work is done on the trolley?
                                (b) How much work is done by the girl?

                                Solution

                                (a) Given,
                                Mass of the girl = 35 kg;
                                Mass of the trolley = 5 kg;
                                Initial velocity(u) = 4 ms-1 ;
                                Final velocity (v) = 0 (as it comes to rest) ;
                                distance (s) = 16 m
                                By using equation of motion, we get
                                v2 - u= 2as
                                ⇒ v2 = u2 + 2as
                                ⇒ 0 = (4)2 + 2 × a × (16)
                                ⇒ 0 = 16 + 32a
                                ⇒ −16 = 32a
                                ⇒ −16/32 = a
                                ⇒ −1/2 = a
                                ⇒ a = − 0.5 m/s2

                                (Acceleration is negative, therefore, retardation)
                                Total mass of the trolley = mass of girl + mass of trolley
                                = 36 + 5 = 40 kg
                                Force (frictional) acting on the trolley = ma
                                = 40 × (-0.5) = –20 N
                                Work done on the trolley = Fs
                                = 20 N × 16 m
                                = 320 J
                                (b) As the girl does not move w.r.t. the trolley, the work done by the girl = 0.


                                24. Four men lift a 250 kg box to a height of 1 m and hold it without raising or lowering it.
                                (a) How much work is done by the men in lifting the box?
                                (b) How much work do they do in just holding it?
                                (c) Why do they get tired while holding it? (g = 10 ms–2)

                                Solution

                                (a) Given,
                                mass = 250 kg
                                height(s) = 1 m
                                g = 10 ms-2
                                F = mg(g = gravity)
                                ⇒ F = 250 kg × g (g = 10 ms-2) = 2500 N
                                ⇒ s = 1 m
                                ⇒ W = F.s = 2500 × 1 N m = 2500 J

                                (b) The men did not do any work in just holding it because there is no displacement while holding the box. Therefore, the work done is zero.

                                (c) There is no displacement while holding the box. Therefore, the work done is zero.


                                25. What is power? How do you differentiate kilowatt from kilowatt hour? The Jog Falls in Karnataka state are nearly 20 m high. 2000 tonnes of water falls from it in a minute. Calculate the equivalent power if all this energy can be utilized? (g = 10 ms–2)

                                Solution

                                Power is defined as the rate of doing work. Power measures the rate of work done.
                                Power = work/time

                                Difference between kilowatt and kilowatt hour

                                Kilowatt

                                Kilowatt hour

                                kW defines how much energy a device uses or generates in a given amount of time.

                                kWh defines how much energy that device actually uses or generates.

                                kW is a measure of power.

                                kWh is a measure of energy.

                                kilowatt hour = kilowatt × hour
                                or, kWh = kW × hr

                                kilowatt = kilowatt hour/hour
                                or, kW = kWh/ hr

                                Given,
                                h = 20 m, and
                                mass, m = 2000 × 103 kg = 2 × 106 kg


                                26. How is the power related to the speed at which a body can be lifted? How many kilograms will a man working at the power of 100 W, be able to lift at constant speed of 1 m s–1 vertically? g = 10 m/s–2 .

                                Solution

                                Power = work/time
                                Work = force × displacement
                                Force = mass ×acceleration
                                Acceleration = velocity/time
                                Therefore,
                                Power = velocity × mass × distance/time × time

                                Given,
                                Power, P = 100W
                                Velocity, v = 1m/s
                                Time = 1s
                                Displacement, s = 1m
                                Acceleration, a = 10m/s2
                                From equation,
                                P = m × a × s/t
                                ⇒ 100 = m×10×1
                                ⇒ m = 10kg


                                27. Define watt. Express kilowatt in terms of joule per second. A 150 kg car engine develops 500 W for each kg. What force does it exert in moving the car at a speed of 20 m s–1?

                                Solution

                                One watt is defined as the energy consumption rate of one joule per second. The power is said to be one watt, when a work of 1 joule is done in 1 s
                                1 W =1 J/ 1 s
                                One watt is also defined as the current flow of one ampere with voltage of one volt.
                                1 kilowatt = 1000 Js-1
                                Given,
                                m = 150 kg;
                                power = 500 w/kg:
                                speed = 20 m/s
                                Power is 500 W/kg.
                                So, total power developed by the engine of 150 kg.
                                Total power = 150 × 500 = 7.5 × 104 W
                                As Power = force × speed
                                Force = Power/Speed = (7.5 × 104)/20
                                = 3.75 × 103 N


                                27. Compare the power at which each of the following is moving upwards against the force of gravity? (given g = 10 ms–2)
                                (i) A butterfly of mass 1.0 g that flies upward at a rate of 0.5 ms–1.
                                (ii) A 250 g squirrel climbing up on a tree at a rate of 0.5 ms–1.

                                Solution

                                (i) Given,
                                mass of butterfly = 1 g = (1/1000) kg :
                                g = 10 ms-2 ;
                                speed (v) = 0.5 ms-1
                                Power = force × speed
                                But, Force = mg
                                Therefore, Power = mg × v
                                P = 1/1000 × 10 × 0.5
                                ⇒ P = 0.5/100 = 5 × 103 W

                                (ii) Given,
                                mass of squirrel = 250 g = 250/1000 g = ¼ kg
                                g = 10 ms-2 ;
                                speed(v) = 0.5 ms-1
                                Power = force × speed
                                But, Force = mg
                                Therefore, Power = mg × v
                                P = ¼ × 10 × 0.5
                                ⇒ P = 1.25 W
                                Thus, the power with which the squirrel is climbing is more than that of a butterfly flying.

                                NCERT Exemplar Chapter 12 Sound Class 9 Science Solutions

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                                NCERT Exemplar Class 9 Science Chapter 12 Sound Solutions

                                NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound covers all the important questions and answers as well as advanced level questions. It helps in learning about the production of sound, vibrations, production of human voice, propagation of sound, medium, characteristics of a sound waves, longitudinal waves, Reflection of sound, echo, reverberation, multiple reflections of sound.

                                The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 12 Sound is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like applications of ultrasound, ultrasound for medical use, structure of the human ear, audible frequencies, concept and diagram of the human ear.


                                Chapter Name

                                Chapter 12 Sound

                                Book Title

                                NCERT Exemplar for Class 9 Science

                                Related Study

                                • NCERT Solutions for Class 9 Science Chapter 12 Sound
                                • Revision Notes for Class 9 Science Chapter 12 Sound
                                • MCQ for Class 9 Science Chapter 12 Sound
                                • Important Questions for Class 9 Science Chapter 12 Sound

                                Topics Covered

                                • MCQ
                                • Short Answers Questions
                                • Long Answers Questions

                                NCERT Exemplar Solutions for Chapter 12 Sound Class 9 Science

                                Multiple Choice Questions

                                1. Note is a sound :

                                (a) of mixture of several frequencies.
                                (b) of mixture of two frequencies only.
                                (c) of a single frequency.
                                (d) always unpleasant to listen.

                                Solution

                                (c) of a single frequency.


                                2. A key of a mechanical piano struck gently and then struck again but much harder this time. In the second case :
                                (a) sound will be louder but pitch will not be different.
                                (b) sound will be louder and pitch will also be higher.
                                (c) sound will be louder but pitch will be lower.
                                (d) both loudness and pitch will remain unaffected.

                                Solution

                                (a) sound will be louder but pitch will not be different.
                                Pitch depends on the frequency and loudness depends on force by which the key is pressed.


                                3. In SONAR, we use :
                                (a) ultrasound waves
                                (b) infrasonic waves
                                (c) radio waves
                                (d) audible sound waves

                                Solution

                                (a) ultrasound waves


                                4. Sound travels in air if :
                                (a) particles of medium travel from one place to another.
                                (b) there is no moisture in the atmosphere.
                                (c) disturbance moves.
                                (d) both particles as well as disturbance travel from one place to another.

                                Solution

                                (c) disturbance moves.
                                Sound waves propagate by vibrating. Disturbances created by vibration of particles move from one place to another.


                                5. When we change feeble sound to loud sound we increase its :
                                (a) frequency
                                (b) amplitude
                                (c) velocity
                                (d) wavelength

                                Solution

                                (b) amplitude
                                Loudness of sound is proportional to amplitude. When the amplitude of a feeble sound increases, it changes to loud sound.


                                6. In the curve, half the wavelength is:

                                (a) AB
                                (b) BD
                                (c) DE
                                (d) AE

                                Solution

                                (b) BD
                                Wavelength is the distance between two consecutive troughs. In the graph half the wavelength is BD.


                                7. Earthquake produces which kind of sound before the main shock wave begins?
                                (a) Ultrasound
                                (b) Infrasonic
                                (c) Audible
                                (d) Inaudible

                                Solution

                                (b) Infrasonic
                                Few animals sense the earthquake and start behaving abnormally before earthquake due to infrared rays.


                                8. Infrasound can be heard by :
                                (a) Dog
                                (b) Bat
                                (c) Rhinoceros
                                (d) Human being

                                Solution

                                (c) Rhinoceros
                                Infrasound has a frequency less than 20 Hz. Rhinoceroses communicate using infrasound waves of frequency of 5 Hz.


                                9. Before playing the orchestra in a musical concert, a sitarist tries to adjust the tension and pluck the string suitably. By doing so, he is adjusting :
                                (a) intensity of sound only.
                                (b) amplitude of sound only.
                                (c) frequency of the sitar string with the frequency of other musical instruments.
                                (d) loudness of sound.

                                Solution

                                (c) frequency of the sitar string with the frequency of other musical instruments.

                                Before beginning to play instruments, the artists adjust the frequencies. This is because the musical instruments should be tuned in with other musical instruments to produce pleasant music.


                                Short Answer Questions

                                10. The given graph (fig.) shows the displacement versus time relation for a disturbance travelling with velocity of 1500 ms-1. Calculate the wavelength of the disturbance.

                                Solution

                                From the graph
                                Time period, T = 2 × 10-6 s
                                Frequency, ν = 1/T = 5×105 Hz
                                v = 1500 ms-1

                                Wavelength, λ = Velocity(v)/Frequency(ν)
                                = 1500/(5× 105
                                )
                                = 3 × 10-3 m


                                11. Which of the above two graphs (a) and (b) representing the human voice is likely to be the male voice ? Give reason for your answer.

                                Solution

                                Usually, the pitch of a male voice is lighter than the pitch of female. Thus, the graph a represents male voice.


                                12. A girl is sitting in the middle of a park of dimension 12 m × 12 m. On the left side of it there is a building adjoining the park and on right side of the park, there is a road adjoining the park. A sound is produced on the road by a cracker. Is it possible for the girl to hear the echo of this sound? Explain your answer.

                                Solution

                                If the gap between the original sound and reflected sound received by the listener is around 0.1 second, only then the echo can be heard. The minimum distance travelled by the reflected sound wave for distinctly listening the echo is
                                Distance = Sound Velocity × time interval
                                = 344×0.1
                                = 34.4 m
                                In this case, the distance travelled by the sound reflected from the building and then reaching to the girl will be (6 + 6) = 12 m, which is much smaller than the required distance. Therefore, no echo can be heard.


                                13. Why do we hear the sound produced by the humming bees while the sound of vibrations of pendulum is not heard?

                                Solution

                                Humming bees produce the sound by beating their wings. The frequency of this sound lies in the range of 20Hz to 20000 Hz which is audible. On the other hand, pendulum produces sound less than 20 Hz. This is below the audible range. This is the reason why we don’t hear the sound of pendulum vibrations.


                                14. If any explosion takes place at the bottom of a lake, what type of shock waves in water will take place?

                                Solution

                                Longitudinal waves will take place in water if any explosion takes place at the bottom of a lake.


                                15. Sound produced by a thunderstorm is heard 10 s after the lightning seen. Calculate the approximate distance of the thunder cloud. (Given speed of sound = 340 ms-1)

                                Solution

                                Speed of sound = 340 ms-1
                                time = 10 s
                                We know,
                                speed = distance/time
                                Therefore,
                                distance = speed × time
                                = 340 × 10
                                = 3400 m


                                16. For hearing the loudest ticking sound heard by the ear, find the angle x in the figure given below.

                                Solution

                                Angle of incidence is always equal to the angle of reflection.
                                Angle of incidence = 90° -50° = 40°
                                Angle of reflection= angle of incidence =40°
                                Hence angle x is 40°.


                                17. Why is ceiling and wall behind the stage of good conference halls or concert halls made curved ?

                                Solution

                                Ceiling and wall behind the stage of good conference halls or concert halls are curved. This ensures that the reflected sound reaches to the audience equally.


                                Long Answer Questions

                                18. Represent graphically by two separate diagrams in each case :
                                (a) Two sound waves having the same amplitude but different frequencies.
                                (b) Two sound waves having the same frequency but different amplitudes.
                                (c) Two sound waves having different amplitudes and also different wavelengths.

                                Solution


                                19. Draw a curve showing density or pressure variations with respect to distance for a disturbance produced by sound. Mark the position of compression and rarefaction on this curve. Also define wavelengths and time period using this curve.

                                Solution

                                Wavelength is the distance between two consecutive compression and rarefaction. Time period is the time taken to travel the distance between any two consecutive compression or refraction from a fixed point.


                                20. Establish the relationship between speed of sound, its wavelength and frequency. If velocity of sound in air is 340 ms-1 , calculate:
                                (i) wavelength when frequency is 256 Hz.
                                (ii) frequency when wavelength is 0.85 m.

                                Solution

                                Relationship between sound speed, wavelength and frequency:

                                Speed, v= wavelength × frequency

                                NCERT Exemplar Chapter 13 Why do We Fall ill Class 9 Science Solutions

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                                NCERT Exemplar Class 9 Science Chapter 13 Why do We Fall ill Solutions

                                NCERT Exemplar Solutions for Class 9 Science Chapter 13 Why do We Fall ill covers all the important questions and answers as well as advanced level questions. It helps in learning about the human body and its capabilities, health, disease, nutritious diet, types of diseases, acute disease, chronic diseases, antibiotics, viral infections and means of spread of infectious diseases.

                                The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 13 Why do We Fall ill is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like principles of treatment, symptoms, microbes causing diseases, Principles of Prevention, physical surroundings, Public Health hygiene, General and specific ways of prevention.


                                Chapter Name

                                Chapter 13 Why do We Fall ill

                                Book Title

                                NCERT Exemplar for Class 9 Science

                                Related Study

                                • NCERT Solutions for Class 9 Science Chapter 13 Why do We Fall ill
                                • Revision Notes for Class 9 Science Chapter 13 Why do We Fall ill
                                • MCQ for Class 9 Science Chapter 13 Why do We Fall ill
                                • Important Questions for Class 9 Science Chapter 13 Why do We Fall ill

                                Topics Covered

                                • MCQ
                                • Short Answers Questions
                                • Long Answers Questions

                                NCERT Exemplar Solutions for Chapter 13 Why do We Fall ill Class 9 Science

                                Multiple Choice Questions

                                1. Which one of the following is not a viral disease?
                                (a) Dengue
                                (b) AIDS
                                (c) Typhoid
                                (d) Influenza

                                Solution

                                (c) Typhoid
                                Typhoid is caused by a bacterium called Salmonella typh.


                                2. Which one of the following is not a bacterial disease?
                                (a) Cholera
                                (b) Tuberculosis
                                (c) Anthrax
                                (d) Influenza

                                Solution

                                (d) Influenza
                                Cholera is caused by Vibrio cholera. Tuberculosis is caused by Mycobacterium tuberculosis. Anthrax is caused by Bacillus anthracis. Influenza is caused by Influenza virus.


                                3. Which one of the following disease is not transmitted by mosquito?
                                (a) Brain fever
                                (b) Malaria
                                (c) Typhoid
                                (d) Dengue

                                Solution

                                (c) Typhoid
                                Typhoid is transmitted through contaminated food and water.


                                4. Which one of the following disease is caused by bacteria?
                                (a) Typhoid
                                (b) Anthrax
                                (c) Tuberculosis
                                (d) Malaria

                                Solution

                                (d) Malaria
                                Malaria is caused by protozoan called Plasmodium vivax.


                                5. Which one of the following diseases is caused by protozoans?
                                (a) Malaria
                                (b) Influenza
                                (c) AIDS
                                (d) Cholera

                                  Solution

                                  (a) Malaria
                                  Malaria is caused by protozoan Plasmodium vivax.


                                  6. Which one of the following has a long term effect on the health of an individual?
                                  (a) Common cold
                                  (b) Chickenpox
                                  (c) Chewing tobacco
                                  (d) Stress

                                  Solution

                                  (c) Chewing tobacco
                                  Common cold and chicken pox are acute disease and they do not produce any long term effects. Stress is manageable and does not have long term effect. Chewing tobacco is dangerous as it produces a long term effect.


                                  7. Which of the following can make you ill if you come in contact with an infected person?
                                  (a) High blood pressure
                                  (b) Genetic abnormalities
                                  (c) Sneezing
                                  (d) Blood cancer

                                  Solution

                                  (c) Sneezing
                                  Sneezing is a symptom of common and other respiratory diseases. These diseases are transmitted through inhalation of contaminated air. Other diseases in the question are not infectious disease.


                                  8. AIDS cannot be transmitted by :
                                  (a) Sexual contact
                                  (b) Hugs
                                  (c) Breast feeding
                                  (d) Blood transfusion

                                  Solution

                                  (b) Hugs
                                  AIDS is transmitted by exchange of body fluids. There is no exchange of body fluids when we hug each other.


                                  9. Making anti-viral drugs is more difficult than making anti-bacterial medicines because :
                                  (a) Viruses make use of host machinery
                                  (b) Viruses are on the border line of living and non- living
                                  (c) Viruses have very few biochemical mechanisms of their own
                                  (d) Viruses have a protein coat

                                    Solution

                                    (c) Viruses have very few biochemical mechanisms of their own
                                    Antibacterial drugs stop certain biochemical reaction in the bacteria. This kills the bacteria. In viruses there are a few biochemical processes which are difficult to target, this makes anti-viral drugs difficult to make.


                                    10. Which one of the following causes kala-azar?
                                    (a) Ascaris
                                    (b) Trypanosoma
                                    (c) Leishmania
                                    (d) Bacteria

                                    Solution

                                    (c) Leishmania


                                    11. If you live in a overcrowded and poorly ventilated house, it is possible that you may suffer from which of the following diseases?
                                    (a) Cancer
                                    (b) AIDS
                                    (c) Air borne diseases
                                    (d) Cholera

                                    Solution

                                    (c) Air borne diseases


                                    12. Which disease is not transmitted by mosquitoes?
                                    (a) Dengue
                                    (b) Malaria
                                    (c) Brain fever or encephalitis
                                    (d) Pneumonia

                                    Solution

                                    (d) Pneumonia
                                    Pneumonia is transmitted through air and not mosquitos.


                                    13. Which one of the following is not important for individual health?
                                    (a) Living in clean space
                                    (b) Good economic condition
                                    (c) Social equality and harmony
                                    (d) Living in a large and well furnished house

                                    Solution

                                    (d) Living in a large and well furnished house


                                    14. Choose the wrong statement :
                                    (a) High blood pressure is caused by excessive weight and lack of exercise
                                    (b) Cancers can be caused by genetic abnormalities
                                    (c) Peptic ulcers are caused by eating acidic food
                                    (d) Acne is not caused by staphylococci

                                    Solution

                                    (c) Peptic ulcers are caused by eating acidic food
                                    Peptic ulcers are caused by a bacterium called Helicobacter pylori.


                                    15. We should not allow mosquitoes to breed in our surroundings because they :
                                    (a) multiply very fast and cause pollution
                                    (b) are vectors for many diseases
                                    (c) bite and cause skin diseases
                                    (d) are not important insects

                                    Solution

                                    (b) are vectors for many diseases
                                    Mosquitos are dangerous as they spread disease like Malaria, Dengue, Japanese encephalitis etc.


                                    16. You are aware of Polio Eradication Programme in your city. Children are vaccinated because vaccination :
                                    (a) kills the polio causing microorganisms
                                    (b) prevents the entry of polio causing organism
                                    (c) creates immunity in the body
                                    (d) all of the above

                                    Solution

                                    (c) creates immunity in the body
                                    Vaccines create immunity against the pathogens against which it is designed.


                                    17. Viruses, which cause hepatitis, are transmitted through :
                                    (a) air
                                    (b) water
                                    (c) food
                                    (d) personal contact

                                      Solution

                                      (b) water


                                      18. Vectors can be defined as :
                                      (a) Animals carry the infecting agents from sick person to another healthy person
                                      (b) Microorganisms which cause many diseases
                                      (c) Infected person
                                      (d) Diseased plants

                                        Solution

                                        (a) Animals carry the infecting agents from sick person to another healthy person


                                        Short Answer Questions

                                        19. Give two examples for each of the following :
                                        (i) Acute diseases
                                        (ii) Chronic diseases
                                        (iii) Infectious diseases
                                        (iv) Non-infectious diseases

                                        Solution

                                        (i) Viral fever and Typhoid
                                        (ii) Diabetes, Cancer and Tuberculosis
                                        (iii) Smallpox and Chickenpox
                                        (iv) Cancer, Epilepsy and Goiter


                                        20. Name two diseases caused by Protozoa. What are their causal organisms?

                                        Solution

                                        1. Malaria is caused by Plasmodium
                                        2. Kala-azar is caused by Leishmania


                                        21. Which bacterium causes peptic ulcers? Who discovered the peptic ulcer’s pathogen for the first time?

                                        Solution

                                        Peptic ulcers are caused by bacterium Helicobacter pylori. This pathogen was first discovered by Barry J. Marshal and Robin Warren.


                                          22. What is an antibiotic? Give two examples.

                                          Solution

                                          Antibiotic is a chemical agent that kills the bacteria or fungi by blocking their biochemical pathway.
                                          Example: Penicillin, Azithromycin and Streptomycin.


                                          23. Fill in the blanks :
                                          (a) Pneumonia is an example of _____ disease.
                                          (b) Many skin diseases are caused by _____
                                          (c) Antibiotics commonly block biochemical pathways important for the growth of _____
                                          (d) Living organisms carrying the infecting agents from one person to another are called _____

                                          Solution

                                          (a) Communicable
                                          (b) Fungi
                                          (c) Bacteria
                                          (d) Vectors


                                          24. Name the target organs for the following diseases :
                                          (a) Hepatitis targets
                                          (b) Fits or unconsciousness targets
                                          (c) Pneumonia targets
                                          (d) Fungal disease targets

                                          Solution

                                          (a) Hepatitis targets liver.

                                          (b) Fits or unconsciousness targets brain.

                                          (c) Pneumonia targets lungs.

                                          (d) Fungal disease targets skin.


                                          25. Who discovered ‘vaccine’ for the first time? Name two diseases which can be prevented by using vaccines.

                                          Solution

                                          Edward Jenner discovered vaccine. Polio and Diptheria are two diseases which can be prevented by using vaccines.


                                          26. Fill in the blanks :
                                          (a) _____ disease continues for many days and _____causes on body.
                                          (b) _____disease continues for a few days and causes no long term effect on body.
                                          (c) _____is defined as physical, mental and social well-being and comfort.
                                          (d) Common cold is_____ disease.
                                          (e) Many skin diseases are caused by_____

                                          Solution

                                          (a) Chronic, long term effect
                                          (b) Acute
                                          (c) Health
                                          (d) communicable (infectious)
                                          (e) fungi


                                          27. Classify the following diseases as infectious or non- infectious.
                                          (a) AIDS
                                          (b) Tuberculosis
                                          (c) Cholera
                                          (d) High blood pressure
                                          (e) Heart disease
                                          (f) Pneumonia
                                          (g) Cancer

                                          Solution

                                          (a) AIDS: Infectious disease

                                          (b) Tuberculosis: Infectious disease

                                          (c) Cholera: Infectious disease

                                          (d) High blood pressure: non-Infectious disease

                                          (e) Heart disease non: Infectious disease

                                          (f) Pneumonia: Infectious disease

                                          (g) Cancer: non-Infectious disease


                                          28. Name any two groups of microorganisms from which antibiotics could be extracted.

                                          Solution

                                          Two groups of microorganisms from which antibiotics could be extracted are:

                                          1. Bacteria
                                          2. Fungi


                                          29. Name any three diseases transmitted through vectors.

                                          Solution

                                          Three diseases transmitted through vectors are:

                                          1. Malaria
                                          2. Typhoid
                                          3. Dengue

                                          Long Answer Questions

                                          30. Explain giving reasons :
                                          (i) Balanced diet is necessary for maintaining healthy body.
                                          (ii) Health of an organism depends upon the surrounding environmental conditions.
                                          (iii) Our surrounding area should be free of stagnant water.
                                          (iv) Social harmony and good economic conditions are necessary for good health.

                                          Solution

                                          (i) Our body is made up of tiny cells which are made up of variety of organic and inorganic substances. A balanced diet is the one which consists of all the essential nutrients in an appropriate quantity. It gives our body proper nutrition required to function properly. It provides all the necessary raw materials such as carbohydrates, proteins, fats, minerals, and vitamins which are essential for proper functioning of all the body system.

                                          (ii) Health is defined as physical, mental and social well-being and comfort. Health is always dependent on our surroundings. Good environmental conditions help us to prevent diseases, increase the life span and improve health standards. Therefore, surrounding environmental conditions is also a necessary factor for a healthy individual. Following points should be maintained to keep the surrounding healthy :

                                          1. Proper removal and disposal of garbage.
                                          2. Proper disposal of sewage.
                                          3. Drinking water should be pure.
                                          4. Timely pest control to control communicable diseases.

                                          (iii) Stagnant water is not flowing and often foul-smelling water. It can become a breeding ground for the mosquitoes. Mosquitoes transmit various infectious water borne diseases like Malaria, Dengue, Elephantiasis, chikungunya etc. So, it is required to keep our surrounding area free from stagnant water to stop the spread of diseases.

                                          (iv) Human beings are social organisms. Social harmony and good economic conditions are also the necessary elements for good health. Proper earning helps in providing adequate and complete balanced diet, clean clothes and hygienic conditions. When we get caught by diseases, we need money and our economic status in order to get proper treatment.
                                          Good social environment is important factor in maintaining good health. If people mistreat each other, they could be mentally ill and they cannot lead a healthy life.


                                          31. What is a disease? How many types of diseases have you studied? Give examples.

                                          Solution

                                          The literal meaning of the term ‘disease’ is disturbed ease or being uncomfortable. Hence, disease is a condition when functioning of the body changes or gets disrupted due to the discomfort or impairment.

                                          Type of diseases:

                                          1. Acute diseases: Diseases that last for only very short periods of time.
                                            Example: Typhoid, flu.
                                          2. Chronic diseases: Diseases that last for a long time even for lifetime.
                                            Example: Asthma, elephantiasis.
                                          3. Congenital diseases: Physiological abnormalities by birth.
                                            Example: Colour blindness.
                                          4. Acquired diseases: Disease that is not present at the time of birth but developed after birth.
                                            Example: AIDS
                                          5. Infectious diseases: Diseases which are caused by microorganisms. These diseases can spread from one person to another by vectors or by other means such as air, food and water.
                                            Example: Tuberculosis, chickenpox, smallpox.
                                          6. Non-infectious diseases: Non-communicable diseases can be genetic or due to malfunctioning of internal body organ without any infection. Their causes vary, but they are not external causes like microbes that can spread in the community.
                                            Example: Cancer, diabetes.


                                          32. What do you mean by disease symptoms? Explain giving two examples.

                                          Solution

                                          When our body encounters a disease, it results in the abnormal functioning of the body. These abnormal functions are known as symptoms.

                                          The most common symptoms observed in various diseases are :

                                          1. Fever
                                          2. Diarrhoea
                                          3. Fatigue
                                          4. Body ache
                                          5. Severe headache/nausea
                                          6. Cold and cough

                                          Few examples are as follows :

                                          • A malaria patient would have symptoms such as high fever with shaking chills, headache, vomiting, nausea, abdominal pain diarrhoea, etc.
                                          • Lesions on the skin can be seen when an individual is infected with chickenpox.
                                          • Swelling of liver and yellowing of eyes can be observed in jaundice.


                                          33. Why is immune system essential for our health?

                                          Solution

                                          The defense mechanism present in our body which is responsible for fighting and destroying infecting microbes is known as immune system. It comprises of different kinds of cells which kill the infecting microorganism. When a pathogen enters our body, antibodies are released in response to the pathogen. These antibodies kill the pathogens and prevent us from falling ill.

                                          Or

                                          Immune system is essential for our health because:

                                          1. Immune system is a defense mechanism which fight against pathogenic microbes.
                                          2. This system plays the main role in distinguishing the acquired and inborn diseases of a body.
                                          3. Having specialized cell to kill infecting microbes and keep our body healthy.
                                          4. Vaccine also helps in acquiring immunity.
                                          5. Instant actions against highly active molecules like sneezing (dust particle), coughing (presence of smoke) are some common reactions observed in our daily life.


                                          34. What precautions will you take to justify “prevention is better than cure”.

                                            Solution

                                            Prevention is better than cure can have the following precious precautionary steps :

                                            1. Maintenance of proper sanitation.
                                            2. Maintaining personal as well as public hygiene
                                            3. Removal of pathogenic vectors.
                                            4. Consumption of proper balanced diet.
                                            5. Using clean water and avoid storage of water to avoid the growth of mosquitos.
                                            6. Educating about the various diseases, cause, symptoms and prevention.
                                            7. Sharing of knowledge of proper medication, suitable tests, and treatment standard.
                                            8. Vaccination for proper immunization.


                                            35. Why do some children fall ill more frequently than others living in the same locality?

                                              Solution

                                              Some children fall ill more frequently than others living in the same locality due to poor immune system. Possible reasons for poor immune system :

                                              1. Lack of balanced diet leading to malnutrition
                                              2. Unhygienic conditions at home
                                              3. Improper sanitation
                                              4. Weak immunity
                                              5. Lack of clean drinking water and hygienic food
                                              6. Using contaminated water
                                              7. Missing vaccinations


                                              36. Why are antibiotics not effective for viral disease?

                                              Solution

                                              Antibiotics act on bacteria and fungi as they block certain biochemical activities necessary for their survival. Viruses have property to use host cell for their own life processes as they can operate only few biochemical cycles by themselves whereas bacteria are capable of performing all the different life processes without the help of host cell. This is the reason why antiviral drugs work differently than the antibiotics.


                                              37. Becoming exposed to or infected with an infectious microbe does not necessarily mean developing noticeable disease. Explain.

                                              Solution

                                              Our immune system of the body fights disease-causing pathogens and clears them by killing through various immune cells, during this infection. Immune system sends active cells that kill and clear the pathogen. Such cells might be capable of completely diminishing the pathogens from the body or could also be incompetent further developing into a disease. Therefore, the development of a disease in individuals varies in reference to even same pathogenic exposure.


                                              38. Give any four factors necessary for a healthy person.

                                              Solution

                                              The essential factors which are necessary for a healthy individual are :
                                              (a) Healthy environment:

                                              1. Good drainage and sanitation system
                                              2. Clean drinking water
                                              3. Clean surrounding
                                              4. Vendor free environment.

                                              (b) Healthy economic conditions :

                                              1. Balanced diet
                                              2. Proper medical facility
                                              3. Timely vaccination

                                              (c) Healthy social harmony :

                                              1. Awareness in society
                                              2. Healthy social environment
                                              3. Green society clean society

                                              (d) Immunisation against such diseases.


                                              39. Why is AIDS considered to be a ‘Syndrome’ and not a disease?

                                              Solution

                                              HIV AIDS is caused by HIV Virus which enters our body through body fluids such as blood and breast milk or by sexual means. HIV virus spreads to the lymph nodes and blocks our immune system. This creates a condition where the body cannot fight minor diseases like common cold.

                                              A person with HIV can get pneumonia from common cold, minor gut infection may turn into severe diarrhoea. In HIV a person will not have specific symptoms rather he will have a series of infections. Hence HIV is called a syndrome, not a disease.

                                              NCERT Exemplar Chapter 14 Natural Resources Class 9 Science Solutions

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                                              NCERT Exemplar Class 9 Science Chapter 14 Natural Resources Solutions

                                              NCERT Exemplar Solutions for Class 9 Science Chapter 14 Natural Resources covers all the important questions and answers as well as advanced level questions. It helps in learning about the natural resources, soil erosion, greenhouse gasses, acid rain, water pollution, rainwater harvesting, greenhouse effect, atmosphere, conservation of rain, acid rain and smog.

                                              The NCERT Exemplar solutions for class 9 science is very important in the examination. NCERT Exemplar Solutions for Class 9 Science Chapter 14 Natural Resources is provided by our experts. They prepared the best solutions which help the students in understanding the solutions in an easy way. This chapters also covers the other topics like Soil Erosion, wrong agricultural practices, immediate condensation, rainfall pattern, pollution of water and soil, deforestation, afforestation and conservation of natural resources.


                                              Chapter Name

                                              Chapter 14 Natural Resources

                                              Book Title

                                              NCERT Exemplar for Class 9 Science

                                              Related Study

                                              • NCERT Solutions for Class 9 Science Chapter 14 Natural Resources
                                              • Revision Notes for Class 9 Science Chapter 14 Natural Resources
                                              • MCQ for Class 9 Science Chapter 14 Natural Resources
                                              • Important Questions for Class 9 Science Chapter 14 Natural Resources

                                              Topics Covered

                                              • MCQ
                                              • Short Answers Questions
                                              • Long Answers Questions

                                              NCERT Exemplar Solutions for Chapter 14 Natural Resources Class 9 Science

                                              Multiple Choice Questions

                                              1. The atmosphere of the Earth is heated by radiations which are mainly :

                                              (a) Radiated by the Sun
                                              (b) Re-radiated by land
                                              (c) Re-radiated by water
                                              (d) Re-radiated by land and water

                                              Solution

                                              (d) Re-radiated by land and water

                                              Heat from the sun reaches the earth and water. This heat is re-radiated by earth and water that heats the atmosphere.


                                              2. If there were no atmosphere around the Earth, the temperature of the Earth will :
                                              (a) Increase
                                              (b) Go on decreasing
                                              (c) Increase during day and decrease during night
                                              (d) Be unaffected

                                              Solution

                                              (c) Increase during day and decrease during night

                                              Atmosphere prevents a lot of heat from the sun to reach the earth’s surface. During the night, the atmosphere also prevents the escape of heat from the earth. Thus atmosphere is responsible for maintaining an ambient temperature on earth.


                                              3. What would happen, if all the oxygen present in the environment is converted to ozone?
                                              (a) We will be protected more
                                              (b) It will become poisonous and kill living forms
                                              (c) Ozone is not stable, hence it will be toxic
                                              (d) It will help harmful Sun radiations to reach Earth and damage many life forms.

                                              Solution

                                              (b) It will become poisonous and kill living forms

                                              All the organisms living on earth require oxygen to live. All the living beings will die if oxygen is converted to ozone.


                                              4. One of the following factors does not lead to soil formation in nature :
                                              (a) The Sun
                                              (b) Water
                                              (c) Wind
                                              (d) Polythene bags

                                              Solution

                                              (d) Polythene bags


                                              5. The two forms of oxygen found in the atmosphere are :
                                              (a) Water and ozone
                                              (b) Water and oxygen
                                              (c) Ozone and oxygen
                                              (d) Water and carbon dioxide

                                              Solution

                                              (c) Ozone and oxygen


                                              6. The process of nitrogen-fixation by bacteria does not take place in the presence of :
                                              (a) Molecular form of hydrogen
                                              (b) Elemental form of oxygen
                                              (c) Water
                                              (d) Elemental form of nitrogen

                                              Solution

                                              (b) Elemental form of oxygen


                                              7. Rainfall patterns depend on :
                                              (a) the underground water table
                                              (b) the number of water bodies in an area
                                              (c) the density pattern of human population in an area
                                              (d) the prevailing season in an area

                                              Solution

                                              (b) the number of water bodies in an area

                                              Water bodies provide water for evaporation which is converted into clouds. Clouds in turn bring rainfall. Lack of water bodies will reduce the humidity which will affect the rainfall pattern.


                                              8. Among the given options, which one is not correct for the use of large amount of fertilizers and pesticides?
                                              (a) They are eco-friendly
                                              (b) They turn the fields barren after some time
                                              (c) They adversely affect the useful component from the soil
                                              (d) They destroy the soil fertility

                                              Solution

                                              (a) They are eco-friendly


                                              9. The nitrogen molecules present in air can be converted into nitrates and nitrites by :
                                              (a) A biological process of nitrogen fixing bacteria present in soil
                                              (b) A biological process of carbon fixing factor present in soil
                                              (c) Any of the industries manufacturing nitrogenous compounds
                                              (d) The plants used as cereal crops in field

                                                Solution

                                                (a) A biological process of nitrogen fixing bacteria present in soil

                                                Cereals cannot fix the nitrogen on their own. They comprise bacteria in their root nodules which will convert nitrates to nitrites. This process is called nitrogen fixation.


                                                10. One of the following processes is not a step involved in the water-cycle operating in nature :
                                                (a) Evaporation
                                                (b) Transpiration
                                                (c) Precipitation
                                                (d) Photosynthesis

                                                  Solution

                                                  (d) Photosynthesis

                                                  Photosynthesis is a part of carbon cycle.


                                                  11. The term “water-pollution” can be defined in several ways. Which of the following statements does not give the correct definition?

                                                  (a) The addition of undesirable substances to water- bodies
                                                  (b) The removal of desirable substances from water- bodies
                                                  (c) A change in pressure of the water bodies
                                                  (d) A change in temperature of the water bodies

                                                    Solution

                                                    (c) A change in pressure of the water bodies

                                                    Physical causes lead to change in pressure on water bodies.


                                                    12. Which of the following is not a greenhouse gas?
                                                    (a) Methane
                                                    (b) Carbon dioxide
                                                    (c) Carbon monoxide
                                                    (d) Ammonia

                                                    Solution

                                                    (d) Ammonia


                                                    13. Which step is not involved in the carbon cycle?
                                                    (a) Photosynthesis
                                                    (b) Transpiration
                                                    (c) Respiration
                                                    (d) Burning of fossil fuels

                                                    Solution

                                                    (b) Transpiration

                                                    Transpiration is the process of exhaling of water by the trees. Transpiration is a part of water cycle.


                                                    14. ‘Ozone-hole’ means :
                                                    (a) A large sized hole in the ozone layer
                                                    (b) Thinning of the ozone layer
                                                    (c) Small holes scattered in the ozone layer
                                                    (d) Thickening of ozone in the ozone layer

                                                    Solution

                                                    (b) Thinning of the ozone layer


                                                    15. Ozone-layer is getting depleted because of :
                                                    (a) excessive use of automobiles
                                                    (b) excessive formation of industrial units
                                                    (c) excessive use of man-made compounds containing both fluorine and chlorine
                                                    (d) excessive deforestation.

                                                    Solution

                                                    (c) excessive use of man-made compounds containing both fluorine and chlorine

                                                    Carbon and fluorine react with ozone and convert it into oxygen. This results in thinning of ozone layer which is called ozone depletion.


                                                    16 Which of the following is a recently originated problem of environment?
                                                    (a) Ozone layer depletion
                                                    (b) Greenhouse effect
                                                    (c) Global warming
                                                    (d) All of these

                                                    Solution

                                                    (d) All of these


                                                    17. When we breathe in air, nitrogen also goes inside along with oxygen. What is the fate of this nitrogen?
                                                    (a) It moves along with oxygen into the cells
                                                    (b) It comes out with the CO2 during exhalation
                                                    (c) It is absorbed only by the nasal cells
                                                    (d) Nitrogen concentration is already more in the cells so it is not at all absorbed.

                                                    Solution

                                                    (b) It comes out with the CO2 during exhalation

                                                    Nitrogen is the most abundant gas in nature and hence during inhalation nitrogen also goes inside our body along with oxygen. But it is not utilized by our body and is exhaled along with carbon dioxide.


                                                    18. Top-soil contains the following :
                                                    (a) Humus and living organisms only
                                                    (b) Humus and soil particles only
                                                    (c) Humus, living organisms and plants
                                                    (d) Humus, living organisms and soil particles.

                                                    Solution

                                                    (d) Humus, living organisms and soil particles.


                                                    19. Choose the correct sequences :
                                                    (a) CO2 in atmosphere → decomposers → organic carbon in animals → organic carbon in plants
                                                    (b) CO2 in atmosphere → organic carbon in plants → organic carbon in animals → inorganic carbon in soil
                                                    (c) Inorganic carbonates in water → organic carbon in plants → organic carbon in animals → scavengers
                                                    (d) Organic carbon in animals → decomposers → CO2 in atmosphere → organic carbon in plants

                                                    Solution

                                                    (b) CO2 in atmosphere → organic carbon in plants → organic carbon in animals → inorganic carbon in soil

                                                    During photosynthesis green plants utilize atmospheric carbon-dioxide to make organic compounds. When animals eat plants, organic compounds reach the animals. When plant or animals die they get decomposed to turn into inorganic carbon.


                                                    20. Major source of mineral in soil is the :
                                                    (a) Parent rock from which soil is formed
                                                    (b) Plants
                                                    (c) Animals
                                                    (d) Bacteria

                                                    Solution

                                                    (a) Parent rock from which soil is formed

                                                    Rock weathering is a process which forms the soil. Rocks are naturally occurring soil aggregates which are rich in minerals.


                                                    21. Total Earth’s surface covered by water is :
                                                    (a) 75%
                                                    (b) 60%
                                                    (c) 85%
                                                    (d) 50%

                                                    Solution

                                                    (a) 75%


                                                    22. Biotic component of biosphere is not constituted by :
                                                    (a) producers
                                                    (b) consumers
                                                    (c) decomposer
                                                    (d) air

                                                    Solution

                                                    (d) air

                                                    Biotic component of biosphere comprises of the living entities, air is a non-living component.


                                                    23. An increase in carbon dioxide content in the atmosphere would not cause :
                                                    (a) More heat to be retained by the environment
                                                    (b) Increase in photosynthesis in plants
                                                    (c) Global warming
                                                    (d) Abundance of desert plants

                                                    Solution

                                                    (d) Abundance of desert plants

                                                    Plants require oxygen, minerals and water to grow and develop.


                                                    24. Oxygen is returned to the atmosphere mainly by :
                                                    (a) Burning of fossil fuel
                                                    (b) Respiration
                                                    (c) Photosynthesis
                                                    (d) Fungi

                                                    Solution

                                                    (c) Photosynthesis

                                                    Plants take CO2 from environment and they return back oxygen by a process called photosynthesis.


                                                    25. Low visibility during cold weather is due to :
                                                    (a) Formation of fossil fuel
                                                    (b) Unburnt carbon particles or hydrocarbons suspended in air
                                                    (c) Lack of adequate power supply
                                                    (d) None of the above

                                                    Solution

                                                    (b) Unburnt carbon particles or hydrocarbons suspended in air

                                                    Unburnt carbon particles and hydrocarbons suspend in air to create smog. The visibility is low in cold weather due to smog.


                                                    26. Growth of lichens on barren rocks is followed by the growth of :
                                                    (a) moss
                                                    (b) ferns
                                                    (c) gymnosperms
                                                    (d) algae

                                                    Solution

                                                    (a) moss

                                                    Lichens release certain enzymes which makes the rock suitable for the growth of moss. Moss in turn leads to the formation of soil which allows growth of more plants.


                                                    27. Marked temperature changes in aquatic environment can affect :
                                                    (a) Breeding of animals
                                                    (b) More growth of aquatic plants
                                                    (c) Process of digestion in animals
                                                    (d) Availability of nutrients

                                                    Solution

                                                    (a) Breeding of animals

                                                    Most aquatic animals are cold-blooded. Their egg and larvae are highly susceptible to temperature changes.


                                                    28. Soil erosion can be prevented by :
                                                    (a) Raising forests
                                                    (b) Deforestation
                                                    (c) Excessive use of fertilizer
                                                    (d) Overgrazing by animals

                                                    Solution

                                                    (a) Raising forests

                                                    The roots of the trees in the forest hold on the soil thus by planting more trees, soil can be prevented from being eroded.


                                                    29. What happens when rain falls on soil without vegetation cover?
                                                    (a) Rain water percolates in soil efficiently
                                                    (b) Rain water causes loss of surface soil
                                                    (c) Rain water leads to fertility of the soil
                                                    (d) Rain water does not cause any change in soil

                                                    Solution

                                                    (b) Rain water causes loss of surface soil

                                                    Roots hold the topsoil and prevent them from getting washed away during the rainfall. When there is no vegetation, the surface soil gets washed away.


                                                    30. Oxygen is harmful for :
                                                    (a) ferns
                                                    (b) nitrogen fixing bacteria
                                                    (c) chara
                                                    (d) mango tree

                                                    Solution

                                                    (b) nitrogen fixing bacteria

                                                    Nitrogen-fixing bacteria have anaerobic respiration and they get killed when exposed to oxygen.


                                                    Short Answer Questions

                                                    31. Rivers from land, add minerals to sea water. Discuss, how?

                                                    Solution

                                                    Water is a universal solvent that can dissolve a large number of substances. When water flows through rocks which have certain soluble minerals, some of them get dissolved in the water in this process. Thus, rivers carry many nutrients from the land to the sea. The salts continue to remain in the sea as no water flows out of the sea.


                                                    32. How can we prevent the loss of topsoil?

                                                    Solution

                                                    Methods of Reducing Soil Erosion :

                                                    1. Afforestation: Plants and trees protect soil erosion by binding soil by roots.
                                                    2. Proper irrigation system: Root systems stabilizes the soil and prevents soil erosion.
                                                    3. Crop rotation: It maintains the fertility water holding capacity of the soil. Thus, it prevents soil erosion.
                                                    4. Contour farming: It prevents soil erosion by reducing sediment and run off and increasing water infiltration.
                                                    5. By preventing excessive grazing by animals.
                                                    6. Prevent deforestation.


                                                    33. How is the life of organisms living in water affected when water gets polluted?

                                                    Solution

                                                    Life of organisms living in water affected in following ways when water gets polluted:

                                                    1. Addition of pollutants like fertilizers and chemicals to the water bodies will cause diseases in aquatic organisms.
                                                    2. Pollutants increase the biological oxygen demand of organisms living in water, since, the dissolved oxygen in the water reduces drastically.
                                                    3. The nutrients in fertilizers cause rapid growth of algae, also known as an algal bloom. This process is known as eutrophication. Algal blooms cover the surface of the water so sunlight does not penetrate as far down as it typically would, reducing the ability of underwater plants to perform photosynthesis and produce oxygen.
                                                    4. Pollution can affect the temperature of water bodies, while aquatic animals can manage a little change in temperature only. Heated water decreases the oxygen content of water thereby leading to the death of aquatic organisms. Similarly, cold water affects eggs and larvae and some invertebrates of the aquatic ecosystem.
                                                    5. The increase in concentration of harmful, non-biodegradable chemical substances in the body of living organisms throughout the trophic levels of a food chain is called biological magnification.


                                                    34. During summer, if you go near the lake, you feel relief from the heat, why?

                                                    Solution

                                                    Due to high temperature in summer, land and water in the lake gets heated. Since land gets heated faster than water, the air over land would also be heated faster than the air over water bodies. Heated air being light starts rising up, thus region of low pressure is created over the land. As a result, the air from the top of the water bodies rushes into this area because air moves from high pressure area towards low pressure area. This wind contains moisture and thus makes us cool and gives us relief.


                                                    35. In coastal area, wind current moves from the sea towards the land during day; but during night it moves from land to the sea. Discuss the reason.

                                                    Solution

                                                    Air above the land gets heated quickly during day and starts rising. This creates a region of low pressure as a result air over sea rushes into this area of low pressure. This movement of air from one region to the other creates winds. During night water cools down slowly, the air above water is warmer than the air on land. So, air moves from land to sea creating winds.


                                                    36. Following are a few organisms
                                                    (a) lichens
                                                    (b) mosses
                                                    (c) mango tree
                                                    (d) cactus.
                                                    Which among the above can grow on stones; and also help in formation of soil? Write the mode of their action for making soil.

                                                    Solution

                                                    • Lichen and mosses can grow on stones and they also help in formation of soil.
                                                    • Lichens grow on rocks and release certain enzymes which favour the growth of mosses.
                                                    • Mosses form clumps of soil from the rock which results in the formation of topsoil.


                                                    37. Soil formation is done by both abiotic and biotic factors. List the names of these factors by classifying them as abiotic and biotic.

                                                    Solution

                                                    1. Abiotic factors making soil: Sun, water and wind
                                                    2. Biotic factors: lichens, mosses and trees


                                                    38. All the living organisms are basically made up of C, N, S, P, H and O. How do they enter the living forms? Discuss.

                                                    Solution

                                                    The living organisms are basically made up of C, N, S, P, H2 and O2. Most of these elements enter in living forms through plants. Plants take up H2 and C by the process of photosynthesis and the other minerals absorb from the soil. They convert them into food. Consumers take in O2 and H2 during respiration and the rest of the minerals are taken through the food. The food prepared by plants is consumed by herbivores and then passes through different levels of food chains for utilization by consumers. At last, the decomposers decompose the dead bodies and the wastes given out by various consumers, thus inorganic nutrient return to the environment. Decomposition or biodegradation results in the breakdown of complex organic materials to forms of carbon that can be used by other organisms. Through the metabolic processes of fermentation and respiration, organic molecules are eventually broken down to CO2 which is returned to the atmosphere. Some bacteria remove N2 from the atmosphere and converts it to ammonia (NH3) by nitrogen fixation process and by symbiotic associations in plants. Other nitrogen-fixing bacteria are free-living in soil and aquatic habitats. Soil also plays important role in biogeochemical cycle which is a main source of recycling of nutrients from atmosphere to soil and then to water.


                                                    39. Why does the percentage of gases like oxygen, nitrogen and carbon dioxide remain almost the same in the atmosphere?

                                                    Solution

                                                    Oxygen, nitrogen and carbon dioxide are used by humans for their growth and development. But these gases are returned to the environment in one or other form. This process of using natural resources and giving them back to nature is called biogeochemical cycle. Because of this cycle the percentage of gases remain almost the same in the atmosphere.


                                                    40. Why does Moon have very cold and very hot temperature variations, e.g., from –190°C to 110°C even though it is at the same distance from the Sun as the Earth is?

                                                    Solution

                                                    The atmosphere of the earth prevents the overheating of the earth during day time. On the other hand, Moon does not have atmosphere. Therefore, it appears to be hotter during day time. Lack of atmosphere in moon also allows heat to escape during the night. This is the reason why the temperature varies on moon surface.


                                                    41. Why do people love to fly kites near the seashore?

                                                    Solution

                                                    The air above land gets heated faster and being light and starts rising up which creates a region of low pressure is created. So, the air from the surrounding areas and from the top of the water bodies moves towards land and fills the low pressure area. This blowing of breeze during day time near the sea shore helps in the flying of kites. Therefore, people love to fly kites near the seashore.


                                                    42. Why does Mathura refinery pose problems to the Taj Mahal?

                                                    Solution

                                                    Mathura refinery releases oxides of sulphur which in turn causes acid rain. Acid rain corrodes the marble of Taj Mahal. Suspended particulate matter (dust and exhaust) are causing the discoloration of the Taj Mahal. This is the reason why Mathura refinery poses a problem to the Taj Mahal.


                                                    43. Why do not lichens occur in Delhi whereas they commonly grow in Manali or Darjeeling?

                                                    Solution

                                                    Delhi occurs in semi-arid area where atmospheric moisture is low. Lichens are sensitive to sulphur dioxide (SO2) and bio-indicators of air pollution. Sulphur dioxide occurs in sufficient quantity in the atmosphere of Delhi due to the pollution caused by large number of vehicles, factories, etc. In Manali and Darjeeling, the atmosphere is humid and sulphur dioxide pollution is comparatively low making it more conducive to lichen growth.


                                                    44. Why does water need conservation even though large oceans surround the land masses?

                                                    Solution

                                                    Water covers nearly 75% of the Earth but most of the water present in the form of sea and ocean. Seawater is not suitable for the livelihood of humans and plants. Freshwater resources are limited and thus they must be conserved. Only three percent of all the water is fresh water, and one percent is available for drinking water. Moreover, most of the sources of water supply are contaminated by industrial waste or sewage. Therefore, it is necessary to conserve water.


                                                    45. There is mass mortality of fishes in a pond. What may be the reasons?

                                                    Solution

                                                    The following can the reasons for the mass mortality of fishes in a pond :

                                                    1. Addition of mercuric compounds and other toxic chemicals.
                                                    2. Blockage of gills by pollutants.
                                                    3. Fertilizer pollution does not make fish grow bigger.
                                                    4. Heated water flowing out of the thermal power plants increase the temperature of the water body. It may also affect the mortality rate of fishes.
                                                    5. Addition of poisonous compounds in water.
                                                    6. It mainly contains organic matter, which is biodegradable. Microorganisms involved in their degradation consume a lot of oxygen and the content of oxygen in the water body decreases leading to the death of fishes.


                                                    46. Lichens are called pioneer colonisers of bare rock. How can they help in formation of soil?

                                                    Solution

                                                    Lichens release enzymes which break the rock into smaller pieces. In this way the lichens help in formation of soil.


                                                    47. (i) “Soil is formed by water.” If you agree to this statement then give reasons.
                                                    (ii) Water helps in the formation of soil in two ways.

                                                    Solution

                                                    (i) Water causes soil formation through the following steps:

                                                    • Water causes wear off of the rocks for a long period of time.
                                                    • Water makes rubbing between the rocks to smaller particles which are deposited as soil.
                                                    • On freezing, water expands to cause breaks in the rock.

                                                    (ii) Fast flowing water often carries big and small particles of rock with it. These rocks rub against other and wear down into smaller and smaller particles. So, soil is found in places far away from its parent-rock.


                                                    48. Fertile soil has lots of humus. Why?

                                                    Solution

                                                    The fertile soil has lots of microorganisms which decompose dead and organic matter and convert then to humus. Humus provides nutrients, absorbs water and makes the soil porous. Earthworms feed on the humus and increase its fertility. Humus causes the soil to become more porous and allows water and air to penetrate deep underground and also gives minerals.


                                                    49. Why step farming is common in hills?

                                                    Solution

                                                    Step farming is used in hills to prevent soil erosion through water currents on the slopes. It is also known as terracing. The mountain is made into steps which slow down the speed of rainwater which prevent damage to crops. Moreover, it allows farmers to cultivate crops on steep slope, and thus provides more usable land.


                                                    50. Why are root nodules useful for the plants ?

                                                    Solution

                                                    Root nodules are the home for nitrogen-fixing bacteria called a Rhizobium. Rhizobium increases soil fertility by fixing atmospheric nitrogen.


                                                    Long Answer Questions

                                                    51. How do fossil fuels cause air pollution?

                                                    Solution

                                                    Fossil fuels like petrol, diesel and kerosene release oxide of sulphur, nitrogen and carbon. Sulphur and nitrogen oxides cause acid rain. Burning fossil fuels releases carbon monoxide which in turn increases the carbon dioxide in the atmosphere. This affects the living organisms. The amount of suspended particles in the air also increases due to burning of fossil fuels. This reduces the visibility.


                                                    52. What are the causes of water pollution? Discuss how you can contribute in reducing water pollution.

                                                    Solution

                                                    Causes of water pollution are as follows:

                                                    • Chemical fertilizers and pesticides consist of poisonous chemical which reach the water body and make it toxic.
                                                    • Dumping of sewage directly into water bodies.
                                                    • Release of industrial effluents into water bodies.

                                                    The steps to betaken in order to reduce water pollution are:

                                                    • Stop connecting sewage lines directly to water bodies.
                                                    • Avoid throwing garbage and other domestic waste into water resource.
                                                    • Prevent dumping of toxic chemical into water body.
                                                    • Avoid washing of clothes near water bodies.


                                                    53. A motorcar, with its glass totally closed, is parked directly under the Sun. The inside temperature of the car becomes very high. Explain, why?

                                                    Solution

                                                    If a closed car is parked directly under the Sun then the inside temperature of the car becomes very high because of greenhouse effect. Infrared radiations emitted by the sun pass through the transparent glass and increases the temperature inside the car. These radiations have smaller wavelength than the radiations emitted by the car. So, the transparent glass remains opaque to the radiations emitted by the car which further increases the temperature.


                                                    54. Justify “Dust is a pollutant”.

                                                    Solution

                                                    Dust is a fine powder of tiny earth particles and waste matter. The dust is carried to the atmosphere by wind. Dust causes allergy in humans and affects the plant growth by blocking the stomata on the leaves. Dust also carries toxic substance from the environment to pollute the water bodies affecting the aquatic life. Therefore, we can say that dust is a pollutant.


                                                    55. Explain the role of the Sun in the formation of soil.

                                                    Solution

                                                    During the afternoon, the Sun heats the rock. which results in the expansion of rocks. During the night, contraction takes place when the rocks cool down. But the rate of contraction is not the same as the rate of expansion. Because of this difference the rocks get cracked and form small rock particles.


                                                    56. Carbon dioxide is necessary for plants. Why do we consider it as a pollutant?

                                                    Solution

                                                    Carbon dioxide is a green-house gas which causes global warming this is the reason why carbon dioxide is considered as a pollutant. This results in change of earth’s climate. Excess of CO2 is also harmful for the plants. Increased CO2 levels changes physiology, growth and chemistry of plants. Increased concentration ofCO2 also causes suffocation, hard breathing and choking issues.

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