NCERT Solutions for Class 7th: Ch 9 Soil Science

August 22, 2015, 12:37 am

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NCERT Solutions for Class 7th: Ch 9 Soil Science

Page No: 105

Exercises

Tick the most suitable answer in questions 1 and 2.
1. In addition to the rock particles, the soil contains
    (i) air and water
    (ii) water and plants
    (iii) minerals, organic matter, air and water
    (iv) water, air and plants

Answer

✓ (iii) minerals, organic matter, air and water

2. The water holding capacity is the highest in
   (i) sandy soil
   (ii) clayey soil
   (iii) loamy soil
   (iv) mixture of sand and loam

Answer

✓ (ii) clayey soil

3. Match the items in Column I with those in Column II:

Column I	Column II
(i) A home for living organisms	(a) Large particles
(ii) Upper layer of the soil	(b) All kinds of soil
(iii) Sandy soil	(c) Dark in colour
(iv) Middle layer of the soil	(d) Small particles and packed tight
(v) Clayey soil	(e) Lesser amount of humus

Answer

Column I	Column II
(i) A home for living organisms	(b) All kinds of soil
(ii) Upper layer of the soil	(c) Dark in colour
(iii) Sandy soil	(a) Large particles
(iv) Middle layer of the soil	(e) Lesser amount of humus
(v) Clayey soil	(d) Small particles and packed tight

4. Explain how soil is formed.

Answer

The soil is formed by the process of weathering in which the rocks break down by the action of wind, water and climate. It is a very slow process and big rocks get converted into soil.

5. How is clayey soil useful for crops?

Answer

Clayey soil is very useful for crops because:
(i) It has very good water retaining capacity.
(ii) This soil is rich in humus and is very fertile.
(iii) It contains useful organic minerals.
These properties of loamy soil is very suitable for growing crops.

6. List the differences between clayey soil and sandy soil.

Answer

Clayey Soil	Loamy Soil
(i) It has much smaller particles.	(i) It has much larger particles.
(ii) It can hold good amount of water.	(ii) It cannot hold water.
(iii) It is fertile.	(iii) It is not fertile.
(iv) Air content is low.	(iv) Air get trapped between the particles.
(iv) Particles are tightly packed	(iv) Particles are loosely packed
(iv) Good for growing various crops.	(iv) Not suitable for growing crops.

7. Sketch the cross section of soil and label the various layers.

Answer

 

8. Razia conducted an experiment in the field related to the rate of percolation. She observed that it took 40 min for 200 mL of water to percolate through the soil sample. Calculate the rate of percolation.

Answer

Amount of water taken = 200 mL

Time taken by water to percolate = 40 min

 ∴ Rate of percolation = Amount of water taken/Time taken by water to percolate

                                    = 200 mL/40 min = 5 mL/min

Page No: 106

9. Explain how soil pollution and soil erosion could be prevented.

Answer

Steps for preventing soil pollution and soil erosion:

(i) Plantation should be encouraged because plant roots firmly bind the soil and help in preventing erosion.

(ii) Methods like crop rotation and mixed farming should be followed.

(iii) Use of organic fertilizers and manure instead of synthetic.

(iv) Pesticides and insecticides should be used in limited quantity and find natural way to prevent it.

(v) Plastic bags should b banned and it doesn't decompose and gives rise to soil pollution.

(vi) Industrial waste shouldn't be dumped directly as it kill necessary micro organisms of soil.

10. Solve the following crossword puzzle with the clues given:

Across
2. Plantation prevents it.
5. Use should be banned to avoid soil pollution.
6. Type of soil used for making pottery.
7. Living organism in the soil.

Down
1. In desert soil erosion occurs through.
3. Clay and loam are suitable for cereals like.
4. This type of soil can hold very little water.
5. Collective name for layers of soil.

Answer

Across
2. Plantation prevents it.→ Erosion
5. Use should be banned to avoid soil pollution. → Polythene
6. Type of soil used for making pottery. → Clay
7. Living organism in the soil. → Earthworm

Down
1. In desert soil erosion occurs through. → Wind
3. Clay and loam are suitable for cereals like. → Wheat
4. This type of soil can hold very little water. → Sandy
5. Collective name for layers of soil. → Profile

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NCERT Solutions for Class 7th: पाठ - 13 एक तिनका (कविता) हिंदी

August 22, 2015, 12:50 pm

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NCERT Solutions for Class 7th:  पाठ - 11  पाठ - 13 एक तिनका (कविता) हिंदी वसंत भाग - II

- अयोध्या सिंह उपाध्याय 'हरिऔध'

पृष्ठ संख्या: 100

प्रश्न अभ्यास

कविता से

1. नीचे दी गई कविता की पंक्तियों को सामान्य वाक्य में बदलिए।

(क) एक दिन जब था मुंडेरे पर खड़ा -
(ख) लाल होकर आँख भी दुखने लगी -
(ग) ऐंठ बेचारी दबे पाँवों भगी -
(घ) जब किसी ढब से निकल तिनका गया -

उत्तर

(क) एक दिन जब मैं अपनी छत की मुंडेर पर खड़ा था।
(ख) आँख में तिनका चले जाने के कारण आँख लाल होकर दुखने लगी।
(ग) बेचारी ऐंठ दबे पावों भागी।
(घ) किसी तरीके से आँख से तिनका निकाला गया।

2. 'एक तिनका' कविता में किस घटना की चर्चा की गई है, जिससे घमंड नहीं करने का संदेश मिलता है?

उत्तर

'एक तिनका' कविता में कवि ने उस दिन की घटना की चर्चा की है जब उसे अपने ऊपर घमंड हो गया और वह अपने को श्रेष्ठ समझने लगा। तभी एक तिनका उसके आँख में घुस गया जिससे उसकी आँखे लाल हो गयीं। बड़े प्रयास करने पर जब तिनका निकला तब लेखक को समझ आई की उसके घमंड को चूर करने के लिए तिनका है। इससे घटना से यह संदेश मिलता है की हमें घमंड नही करना चाहिए।

3. आँख में तिनका पड़ने के बाद घमंडी की क्या दशा हुई?

उत्तर

आँख में तिनका पड़ने के बाद घमंडी की आँखे लाल हो गयीं और दर्द करने लगीं। वह बैचैन हो उठा और कराहने लगा।

4. घमंडी की आँख से तिनका निकालने के लिए उसके आसपास लोगों ने क्या किया?

उत्तर

घमंडी की आँख से तिनका निकालने के लिए उसके आसपास लोगों ने कपड़े की मूँठ बनाकर उसकी आँख पर लगाकर तिनका निकालने का प्रयास किया।

5. 'एक तिनका' कविता में घमंडी को उसकी 'समझ' ने चेतावनी दी -
ऐंठता तू किसलिए इतना रहा,
एक तिनका है बहुत तेरे लिए।
इसी प्रकार की चेतावनी कबीर ने भी दी है -
तिनका कबहूँ न निंदिए, पाँव तले जो होय।
कबहूँ उड़ि आँखिन परै, पीर घनेरी होय।।

• इन दोनों में क्या समानता है और क्या अंतर? लिखिए।

उत्तर

तिनके का प्रयोग दोनों काव्यांश में उदहारण देने के लिए किया गया है। यह समानता है।
पहले काव्यांश में कवि हरिऔधजी जी ने हमें घमंड न करने की सीख दी है तथा दूसरे काव्यांश में कबीरजी ने हमें किसी को भी तुच्छ न समझने की सीख दी है। यह दोनों में अंतर है।

पृष्ठ संख्या: 101

भाषा की बात

1. 'किसी ढब से निकलना' का अर्थ है किसी ढंग से निकलना। 'ढब से' जैसे कई वाक्यांशों से आप परिचित होंगे, जैसे - धम से वाक्यांश है लेकिन ध्वनियों में समानता होने के बाद भी ढब से और धम से जैसे वाक्यांशों के प्रयोग में अंतर है। 'धम से', 'छप से', इत्यादि का प्रयोग ध्वनि द्वारा क्रिया को सूचित करने के लिए किया जाता है। नीचे कुछ ध्वनि द्वारा क्रिया को सूचित करने वाले वाक्यांश और कुछ अधूरे वाक्य दिए गए हैं।
उचित वाक्यांश चुनकर वाक्यों के खाली स्थान भरिए - 

छप से, टप से, थर्र से, फुर्र से, सन् से 

क) मेढ़क पानी में......कूद गया।
ख) नल बंद होने के बाद पानी की एक बूँद......चू गई।
ग) शोर होते ही चिड़िया......उड़ी।
घ) ठंडी हवा.......गुजरी, मैं ठंड में..... काँप गया।
उत्तर

क) मेढ़क पानी में छप सेकूद गया।
ख) नल बंद होने के बाद पानी की एक बूँद टप सेचू गई।
ग) शोर होते ही चिड़िया फुर्र सेउड़ी।
घ) ठंडी हवा सन् सेगुजरी, मैं ठंड में थर्र सेकाँप गया।

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NCERT Solutions for Class 7th: Ch 10 Respiration in Organisms Science

August 23, 2015, 1:20 am

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NCERT Solutions for Class 7th: Ch 10 Respiration in Organisms Science

Page No: 118

Exercises

1. Why does an athlete breathe faster and deeper than usual after finishing the race?

Answer

During the run, the demand of energy is high but the supply of oxygen to produce energy is limited. Therefore, anaerobic respiration takes places in the muscle cells to fulfill the demand of energy. After finishing the race, an athlete breathe faster and deeper than usual so that more oxygen is supplied to the cells.

2. List the similarities and differences between aerobic and anaerobic respiration.

Answer

Similarity:
(i) In both aerobic and anaerobic respiration, food is broken down to release energy.
(ii) Both takes place inside cells.
(iii) Both produces byproducts.

Differences:

Aerobic Respiration	Anaerobic Respiration
(i) It takes place in the presence of oxygen.	(i) It takes place in the absence of oxygen.
(ii) Energy is released in higher amount.	(ii) Energy is released in lesser amount.
(iii) Carbon dioxide and water are produced as byproducts.	(iii) Carbon dioxide and water are produced as byproducts.
(iv) It is a slow process.	(iv) It is a fast process.
(v) Examples: Animals and plants cells.	(iv) Examples: Human cells, yeast, Bacteria etc.

3. Why do we often sneeze when we inhale a lot of dust-laden air?

Answer

We often sneeze when we inhale a lot of dust-laden air to expel out these foreign particles. These particles get past the hair in the nasal cavity and  irritate the lining of the cavity which results in sneezing.


4. Take three test-tubes. Fill each of them with water. Label them A, B and C. Keep a snail in test-tube A, a water plant in test-tube B and in C, keep snail and plant both. Which test-tube would have the highest concentration of CO₂ ?

Answer

Test-tube A will have the highest concentration of CO₂ because snail will take in oxygen and gives out CO₂.
In test-tubes B and C, the CO₂ will be utilized by the water plant for synthesizing food and hence there will be less concentration of CO₂ in these.

5. Tick the correct answer:
(a) In cockroaches, air enters the body through
     (i) lungs             (ii) gills              (iii) spiracles             (iv) skin
✓ (iii) spiracles

(b) During heavy exercise, we get cramps in the legs due to the accumulation of
     (i) carbon dioxide              (ii) lactic acid                (iii) alcohol           (iv) water
✓ (ii) lactic acid

(c) Normal range of breathing rate per minute in an average adult person at rest is:
     (i) 9 - 12               (ii) 15 - 18             (iii) 21 - 24               (iv) 30 - 33
✓ (ii) 15 - 18

(d) During exhalation, the ribs
     (i) move outwards             (ii) move downwards          (iii) move upwards         (iv) do not move at all
✓ (ii) move downwards

6. Match the items in Column I with those in Column II:

Column I	Column II
(a) Yeast	(i) Earthworm
(b) Diaphragm	(ii) Gills
(c) Skin	(iii) Alcohol
(d) Leaves	(iv) Chest cavity
(e) Fish	(v) Stomata
(f) Frog	(vi) Lungs and skin
-	(vii) Tracheae

Answer

Column I	Column II
(a) Yeast	(iii) Alcohol
(b) Diaphragm	(iv) Chest cavity
(c) Skin	(i) Earthworm
(d) Leaves	(v) Stomata
(e) Fish	(ii) Gills
(f) Frog	(vi) Lungs and skin

Page No: 119

7. Mark T if the statement is true and F if it is false:
(i) During heavy exercise the breathing rate of a person slows down. (T/ F)
(ii) Plants carry out photosynthesis only during the day and respiration only at night. (T/ F)
(iii) Frogs breathe through their skins as well as their lungs. (T/ F)
(iv) The fishes have lungs for respiration. (T/ F)
(v) The size of the chest cavity increases during inhalation. (T/ F)

Answer

(i) F
(ii) F
(iii) T
(iv) F
(v) T

8. Given below is a square of letters in which are hidden different words related to respiration in organisms. These words may be present in any direction - upwards, downwards, or along the diagonals. Find the words for your respiratory system. Clues about those words are given below the square.

(i) The air tubes of insects
(ii) Skeletal structures surrounding chest cavity
(iii) Muscular floor of chest cavity
(iv) Tiny pores on the surface of leaf
(v) Small openings on the sides of the body of an insect
(vi) The respiratory organs of human beings
(vii) The openings through which we inhale
(viii) An anaerobic organism
(ix) An organism with tracheal system

Answer

(i) The air tubes of insects → Trachea
(ii) Skeletal structures surrounding chest cavity → Ribs
(iii) Muscular floor of chest cavity → Diaphragm
(iv) Tiny pores on the surface of leaf → Stomata
(v) Small openings on the sides of the body of an insect → Spiracles
(vi) The respiratory organs of human beings → Lungs
(vii) The openings through which we inhale → Nostrils
(viii) An anaerobic organism → Yeast
(ix) An organism with tracheal system → Ant

9. The mountaineers carry oxygen with them because:
(a) At an altitude of more than 5 km there is no air.
(b) The amount of air available to a person is less than that available on the ground.
(c) The temperature of air is higher than that on the ground.
(d) The pressure of air is higher than that on the ground.

Answer

The mountaineers carry oxygen with them because (b) The amount of air available to a person is less than that available on the ground.

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NCERT Solutions for Class 9th: Ch 2 Polynomials Maths (Part -2)

August 23, 2015, 4:31 pm

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NCERT Solutions for Class 9th: Ch 2 Polynomials Maths  (Part -2)

Page No: 43

Exercise 2.4

1. Determine which of the following polynomials has (x + 1) a factor:
(i) x³ + x² + x + 1

(ii) x⁴ + x³ + x² + x + 1
(iii) x⁴ + 3x³ + 3x² + x + 1 

(iv) x³ - x² - (2 + √2)x + √2

Answer

(i) If (x + 1) is a factor of p(x) = x³ + x² + x + 1, p(-1) must be zero.
Here, p(x) = x³ + x² + x + 1
p(-1) = (-1)³ + (-1)² + (-1) + 1
= -1 + 1 - 1 + 1 = 0
Therefore, x + 1 is a factor of this polynomial

(ii) If (x + 1) is a factor of p(x) = x⁴ + x³ + x² + x + 1, p(-1) must be zero.
Here, p(x) = x⁴ + x³ + x² + x + 1
p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
= 1 - 1 + 1 - 1 + 1 = 1

As, p(-1) ≠ 0
Therefore, x + 1 is not a factor of this polynomial

(iii)If (x + 1) is a factor of polynomial p(x) = x⁴ + 3x³ + 3x² + x + 1, p(- 1) must be 0.
p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
= 1 - 3 + 3 - 1 + 1 = 1
As, p(-1) ≠ 0
Therefore, x + 1 is not a factor of this polynomial.

(iv) If (x + 1) is a factor of polynomial

p(x) = x³ - x² - (2 + √2)x + √2, p(- 1) must be 0.

p(-1) =  (-1)³ -  (-1)² -  (2 + √2) (-1) + √2
= -1 - 1 + 2 + √2 + √2
=2√2
As, p(-1) ≠ 0
Therefore,, x + 1 is not a factor of this polynomial.

2. Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) p(x) = 2x³ + x² - 2x - 1, g(x) = x + 1
(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
(iii) p(x) = x³ - 4 x² + x + 6, g(x) = x - 3

Answer

(i) If g(x) = x + 1 is a factor of given polynomial p(x), p(- 1) must be zero.
p(x) = 2x³ + x² - 2x - 1
p(- 1) = 2(-1)³ + (-1)² - 2(-1) - 1
= 2(- 1) + 1 + 2 - 1 = 0
Hence, g(x) = x + 1 is a factor of given polynomial.

(ii) If g(x) = x + 2 is a factor of given polynomial p(x), p(- 2) must be 0.
p(x) = x³ +3x² + 3x + 1
p(-2) = (-2)³ + 3(- 2)² + 3(- 2) + 1
= -8 + 12 - 6 + 1
= -1

As, p(-2) ≠ 0

Hence g(x) = x + 2 is not a factor of given polynomial.

(iii) If g(x) = x - 3 is a factor of given polynomial p(x), p(3) must be 0.
p(x) = x³ - 4x² + x + 6
p(3) = (3)³ - 4(3)² + 3 + 6
= 27 - 36 + 9 = 0
Therefore,, g(x) = x - 3 is a factor of given polynomial.

Page No: 44

3. Find the value of k, if x - 1 is a factor of p(x) in each of the following cases:(i) p(x) = x² + x + k
(ii) p(x) = 2x² + kx +  √2
(iii) p(x) = kx² - √2x + 1
(iv) p(x) = kx² - 3x + k

Answer

(i) If x - 1 is a factor of polynomial p(x) = x² + x + k, then

p(1) = 0
⇒ (1)² + 1 + k = 0
⇒ 2 + k = 0
⇒ k = - 2

Therefore, value of k is -2.

(ii) If x - 1 is a factor of polynomial p(x) = 2x² + kx +  √2, then
p(1) = 0
⇒ 2(1)² + k(1) + √2 = 0
⇒ 2 + k + √2 = 0
⇒ k = -2 - √2 = -(2 + √2)

Therefore, value of k is -(2 + √2).

(iii) If x - 1 is a factor of polynomial p(x) = kx² - √2x + 1, then
p(1) = 0
⇒ k(1)² - √2(1) + 1 = 0
⇒ k - √2 + 1 = 0
⇒ k = √2 - 1

Therefore, value of k is √2 - 1.

(iv) If x - 1 is a factor of polynomial p(x) = kx² - 3x + k, then
p(1) = 0
⇒ k(1)² + 3(1) + k = 0
⇒ k - 3 + k = 0
⇒ 2k - 3 = 0
⇒ k = 3/2

Therefore, value of k is 3/2.

4. Factorise:
(i) 12x² + 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x - 6
(iv) 3x² - x - 4

Answer

(i) 12x² + 7x + 1
= 12x² - 4x - 3x+ 1                  
= 4x (3x - 1) - 1 (3x - 1)
= (3x - 1) (4x - 1)

(ii) 2x² + 7x + 3
= 2x² + 6x + x + 3
= 2x (x + 3) + 1 (x + 3)
=  (x + 3) (2x + 1)

(iii) 6x² + 5x - 6
= 6x² + 9x - 4x - 6

 = 3x (2x + 3) - 2 (2x + 3)
= (2x + 3) (3x - 2)

(iv) 3x² - x - 4
= 3x² - 4x + 3x - 4
= x (3x - 4) + 1 (3x - 4)
= (3x - 4) (x + 1)

5. Factorise:
(i) x³ - 2x² - x + 2

(ii) x³ - 3x² - 9x - 5
(iii) x³ + 13x² + 32x + 20 

(iv) 2y³ + y² - 2y - 1

Answer

(i) Let p(x) = x³ - 2x² - x + 2
Factors of 2 are ±1 and ± 2
By trial method, we find that
p(1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) = x³ - 2x² - x + 2
p(-1) = (-1)³ - 2(-1)²- (-1) + 2 = -1 -2 + 1 + 2 = 0
Therefore, (x+1) is the factor of  p(x)

 

Now, Dividend = Divisor × Quotient + Remainder

(x+1) (x² - 3x + 2)

= (x+1) (x² - x - 2x + 2)

= (x+1) {x(x-1) -2(x-1)}

= (x+1) (x-1) (x+2)

(ii) Let p(x) = x³ - 3x² - 9x - 5
Factors of 5 are ±1 and ±5
By trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
Now,
p(x) = x³ - 2x² - x + 2
p(5) = (5)³ - 3(5)²- 9(5) - 5 = 125 - 75 - 45 - 5 = 0
Therefore, (x-5) is the factor of  p(x)

Now, Dividend = Divisor × Quotient + Remainder

(x-5) (x² + 2x + 1)

= (x-5) (x² + x + x + 1)

= (x-5) {x(x+1) +1(x+1)}

= (x-5)(x+1) (x+1)

(iii) Let p(x) = x³ + 13x² + 32x + 20
Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x) =  x³ + 13x² + 32x + 20
p(-1) = (-1)³ + 13(-1)²+ 32(-1) + 20 = -1 + 13 - 32 + 20 = 0
Therefore, (x+1) is the factor of  p(x)

 

Now, Dividend = Divisor × Quotient + Remainder

(x+1) (x² + 12x + 20)

= (x+1) (x² + 2x + 10x + 20)

= (x-5) {x(x+2) +10(x+2)}

= (x-5)(x+2) (x+10)

(iv) Let p(y) = 2y³ + y² - 2y - 1
Factors of ab = 2× (-1) = -2 are ±1 and ±2
By trial method, we find that
p(1) = 0
So, (y-1) is factor of p(y)
Now,
p(y) =  2y³ + y² - 2y - 1
p(1) = 2(1)³ + (1)²- 2(1) - 1 = 2 +1 - 2 - 1 = 0
Therefore, (y-1) is the factor of  p(y)

 

 Now, Dividend = Divisor × Quotient + Remainder

(y-1) (2y² + 3y + 1)

= (y-1) (2y² + 2y + y + 1)

= (y-1) {2y(y+1) +1(y+1)}

= (y-1)(2y+1) (y+1)

Page No: 48

Exercise 2.5

1. Use suitable identities to find the following products:
    (i) (x + 4) (x + 10)                     (ii) (x + 8) (x– 10)                      (iii) (3x + 4) (3x– 5)
    (iv) (y² + 3/2) (y² - 3/2)             (v) (3 - 2x) (3 + 2x)

Answer

(i) Using identity, (x + a) (x + b) = x² + (a + b) x + ab
In (x + 4) (x + 10), a = 4 and b = 10
Now,
(x + 4) (x + 10) = x² + (4 + 10)x + (4 × 10)
                         = x² + 14x + 40

(ii) (x + 8) (x– 10)
Using identity, (x + a) (x + b) = x² + (a + b) x + ab
Here, a = 8 and b = –10
(x + 8) (x– 10) = x² + {8 +(– 10)}x + {8×(– 10)}
                         = x² + (8 – 10)x– 80
                         = x²– 2x– 80

(iii) (3x + 4) (3x– 5)
Using identity, (x + a) (x + b) = x² + (a + b) x + ab
Here, x = 3x , a = 4 and b = -5
(3x + 4) (3x– 5) = (3x) ² + {4 + (-5)}3x + {4×(-5)}
                           = 9x² + 3x(4 - 5) - 20
                           = 9x² - 3x - 20

(iv) (y² + 3/2) (y² - 3/2)
Using identity, (x + y) (x -y) = x² - y²
Here, x =y²and y = 3/2
(y² + 3/2) (y² - 3/2) = (y²)² - (3/2)²
                                         = y⁴- 9/4

(v) (3 - 2x) (3 + 2x)
Using identity, (x + y) (x -y) = x² - y²
Here, x = 3 and y = 2x
(3 - 2x) (3 + 2x) = 3² - (2x)²
                                   =  9- 4x²

2. Evaluate the following products without multiplying directly:
    (i) 103 × 107               (ii) 95 × 96               (iii) 104 × 96


Answer

(i) 103 × 107 = (100 + 3) (100 + 7)
Using identity, (x + a) (x + b) = x² + (a + b) x + ab

Here, x = 100, a = 3 and b = 7
103 × 107 = (100 + 3) (100 + 7) = (100)² + (3 + 7)10 + (3 × 7)
                 = 10000 + 100 + 21
                 = 11021

(ii) 95 × 96 = (90 + 5) (90 + 4)
Using identity, (x + a) (x + b) = x² + (a + b) x + ab
Here, x = 90, a = 5 and b = 4
95 × 96 = (90 + 5) (90 + 4) = 90² + 90(5 + 6) + (5 × 6)
             = 8100 + (11 × 90) + 30
             = 8100 + 990 + 30 = 9120

(iii) 104 × 96 = (100 + 4) (100 - 4)
Using identity, (x + y) (x -y) = x² - y²
Here, x = 100 and y = 4
104 × 96 = (100 + 4) (100 - 4) = (100)² -(4)² = 10000 - 16 = 9984

3. Factorise the following using appropriate identities:
   (i) 9x² + 6xy + y²                 (ii) 4y² - 4y + 1              (iii) x² - y²/100

Answer

(i) 9x² + 6xy + y²  = (3x) ² + (2×3x×y) + y²
Using identity, (a + b)² = a² + 2ab + b²
Here, a = 3x and b = y
9x² + 6xy + y²  = (3x) ² + (2×3x×y) + y² = (3x + y)² =(3x + y) (3x + y)

(ii) 4y² - 4y + 1 = (2y)² - (2×2y×1) + 1²
Using identity, (a - b)² = a² - 2ab + b²
Here, a = 2y and b = 1
4y² - 4y + 1 = (2y)² - (2×2y×1) + 1² = (2y - 1)² =(2y - 1) (2y - 1)

(iii) x² - y²/100 = x² - (y/10)²
Using identity, a² - b² = (a + b) (a - b)
Here, a = x and b = (y/10)
x² - y²/100 = x² - (y/10)² = (x- y/10) (x+ y/10)

Page No: 49

4. Expand each of the following, using suitable identities:
    (i) (x + 2y + 4z)²                     (ii) (2x– y + z)²                   (iii) (–2x + 3y + 2z)²
    (iv) (3a– 7b– c)²                         (v) (–2x + 5y– 3z)²                   (vi) [1/4 a - 1/2 b + 1]² 

Answer

(i) (x + 2y + 4z)²
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 
Here, a = x, b = 2y and c = 4z
(x + 2y + 4z)² = x² + (2y)² + (4z)² + (2×x×2y) + (2×2y×4z) + (2×4z×x)
                      = x² + 4y² + 16z² + 4xy + 16yz + 8xz

(ii)  (2x– y + z)²
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 
Here, a = 2x, b = -y and c = z
(2x– y + z)² = (2x)² + (-y)² + z² + (2×2x×-y) + (2×-y×z) + (2×z×2x)
                     = 4x² + y² + z² - 4xy - 2yz + 4xz

(iii) (–2x + 3y + 2z)²
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 
Here, a = -2x, b = 3y and c = 2z
(–2x + 3y + 2z)² = (-2x)² + (3y)² + (2z)² + (2×-2x×3y) + (2×3y×2z) + (2×2z×-2x)
                     = 4x² + 9y² + 4z² - 12xy + 12yz - 8xz

(iv) (3a– 7b– c)²
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 
Here, a = 3a, b = -7band c = -c
(3a – 7b – c)²= (3a)² + (-7b)² + (-c)² + (2×3a×-7b) + (2×-7b×-c) + (2×-c×3a)
                     = 9a² + 49b² + c² - 42ab + 14bc - 6ac

(v) (–2x + 5y– 3z)² 
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 
Here, a = -2x, b = 5y and c = -3z
(–2x + 5y – 3z)²= (-2x)² + (5y)² + (-3z)² + (2×-2x×5y) + (2×5y×-3z) + (2×-3z×-2x)
                     = 4x² + 25y² + 9z² - 20xy - 30yz + 12xz

(vi) [1/4 a - 1/2 b + 1]²
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca 
Here, a = 1/4 a, b = -1/2 band c = 1
[1/4 a - 1/2 b + 1]²= (1/4 a)² + (-1/2 b)² + 1² + (2×1/4 a×-1/2 b) + (2×-1/2 b×1) + (2×1×1/4 a)
                                = 1/16 a² + 1/4 b² + 1 - 1/4 ab - b + 1/2 a

5. Factorise:
(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz
(ii) 2x² + y² + 8z² - 2√2 xy + 4√2yz - 8xz

Answer

(i) 4x² + 9y² + 16z² + 12xy - 24yz - 16xz
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
4x² + 9y² + 16z² + 12xy - 24yz - 16xz
= (2x)² + (3y)² + (-4z)² + (2×2x×3y) + (2×3y×-4z) + (2×-4z×2x)
= (2x + 3y - 4z)²
=  (2x + 3y - 4z) (2x + 3y - 4z)

(ii) 2x² + y² + 8z² - 2√2 xy + 4√2yz - 8xz
Using identity, (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
2x² + y² + 8z² - 2√2 xy + 4√2 yz - 8xz
= (-√2x)² + (y)² + (2√2z)² + (2×-√2x×y) + (2×y×2√2z) + (2×2√2z×-√2x)
= (-√2x + y + 2√2z)²
=  (-√2x + y + 2√2z)(-√2x + y + 2√2z)

6. Write the following cubes in expanded form:
    (i) (2x + 1)³                (ii) (2a – 3b)³                (iii) [3/2 x + 1]³          (iv) [x - 2/3 y]³

Answer

(i)(2x + 1)³
Using identity, (a + b)³ = a³ + b³ + 3ab(a + b)
(2x + 1)³ = (2x)³ + 1³ + (3×2x×1)(2x + 1)
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 12x² + 6x + 1

(ii) (2a – 3b)³
Using identity, (a - b)³ = a³ - b³ - 3ab(a - b)
(2a – 3b)³ = (2a)³ - (3b)³ - (3×2a×3b)(2a - 3b)
= 8a³ - 27b³ - 18ab(2a - 3b)
= 8a³ - 27b³ - 36a²b + 54ab²

(iii) [3/2 x + 1]³
Using identity, (a + b)³ = a³ + b³ + 3ab(a + b)
[3/2 x + 1]³ = (3/2 x)³ + 1³ + (3×3/2 x×1)(3/2 x + 1)
= 27/8 x³ + 1 + 9/2 x(3/2 x + 1)
= 27/8 x³ + 1 + 27/4 x² + 9/2 x
= 27/8 x³ + 27/4 x² + 9/2 x + 1

(iv) [x - 2/3 y]³
Using identity, (a - b)³ = a³ - b³ - 3ab(a - b)
[x - 2/3 y]³ = (x)³ - (2/3 y)³ - (3×x×2/3 y)(x - 2/3 y)
= x³ - 8/27y³ - 2xy(x - 2/3 y)
= x³ - 8/27y³ - 2x²y + 4/3xy²

7. Evaluate the following using suitable identities:
    (i) (99)³            (ii) (102)³            (iii) (998)³ 

Answer

(i) (99)³ = (100 - 1)³
Using identity, (a - b)³ = a³ - b³ - 3ab(a - b)
(100 - 1)³ = (100)³ - 1³ - (3×100×1)(100 - 1)
= 1000000 - 1 - 300(100 - 1)
=1000000 - 1 - 30000 + 300
= 970299

(ii) (102)³ = (100 + 2)³
Using identity, (a + b)³ = a³ + b³ + 3ab(a + b)
(100 + 2)³ = (100)³ + 2³ + (3×100×2)(100 + 2)
= 1000000 + 8 + 600(100 + 2)
=1000000 + 8 + 60000 + 1200
= 1061208

(iii) (998)³
Using identity, (a - b)³ = a³ - b³ - 3ab(a - b)
(1000 - 2)³ = (1000)³ - 2³ - (3×1000×2)(1000 - 2)
= 100000000 - 8 - 6000(1000 - 2)
=100000000 - 8- 600000 + 12000
= 994011992

8. Factorise each of the following:
(i) 8a³ + b³ + 12a²b + 6ab²                          (ii) 8a³ - b³ - 12a²b + 6ab²
(iii) 27 - 125a³ - 135a + 225a²                      (iv) 64a³ - 27b³ - 144a²b + 108ab²
(v) 27p³ - 1/216 - 9/2 p² + 1/4 p

Answer


(i) 8a³ + b³ + 12a²b + 6ab²
Using identity, (a + b)³ = a³ + b³ + 3a²b + 3ab²
8a³ + b³ + 12a²b + 6ab²
= (2a)³ + b³ + 3(2a)²b +3(2a)(b)²
= (2a + b)³
= (2a + b)(2a + b)(2a + b)

(ii) 8a³ - b³ - 12a²b + 6ab²
Using identity, (a - b)³ = a³ - b³ - 3a²b + 3ab²
8a³ - b³ - 12a²b + 6ab²= (2a)³ - b³ - 3(2a)²b +3(2a)(b)²
= (2a - b)³
= (2a - b)(2a - b)(2a - b)

(iii) 27 - 125a³ - 135a + 225a²
Using identity, (a - b)³ = a³ - b³ - 3a²b + 3ab²
27 - 125a³ - 135a + 225a²=3³ - (5a)³ - 3(3)²(5a) +3(3)(5a)²
= (3 - 5a)³
= (3 - 5a)(3 - 5a)(3 - 5a)

(iv) 64a³ - 27b³ - 144a²b + 108ab²
Using identity, (a - b)³ = a³ - b³ - 3a²b + 3ab²
64a³ - 27b³ - 144a²b + 108ab²=(4a)³ - (3b)³ - 3(4a)²(3b) +3(4a)(3b)²
= (4a - 3b)³
= (4a - 3b)(4a - 3b)(4a - 3b)

(v) 27p³ - 1/216 - 9/2 p² + 1/4 p

 Using identity, (a - b)³ = a³ - b³ - 3a²b + 3ab²
 27p³ - 1/216 - 9/2 p² + 1/4 p
= (3p)³ - (1/6)³ - 3(3p)²(1/6) +3(3p)(1/6)²
= (3p - 1/6)³
= (3p - 1/6)(3p - 1/6)(3p - 1/6)

9. Verify : (i) x³ + y³ = (x + y) (x² - xy + y²)             (ii) x³ - y³ = (x - y) (x² + xy + y²)

Answer

(i) x³ + y³ = (x + y) (x² - xy + y²)
We know that,
(x + y)³ = x³ + y³ +3xy(x + y)
⇒ x³ + y³ = (x + y)³ -3xy(x + y)
⇒ x³ + y³ = (x + y)[(x + y)² -3xy]                  {Taking (x+y) common}
⇒ x³ + y³ = (x + y)[(x² + y² + 2xy) -3xy]
⇒ x³ + y³ = (x + y)(x² + y² - xy)

(ii) x³ - y³ = (x - y) (x² + xy + y² )
We know that,
(x - y)³ = x³ - y³ -3xy(x - y)
⇒ x³ - y³ = (x - y)³ +3xy(x - y)
⇒ x³ + y³ = (x - y)[(x - y)² +3xy]                     {Taking (x-y) common}
⇒ x³ + y³ = (x - y)[(x² + y² - 2xy) +3xy]
⇒ x³ + y³ = (x + y)(x² + y² + xy)

10. Factorise each of the following:
      (i) 27y³ + 125z³                    (ii) 64m³ - 343n³

Answer

(i) 27y³ + 125z³
Using identity, x³ + y³ = (x + y) (x² - xy + y²)
27y³ + 125z³ = (3y)³ + (5z)³
= (3y + 5z) {(3y)² - (3y)(5z) + (5z)²}
=(3y + 5z) (9y² - 15yz + 25z)²

(ii) 64m³ - 343n³
Using identity,x³ - y³ = (x - y) (x² + xy + y² )
64m³ - 343n³= (4m)³ - (7n)³
= (4m + 7n) {(4m)² + (4m)(7n) + (7n)²}
=(4m + 7n) (16m² + 28mn + 49n)²

11. Factorise : 27x³ + y³ + z³ - 9xyz

Answer


27x³ + y³ + z³ - 9xyz = (3x)³ + y³ + z³ - 3×3xyz
Using identity,x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz)
27x³ + y³ + z³ - 9xyz
= (3x + y + z) {(3x)² + y² + z² - 3xy - yz - 3xz}
= (3x + y + z) (9x² + y² + z² - 3xy - yz - 3xz)

12. Verify that: x³+ y³ + z³ - 3xyz = 1/2(x + y + z) [(x - y)² + (y - z)² + (z - x)²]

Answer

We know that,
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz)
⇒ x³ + y³ + z³ - 3xyz = 1/2×(x + y + z) 2(x² + y² + z² - xy - yz - xz)
= 1/2(x + y + z) (2x² + 2y² + 2z² - 2xy - 2yz - 2xz)
= 1/2(x + y + z) [(x² + y² -2xy) + (y² +z²- 2yz) + (x² + z² - 2xz)]
= 1/2(x + y + z) [(x - y)² + (y - z)² + (z - x)²]

13. If x + y + z = 0, show that x³ +y³ + z³ = 3xyz.

Answer

We know that,
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - xz)
Now put (x + y + z) = 0, 
x³ + y³ + z³ - 3xyz = (0)(x² + y² + z² - xy - yz - xz) 
⇒ x³ + y³ + z³ - 3xyz = 0


14. Without actually calculating the cubes, find the value of each of the following:
     (i) (-12)³ + (7)³ + (5)³
     (ii) (28)³ + (–15)³ + (-13)³

Answer

(i) (-12)³ + (7)³ + (5)³
 Let x =-12, y = 7 and z = 5
We observed that, x + y + z = -12 + 7 + 5 = 0

We know that if,
x + y + z = 0, then x³ +y³ + z³ = 3xyz
(-12)³ + (7)³ + (5)³ = 3(-12)(7)(5) = -1260

(ii)(28)³ + (–15)³ + (-13)³
 Let x =28, y = -15 and z = -13
We observed that, x + y + z = 28 - 15 - 13 = 0

We know that if,
x + y + z = 0, then x³ +y³ + z³ = 3xyz
(28)³ + (–15)³ + (-13)³ = 3(28)(-15)(-13) = 16380

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: 
(i) Area : 25a² - 35a + 12
(ii) Area : 35y² + 13y - 12

Answer


(i) Area : 25a² - 35a + 12

Since, area is product of length and breadth therefore by factorizing the given area, we can know the length and breadth of rectangle.
25a² - 35a + 12
= 25a² - 15a -20a + 12
= 5a(5a - 3) - 4(5a - 3)
= (5a - 4)(5a - 3)
Possible expression for length = 5a - 4
Possible expression for breadth = 5a - 3

(ii)Area : 35y² + 13y - 12
35 y² + 13y - 12
= 35y² - 15y + 28y - 12
= 5y(7y - 3) + 4(7y - 3)
= (5y + 4)(7y - 3)
Possible expression for length = (5y + 4)
Possible expression for breadth = (7y - 3)

Page No: 50

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume : 3x² - 12x
(ii) Volume : 12ky² + 8ky - 20k

Answer

(i) Volume : 3x² - 12x
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid.
3x² - 12x
= 3x(x - 4)
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height= (x - 4)

(ii) Volume : 12ky² + 8ky - 20k
Since, volume is product of length, breadth and height therefore by factorizing the given volume, we can know the length, breadth and height of the cuboid.
12ky² + 8ky - 20k
= 4k(3y² + 2y - 5)
= 4k(3y² +5y - 3y - 5)
= 4k[y(3y +5) - 1(3y + 5)]
= 4k (3y +5) (y - 1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height= (y - 1)

 

Polynomials Maths (Part-1) From Exercise 2.1 to Exercise 2.3

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NCERT Solutions for Class 7th: Ch 11 Transportation in Animals and Plants Science

Page No: 131

Exercises

1. Match structures given in Column I with functions given in Column II.

Column I	Column II
(i) Stomata	(a) Absorption of water
(ii) Xylem	(b) Transpiration
(iii) Root hairs	(c) Transport of food
(iv) Phloem	(d) Transport of water
-	(e) Synthesis of carbohydrates

Answer

Column I	Column II
(i) Stomata	(b) Transpiration
(ii) Xylem	(d) Transport of water
(iii) Root hairs	(a) Absorption of water
(iv) Phloem	(c) Transport of food

2. Fill in the blanks.
(i) The blood from the heart is transported to all parts of the body by the __________.
(ii) Haemoglobin is present in __________ cells.
(iii) Arteries and veins are joined by a network of __________.
(iv) The rhythmic expansion and contraction of the heart is called __________.
(v) The main excretory product in human beings is __________.
(vi) Sweat contains water and __________.
(vii) Kidneys eliminate the waste materials in the liquid form called __________.
(viii) Water reaches great heights in the trees because of suction pull caused by __________.

Answer

(i) The blood from the heart is transported to all parts of the body by the arteries.

(ii) Haemoglobin is present in red blood cells.

(iii) Arteries and veins are joined by a network of capillaries.

(iv) The rhythmic expansion and contraction of the heart is called heart beat.

(v) The main excretory product in human beings is urea.

(vi) Sweat contains water and salts.

(vii) Kidneys eliminate the waste materials in the liquid form called urine.

(viii) Water reaches great heights in the trees because of suction pull caused by transpiration.

Page No: 132

3. Choose the correct option:
(a) In plants, water is transported through
     (i) xylem            (ii) phloem           (iii) stomata                (iv) root hair
► (i) xylem

(b) Water absorption through roots can be increased by keeping the plants
      (i) in the shade           (ii) in dim light              (iii) under the fan             (iv) covered with a polythene bag
 ► (iii) under the fan

4. Why is transport of materials necessary in a plant or in an animal? Explain.

Answer

(i) Transport of materials in a plant or in a animal is necessary for carrying out metabolic activities.
(ii) Transportation of materials help in the supply of nutrients and energy to each every parts of animals and plants need energy which they get from the transported materials.
(iii) Also, the waste materials produced during metabolic activities are toxic and hence need to be removed from the body by transportation.

5. What will happen if there are no platelets in the blood?

Answer

Platelets help in the clotting of blood at the time of injury. If there would be no platelets, then there would be no clotting of blood and person may die due to excess flow of blood from the body.

6. What are stomata? Give two functions of stomata.

Answer

The tiny holes or openings present under the leaves of the plants is called stomata.
Two functions of stomata:
(i) It helps in breathing of the plants.
(ii) It helps in exchange of gases takes place inside the plants cells.

7. Does transpiration serve any useful function in the plants? Explain.

Answer

(i) The excess water absorbed by the root system of the plants lost in the form of water vapour to their surroundings by the process of transpiration.
(ii) It also in transport of absorbed water to the leaves of plants from the roots for photosynthesis and helping the plants keeping erect.
(iii) It also produces cooling effect for the plants.

8. What are the components of blood?

Answer

The main components of blood are:
(i) Plasma
(ii) Red blood cells (RBC)
(iii) White blood cells (WBC)
(iv) Platelets

9. Why is blood needed by all the parts of a body?

Answer

Blood is needy by all the parts of a body because:
(i) It carries oxygen to all the parts of the body and also carries carbon dioxide back to the lungs.
(ii) It carries digested food to various parts of the body for absorption.
(iii) It contains platelets which help in the clotting of blood.
(iv) It helps in maintaining constant body temperature.
(v) It transports hormones and help in fighting the body with germs and bacteria.

10. What makes the blood look red?

Answer

The presence of red pigment called haemoglobin in red blood cells (RBC) makes the blood look red.

11. Describe the function of the heart.

Answer

Functions of the heart:
(i) It helps in the circulation of oxygen rich blood throughout the body by the pumping.
(ii)It receives oxygenated blood from the lungs.
(iii) It also pumps back the blood carrying carbon dioxide to the lungs.
(iv) It shows rhythmic contraction and relaxation for movement of blood.

12. Why is it necessary to excrete waste products?

Answer

The waste materials produced during the metabolic activities are toxic to the body and must not be accumulated inside and therefore it has to be excreted out from the body by the process of excretion.

13. Draw a diagram of the human excretory system and label the various parts.

Answer

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NCERT Solutions for Class 11th: Ch 2 Methods of Enquiry in Psychology

August 24, 2015, 5:44 pm

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NCERT Solutions for Class 11th: Ch 2 Methods of Enquiry in Psychology 

Page No: 42

Review Questions

1. What are the goals of scientific enquiry?

Answer

The goals of scientific enquiry are:

→ Description: It is important in scientific enquiry to describe a behaviour or a phenomenon as accurately as possible which helps in its proper understanding.

→ Prediction: The second goal of scientific enquiry is understanding of a particular behaviour in relationship to other behaviours, events or phenomena. It tries to predict their occurrences under certain conditions with a margin of error. Prediction becomes more accurate with the increase in the number of persons observed.

→ Explanation: The third goal of psychological enquiry is to know the causal factors or determinants of behaviour and the conditions where the behaviour does not occur.

→ Control: If a person able to explain why a particular behaviour occurs, person can control that behaviour by making changes in its antecedent conditions. Control refers to three things: making a particular behaviour happen, reducing it, or enhancing it.

→ Application: The final goal of the scientific enquiry is to bring out positive changes in the lives of people through application of a particular behaviour.

2. Describe the various steps involved in conducting a scientific enquiry.

Answer

The various steps involved in conducting a scientific enquiry are: 

→ Conceptualising a Problem: The researcher have to select a theme or topic for study. Then narrows down the focus and develops specific research questions or problems for the study. This is done on the basis of review of past research, observations, and personal experiences. Next, they have to prepare a hypothesis or a tentative solution of the problem.

→ Collecting Data: The second step in scientific research is to collect data. Data collection requires developing a research design or a blueprint of the entire study. It requires taking decisions about the following four aspects: participants in the study, methods of data collection, tools to be used in research, and procedure for data collection.

→ Drawing Conclusions : The next step is to analyse data so collected through the use of statistical procedures to understand what the data mean. This can be achieved through graphical representations such as preparation of pie-chart, bar-diagram, etc. and by the use of different statistical methods.It helps to verify the hypothesis and draw conclusions by putting them into an appropriate context.

→ Revising Research Conclusions:The existing hypothesis is finally confirmed on the basis of revision of data else, a new hypothesis is stated and tested by new data. The research may also be revised by other researchers, hence making it a continuous process.

3. Explain the nature of psychological data.

Answer

The nature of psychological data are:

→ These are approximate the reality to some extent and provide an opportunity to verify or falsify our ideas, hunches, notions, etc.

→ These are are not independent of the physical or social context, the persons involved, and the time when the behaviour occurs.

→ The method of data collection such as survey, interview, experiment, etc. used and the characteristics of respondents such as, individual or group, young or old, male or female, rural or urban, etc. also influence the nature and quality of data.

4. How do experimental and control groups differ? Explain with the help of an example.

Answer

Experimental groups differ from control groups as independent variable manipulation occurs in an experimental group whereas it is absent in a control group. For example, in the study by Latane and Darley, there were two experimental groups and one control group. The participants in the study were sent to three types of rooms.
In one room no one was present (control group). In the other two rooms, two persons were already seated (experimental groups).
The independent variable, in this study, was the absence or presence of other persons sitting in the room. The remaining factors in the experiment were the same for both kinds of groups. In experimental groups, two persons were present with the real participant while in the control group, participant was alone. Therefore, it can be said that the manipulated variable is absent in control group.

5. A researcher is studying the relationship between speed of cycling and the presence of people. Formulate a relevant hypothesis and identify the independent and dependent variables.

Answer

A study of relationship between speed of cycling and the presence of people.

Hypothesis− As the speed of cycling increases people tend to move away fast.
Field experiment− Two market places
A boy is asked to ride a bicycle with different speeds in the market.
Market 1− It is observed that when the boy passes through the market street with high speed on the bicycle, people surrounding him will get away quickly in order to protect themselves from getting hit by the cycle.
Market 2− It is observed that when the boy passes through the market street with normal speed on the bicycle people around him will get away normally and slowly to give him the way as compared to the people of market 1.
Conclusion− When the speed of the cycle is high people move away from it quickly and when the speed of cycle is normal people will move away slowly in comparison.
Revision of research conclusion− The conclusion has matched the hypothesis. Therefore, the hypothesis is correct.
Independent variable− Speed of cycle
Dependent variable− Movement of people

6. Discuss the strengths and weaknesses of experimental method as a method of enquiry.

Answer

The strengths of experimental method as a method of enquiry are:
→ It makes possible to determine whether changes in the independent variable cause subsequent changes in the dependent variable.
→ The extraneous variables can be minimised.
→ It is used to minimise the sequence effect with the help of counter-balancing technique.
→ It eliminates any potential systematic differences between groups through random assignment of participants to different groups

The weaknesss of experimental method as a method of enquiry are:
→ A highly controlled laboratory only simulate situations that exist in the outside world.

→ The experiments may produce results that do not generalise well, or apply to real situations.
→ It is not always feasible to study a particular problem experimentally.
→ It is difficult to know and control all the relevant variables.

7. Dr. Krishnan is going to observe and record children’s play behaviour at a nursery school without attempting to influence or control the behaviour. Which method of research is involved? Explain the process and discuss its merits and demerits.

Answer

Non-participant observation method is involved in research of Dr. Krishnan here. He can install a video camera to record children’s play behaviour at a nursery school or sit in a corner of the class without interfering or participating in their everyday activities and then analyse and conclude it.

• Merits: The researcher study people and their behaviour in a naturalistic situation, as it occurs.

• Demerits: This method is labour intensive, time consuming, and is susceptible to the observer’s bias. Our observation is influenced by our values and beliefs about the person or the event.

8. Give two examples of the situations where survey method can be used. What are the limitations of this method?

Answer

Two Examples of the situations where survey method can be used are:
→ The attitude of people towards family planning.

→ The attitude towards giving powers to the panchayati raj institutions for running programmes related to health, education, sanitation etc.

The limitations of this method are:
→ People may give inaccurate information because of memory lapses or they may not want to let the researcher know what they really believe about a particular issue.
→ People sometimes offer responses they think the researcher wants to hear.

9. Differentiate between an interview and a questionnaire.

Answer

Interview	Questionnaire
It is a form of interaction in which questions are asked directly to the respondents.	It is a framework in which questions of scientific enquiry are written.
Its questions may vary in their sequence according to the need of the situation.	It consists of a predetermined set of questions.
Researcher and respondents are in face-to-face contact.	Researcher and respondents are not required to be in face-to-face contact.
Number of questions can be increased or decreased.	Number of questions cannot be changed.

10. Explain the characteristics of a standardised test.

Answer

The characteristics of a standardised test are:

• Reliability: It refers to the consistency of scores obtained by an individual on the same test on two different occasions. Test-retest indicates temporal stability and split-half indicates internal consistency of the test.

• Validity: The test has to be devised to measure what it claims to measure in order to be held as valid and usable.

• Norms: The test needs to devise norms or the average performance of the group. It helps in comparison and interpretation of an individual's performance in relation to the overall standards of the group.

11. Describe the limitations of psychological enquiry.

Answer

The limitations of psychological enquiry are:

• Lack of True Zero Point: Psychological measurements do not have a true zero point. For example, no person in this world has zero intelligence.The scores given to an individual in psychological studies, are not absolute in nature; rather, they have relative value.

• Relative Nature of Psychological Tools: Psychological tests are developed keeping in view the salient features of a particular context. For example, a test developed for urban children is not suitable and cannot be applied on tribal children.

• Subjective Interpretation of Qualitative Data: Data from qualitative studies are largely subjective since they involve interpretation on the part of the researcher as well as the person providing data. The interpretations may vary from one individual to the other.

12. What are the ethical guidelines that a psychologist needs to follow while conducting a psychological enquiry?

Answer

The ethical guidelines that a psychologist needs to follow while conducting a psychological enquiry are:

• Voluntary Participation : This principle states that the persons on whom researcher want to conduct the study should have the choice to decide whether to participate or not to participate in the study.

• Informed Consent : It is essential that the participants in a study should understand what will happen to them during the study.

• Debriefing : Once the study is over, the participants are provided with necessary information to complete their understanding of research.

• Sharing the Results of the Study : In psychological research, after collecting information from the participants, we come back to our places of work, analyse the data and draw conclusions. It is obligatory for the researcher to go back to the participants and share the results of the study with them.
• Confidentiality of Data Source : The participants in a study have the right to privacy. The researcher must safeguard their privacy by keeping the information provided by them in strict confidence.

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NCERT Solutions for Class 7th: पाठ - 14 खानपान की बदलती तस्वीर (निबंध) हिंदी

August 25, 2015, 3:04 am

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NCERT Solutions for Class 7th:  पाठ - 11  पाठ - 14 खानपान की बदलती तस्वीर (निबंध) हिंदी वसंत भाग - II

- प्रयाग शुक्ल

पृष्ठ संख्या: 105

प्रश्न अभ्यास

निबंध से

1. खानपान की मिश्रित संस्कृति से लेखक का क्या मतलब है? अपने घर के उदाहरण देकर इसकी व्याख्या करें।

उत्तर

खानपान की मिश्रित संस्कृति से लेखक का मतलब विभिन्न प्रदेशों के खान-पान के मिश्रित रूप से है। आज हमें एक ही घर में हमें कई प्रान्तों के खाने देखने के लिए मिल जाते हैं। उदाहरण के तौर पर मेरा घर दिल्ली में है जहाँ पराठे आदि ज्यादा बनते हैं परन्तु खानपान की मिश्रित संस्कृति की वजह से साम्भर-डोसा, इडली जो की दक्षिण भारत का प्रमुख भोजन है वो भी बनता है।
(आप अपने घर के भोजन को भी उदाहरण के लिए दे सकते हैं। अगर आप उत्तर भारतीय हैं तो आपके घर में दक्षिण भारतीय भोजन भी बनता होगा और दक्षिण भारतीय के घरों में उत्तर भारत के भोजन भी बनते हैं।)

2. खानपान में बदलाव के कौन से फ़ायदे हैं? फिर लेखक इस बदलाव को लेकर चिंतित क्यों है?

उत्तर

खानपान में बदलाव के कई फायदे हैं जैसे हमारी खाने में रूचि बनी रहती है, देश-विदेश के व्यंजन पता चलते हैं, इससे भारत की राष्ट्रीय एकता भी बनी रहती है। साथ ही इससे जल्दी बनने वाले खानों का उपलब्ध होने लगी हैं जिससे समय की भी बचत होती है। हम अपने स्वास्थ्य और स्वाद के अनुसार भी भोजन का चयन कर सकते हैं।
इन सब फायदों के बावजूद लेखक इसलिए चिंतित हैं क्योंकि इसके नुकसान भी हैं जैसे स्थानीय भोजन की लोकप्रियता का कम हो हो रही है साथ ही खाद्य पदार्थों में शुद्धता की कमी होती जा रही है। कुछ लोग उन व्यंजनों का प्रयोग अत्याधिक करने लगे हैं जो केवल स्वाद देते हैं परन्तु स्वास्थ्य के लिए हानिकारक हैं।

3. खानपान के मामले में स्थानीयता का क्या अर्थ है?

उत्तर

खानपान के मामले में स्थानीयता का अर्थ है कि वे व्यंजन जो स्थानीय आधार पर बनते थे। जैसे मुम्बई की पाव-भाजी, दिल्ली के छोले-कुलचे, आगरा के पेठे आदि।

पृष्ठ संख्या: 106

निबंध से आगे

2. यहाँ खाने, पकाने और स्वाद से संबंधित कुछ शब्द दिए गए हैं। इन्हें ध्यान से देखिए और इनका वर्गीकरण कीजिए -
उबालना, तलना, भूनना, सेंकना, दाल, भात, रोटी, पापड़, आलू, बैंगन, खट्टा, मीठा, तीखा, नमकीन, कसैला।

भोजन	कैसे पकाया	स्वाद

उत्तर

भोजन	कैसे पकाया	स्वाद
दाल	उबालना	नमकीन
भात	उबालना	मीठा
रोटी	सेंकना	मीठा
पापड़	तलना	नमकीन
आलू	उबालना	मीठा
बैंगन	भूनना	कसैला

पृष्ठ संख्या: 107

भाषा की बात

1. खानपान शब्द, खान और पान दो शब्दों को जोड़कर बना है। खानपान शब्द में और छिपा हुआ है। जिन शब्दों के योग में और, अथवा, या जैसे योजक शब्द छिपे हों, उन्हें द्वंद्व समास कहते हैं। नीचे द्वंद्व समास के कुछ उदाहरण दिए गए हैं। इनका वाक्यों में प्रयोग कीजिए और अर्थ समझिए -
सीना-पिरोना, भला-बुरा, चलना-फिरना,
लंबा-चौड़ा, कहा-सुनी, घास-फूस।

उत्तर

सीना-पिरोना - सीना-पिरोना की कला हर व्यक्ति के लिए बहुत जरुरी है।
भला-बुरा - मैंने उसे भला-बुरा कह दिया था।
चलना-फिरना - वृद्धावस्था के कारण अब चलना-फिरना कठिन हो गया है।
लंबा-चौड़ा - ये पुल बहुत लम्बा-चौड़ा है।
कहा-सुनी - मेरी रमण से खेल में कहा-सुनी हो गयी।
घास-फूस - उसका घर घास-फुस का है।

पाठ में वापिस जाएँ

NCERT Solutions for Class 7th: Ch 12 Reproduction in Plants Science

August 25, 2015, 6:32 pm

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NCERT Solutions for Class 7th: Ch 12 Reproduction in Plants Science

Page No: 141

Exercises

1. Fill in the blanks:
(a) Production of new individuals from the vegetative part of parent is called _____________.
(b) A flower may have either male or female reproductive parts. Such a flower is called_____________.
(c) The transfer of pollen grains from the anther to the stigma of the same or of another flower of the same kind is known as _____________.
(d) The fusion of male and female gametes is termed as _____________.
(e) Seed dispersal takes place by means of _____________, _____________ and _____________.

Answer

(a) Production of new individuals from the vegetative part of parent is called vegetative propagation.

(b) A flower may have either male or female reproductive parts. Such a flower is called unisexual.

(c) The transfer of pollen grains from the anther to the stigma of the same or of another flower of the same kind is known as pollination.

(d) The fusion of male and female gametes is termed as fertilisation.

(e) Seed dispersal takes place by means of wind, water and animals.

2. Describe the different methods of asexual reproduction. Give examples.

Answer

Different methods of asexual reproduction:
(i) Vegetative propagation: In this asexual reproduction, new plants are produced from roots, stems, leaves and buds of individual plant. Examples: Stem cutting cutting in champa, eye growth in potatoes, bud in case of bryophyllum etc.

(ii) Budding: The bud is a small projection which gradually grows and gets detached from the parent cell and
forms a new yeast cell. The new yeast cell grows, matures and produces more yeast cells. example: Yeast.

(iii) Fragmentation: In this mode of reproduction, the growth and multiplication is done by rapidly breaking down into two or more fragments. Each pieces grow into new individuals whenwater and nutrients are available. Example: Algae

(iv) Spore Formation: This reproduction is done by spores which under favourable condition germinates and develops into a new individual. Examples: Moss and ferns.

3. Explain what you understand by sexual reproduction.

Answer

The mode of reproduction in which two parents are involved for the production of new generation. the reproduction is done by male and female gametes. The stigma contain the male parts in which pollen grains are formed and pistil consists of stigma, style and ovary which contains the female parts. Most of the plants reproduce sexually with the help of flowers and seeds.

4. State the main difference between asexual and sexual reproduction.

Answer

Asexual Reproduction	Sexual reproduction
(i) One parent is involved.	(i) Two parents are involved.
(ii) New generation is identical or true copy of their parent.	(ii) New born are similar to their parents.
(iii) It doesn't require the formation of gametes.	(iii) It requires the formation of gametes.
(iv) Special organs for reproduction are not required.	(iv) Special organs for reproduction are required.
(v) Examples: Potato, Jasmine, Rose, Yeast, Bryophyllum etc.	(v) Examples: Mangoes, coconut, Hibiscus etc.

5. Sketch the reproductive parts of a flower.

Answer

 

6. Explain the difference between self-pollination and cross-pollination.

Answer

Self pollination	Cross Pollination
(i) Transfer of pollen from the stamen to the pistil of the same flower.	(i) Transfer of pollen from the stamen of one flower to the pistil of another flower of the same plant or different plants of the same kind.
(ii) External medium is not required.	(ii) External medium is required.
(iii) It occurs only in bisexual flower.	(iii) It occurs in both unisexual and bisexual flowers.

7. How does the process of fertilisation take place in flowers?

Answer

when pollen grain landed on the stigma of the suitable flower, it makes a tiny through style to the ovary. The pollen grain carries male gamete which met the female gamete in the ovule and fertilsation takes in flowers. The cell which results after fusion of the gametes is called a zygote which further develops into embryo.

8. Describe the various ways by which seeds are dispersed.

Answer

Various ways by which seeds are dispersed are:
(i) Dispersal by wind: Light seeds or hairy seeds and hairy fruit get blown off with the wind to far away places. Examples: Sunflower, maple, drumsticks etc.

(ii) Dispersal by water: Fruits or seeds which develop floating ability in the form of spongy or fibrous
outer coat are carried away with to different places. Example: Coconut.

(iii) Dispersal by animals or birds: Spiny seeds with hooks which get attached to the bodies of animals and
are carried to distant places. Also, the fruits are eaten up by animals and birds and their seeds get dispersed to far away places. Examples: Xanthium, Urena, Mango etc.

(iv)  Dispersal by bursting: Some seeds are dispersed when the fruits burst with sudden jerks and the seeds
are scattered far from the parent plant. Examples: castor and balsam.

(v) Dispersal by human being: They also play an important role in seed dispersal especially during plantation and farming. They also transport fruits which also help in the dispersal of seeds.

9. Match items in Column I with those in Column II:

Column I	Column II
(a) Bud	(i) Maple
(b) Eyes	(ii) Spirogyra
(c) Fragmentation	(iii) Yeast
(d) Wings	(iv) Bread mould
(e) Spores	(v) Potato
-	(vi) Rose

Answer

Column I	Column II
(a) Bud	(iii) Yeast
(b) Eyes	(ii) Potato
(c) Fragmentation	(v) Spirogyra
(d) Wings	(i) Maple
(e) Spores	(iv) Bread mould

10. Tick (✓) the correct answer:

(a) The reproductive part of a plant is the
     (i) leaf            (ii) stem          (iii) root         (iv) flower
✓ (iv) flower

(b) The process of fusion of the male and the female gametes is called
     (i) fertilisation          (ii) pollination            (iii) reproduction           (iv) seed formation
✓ (i) fertilisation

Page No: 142

(c) Mature ovary forms the
    (i) seed            (ii) stamen         (iii) pistil          (iv) fruit
✓ (iv) fruit

(d) A spore producing plant is
     (i) rose          (ii) bread mould           (iii) potato         (iv) ginger
✓ (ii) bread mould

(e) Bryophyllum can reproduce by its
     (i) stem         (ii) leaves          (iii) roots          (iv) flower
✓ (ii) leaves

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NCERT Solutions for Class 7th: Ch 13 Motion and Time Science

August 26, 2015, 6:51 pm

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NCERT Solutions for Class 7th: Ch 13 Motion and Time Science

Page No: 156

Exercises

1. Classify the following as motion along a straight line, circular or oscillatory motion:
(i) Motion of your hands while running.
(ii) Motion of a horse pulling a cart on a straight road.
(iii) Motion of a child in a merry-go-round.
(iv) Motion of a child on a see-saw.
(v) Motion of the hammer of an electric bell.
(vi) Motion of a train on a straight bridge.

Answer

(i)Oscillatory motion

(ii)Straight line motion

(iii)Circular motion 

(iv)Oscillatory motion 

(v)Oscillatory motion 

(vi)Straight line motion

2. Which of the following are not correct?
(i) The basic unit of time is second.
(ii) Every object moves with a constant speed.
(iii) Distances between two cities are measured in kilometres.
(iv) The time period of a given pendulum is not constant.
(v) The speed of a train is expressed in m/h.

Answer

(i)Correct 

(ii) Not correct

Every object may or may not moves with a constant speed.

(iii)Correct

(iv) Not correct

Time period of a given pendulum is always constant because it depends on the length of the pendulum..

 

(v) Not correct

The speed of a train is measured in km/h or in m/s.

3. A simple pendulum takes 32 s to complete 20 oscillations. What is the time period of the pendulum?

Answer

Numbers of oscillations = 20

Time taken = 32 sec

Time period of the pendulum = Time taken/Numbers of oscillations = 32/20 = 1.6 s.

4. The distance between two stations is 240 km. A train takes 4 hours to cover this distance. Calculate the speed of the train.

Answer

Distance between two stations = 240 km
Time taken = 4 hrs
Speed of train = Distance/Time taken = 240/4 = 60 km/hr

5. The odometer of a car reads 57321.0 km when the clock shows the time 08:30 AM. What is the distance moved by the car, if at 08:50 AM, the odometer reading has changed to 57336.0 km? Calculate the speed of the car in km/min during this time. Express the speed in km/h also.

Answer

Initial reading of odometer = 57321.0 km
Final reading of odometer = 57336.0 km
Initial Time = 8:30 AM
Final Time = 8:50 AM
Total distance covered = Final reading of odometer - Initial reading of odometer
                                   =57336.0 - 57321.0 = 15.0 km

Total time taken = Final Time - Initial Time =  8:50 AM - 8:30 AM = 20 minutes
20 minutes = 20/60 hrs = 1/3 hrs

Speed of the car = Total distance covered/Total time taken = 15.0/20 = 0.75 km/min

Speed of the car in km/hr =15/(1/3) = 45 km/hr

6. Salma takes 15 minutes from her house to reach her school on a bicycle. If the bicycle has a speed of
2 m/s, calculate the distance between her house and the school.

Answer

Speed of bicycle = 2 m/s
Time taken = 15 mins = 15 × 60 secs = 900 secs
A/q,
Speed = Distance/Time
⇒ Distance = Speed × Time = 2 × 900 metres = 1800 metres = 1.8 km

7. Show the shape of the distance-time graph for the motion in the following cases:
(i) A car moving with a constant speed.
(ii) A car parked on a side road.

Answer

(i)The graph will be a straight line passing through the origin.

 

(ii) The graph is a straight line parallel to x-axis or time axis. 

 

8. Which of the following relations is correct?
   (i) Speed = Distance × Time           (ii) Speed = Distance/Time
   (iii) Speed = Time/Distance           (iv) Speed = 1/Distance× Time

Answer

(ii) Speed = Distance/Time

Page No: 157

9. The basic unit of speed is:
    (i) km/min                    (ii) m/min
    (iii) km/h                      (iv) m/s

Answer

(iv) m/s

10. A car moves with a speed of 40 km/h for 15 minutes and then with a speed of 60 km/h for the next 15 minutes. The total distance covered by the car is:
(i) 100 km                   (ii) 25 km
(iii) 15 km                   (iv) 10 km

Answer

(ii) 25 km
 
Firstly, the car is moving with 40km/h for 15 minutes.

15 minutes = 15/60 hrs = 1/4 hrs

Distance covered = 40 × 1/4 = 10 km

Secondly, the car is moving with 60km/h for 15 minutes.Distance covered = 60× 1/4 = 15 km 

Total distance covered = 10 km + 15 km = 25 km

11. Suppose the two photographs, shown in Fig. 13.1 and Fig. 13.2, had been taken at an interval of 10 seconds. If a distance of 100 metres is shown by 1 cm in these photographs, calculate the speed of the blue car.

Answer

First, measure the distance with the help of scale and then proceed as given below.
Suppose the distance measured by scale is 2cm.
Multiply 2 with 100 to get the actual distance = 2 × 100 = 200 m.    (1cm = 100m)
Interval time between the photos taken = 10 s.
Speed of the blue car = 200/10 m/s = 20 m/s

12. Fig. 13.15 shows the distance-time graph for the motion of two vehicles A and B. Which one of them is moving faster?

Answer

Vehicle A is moving faster because it has more slope than vehicle B. In distance-time graph, speed is measured by its slope.

13. Which of the following distance-time graphs shows a truck moving with speed which is not constant?

Answer

Option (iii) because the slope of the graph is not a straight line and hence it does not show a uniform motion.

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NCERT Solutions for Class 7th: Ch 14 Electric Current and its Effects Science

August 27, 2015, 5:49 am

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NCERT Solutions for Class 7th: Ch 14 Electric Current and its Effects Science

Page No: 170

Exercises

1. Draw in your notebook the symbols to represent the following components of electrical circuits: connecting wires, switch in the ‘OFF’ position, bulb, cell, switch in the ‘ON’ position, and battery.


Answer

 

2. Draw the circuit diagram to represent the circuit shown in Fig.14.21.

Answer

In the fig., one end of the bulb is connected with the battery while other end is connected with safety pin which is a conductor. However, the other end battery is not connected with the safety pin which was required for the current to flow. Thus, the switch is in off position or circuit is not completed.

3. Fig.14.22 shows four cells fixed on a board. Draw lines to indicate how you will connect their terminals with wires to make a battery of four cells.

Answer

The positive terminals of cell will be connected to negative terminal of the cell and vice versa with the wire to make it a battery.

4. The bulb in the circuit shown in Fig.14.23 does not glow. Can you identify the problem? Make necessary changes in the circuit to make the bulb glow.

Answer

In the given circuit, the positive terminals of both the batteries are joined together so there is no flow of current and hence the bulb didn't glow.
Suggested changes: The current flows from positive terminals therefore the positive terminals of cell will be connected to negative terminal of the cell and vice versa.

5. Name any two effects of electric current.

Answer

Two effects of electric current:
(i) Heating effect
(ii) Magnetic effect

Page No: 171

6. When the current is switched on through a wire, a compass needle kept nearby gets deflected from its north-south position. Explain.

Answer

The current flowing through the wire produces magnetic effect around it which deflect the needle of the compass kept nearby as it is also a piece of magnet.

7. Will the compass needle show deflection when the switch in the circuit shown by Fig.14.24 is closed?

Answer

No, the circuit is not having any source of electricity. Therefore, there will be no flow of electric current through and thus there will be no magnetic effect and neither the deflection of the needle.

8. Fill in the blanks:
(a) Longer line in the symbol for a cell represents its ___________ terminal.
(b) The combination of two or more cells is called a ___________ .
(c) When current is switched ‘on’ in a room heater, it ____________ .
(d) The safety device based on the heating effect of electric current is called a ____________ .

Answer

(a) Longer line in the symbol for cell represents its positive terminal.

(b) The combination of two or more cells is called a battery.

(c) When current is switched 'on' in a room heater, it heats up.

(d) The safety device based on the heating effect of electric current is called a electric fuse.

9. Mark ‘T’ if the statement is true and ‘F’ if it is false:
(a) To make a battery of two cells, the negative terminal of one cell is connected to the negative terminal of the other cell. (T/F)
(b) When the electric current through the fuse exceeds a certain limit, the fuse wire melts and breaks. (T/F)
(c) An electromagnet does not attract a piece of iron. (T/F)
(d) An electric bell has an electromagnet. (T/F)

Answer

(a) F
(b) T
(c) F
(d) T

10. Do you think an electromagnet can be used for separating plastic bags from a garbage heap? Explain.

Answer

No, an electromagnet can be used for separating plastic bags from a garbage heap. Since plastic bags are non-magnetic material and electromagnet is used to seperate only the magnetic materials.

11. An electrician is carrying out some repairs in your house. He wants to replace a fuse by a piece of wire. Would you agree? Give reasons for your response.

Answer

No, I will not agree with the electrician.
The wire use for making fuse have low melting point and therefore any wire cannot be used to replace the fuse. In most cases, the ordinary wires have high melting point.

12. Zubeda made an electric circuit using a cell holder shown in Fig. 14.4, a switch and a bulb. When she put the switch in the ‘ON’ position, the bulb did not glow. Help Zubeda in identifying the possible defects in the circuit.

Answer

Possible defect:
(i) Bulb may have fused.
(ii) Rubber band of cell holder may not be tight and connections are loose.
(ii) Terminals of the cells may not be in the correct positions.

Page No: 172

13. In the circuit shown in Fig. 14.25

 

(i) Would any of the bulb glow when the switch is in the ‘OFF’ position?
(ii) What will be the order in which the bulbs A, B and C will glow when the switch is moved to the ‘ON’ position?

Answer

(i) No, none of the bulb will glow when the switch is in the ‘OFF’ position.

(ii) All the bulbs will glow at once because connections are ok.

 

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डायरी का एक पन्ना - पठन सामग्री और सार NCERT Class 10th Hindi

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पठन सामग्री, अतिरिक्त प्रश्न और उत्तर और सार - डायरी का एक पन्ना स्पर्श भाग - 2

सारांश

इस पाठ में लेखक सीताराम सेकसरिया ने 26 जनवरी 1931 को कोलकाता में मनाए गए स्वतंत्रता दिवस का विवरण प्रस्तुत किया है। लेखक ने बताया है की भारत में स्वतंत्रता दिवस पहली बार 26 जनवरी 1930 में मनाया गया था परन्तु उस साल कोलकाता की स्वतंत्रता दिवस में ज्यादा हिस्सेदारी नही थी परन्तु इस साल पूरी तैयारियाँ की गई थीं। केवल प्रचार में दो हजार रूपए खर्च किये गए थे। लोगों को घर-घर जाकर समझाया गया।

बड़े बाजार के प्रायः मकानों पर तिरंगा फहराया गया था। कलकत्ता के हर भाग में झंडे लगाये गए थे, ऐसी सजावट पहले कभी नही हुई थी। पुलिस भी प्रत्येक मोड़ में तैनात होकर अपनी पूरी ताकत से गश्त दे रही थी। घुड़सवारों का भी प्रबंध था।

मोनुमेंट के नीचे जहाँ सभा होने वाली थी उस जगह को पुलिस ने सुबह छः बजे ही घेर लिया फिर भी कई जगह सुबह में ही झंडा फहराया गया। श्रद्धानंद पार्क में बंगाल प्रांतीय विद्यार्थी संघ के मंत्री अविनाश बाबू ने जब झंडा गाड़ा तब उन्हें पकड़ लिया। तारा सुंदरी पार्क में बड़ा-बाजार कांग्रेस कमेटी के युद्ध मंत्री हरिश्चंद्र सिंह को झंडा फहराने से पहले ही गिरफ्तार कर लिया गया। वहाँ मारपीट भी हुई जिसमे दो-चार लोगों के सिर फट गए तथा गुजरात सेविका संघ की ओर से निकाले गए जुलुस में कई लड़कियों को गिरफ्तार किया गया।

मारवाड़ी बालिका विद्यालय की लड़कियों ने 11 बजे झंडा फहराया। जगह- जगह उत्सव और जुलुस के फोटो उतारे गए। दो-तीन कई आदमियों को पकड़ लिया गया जिनमें पूर्णोदास और परुषोत्तम राय प्रमुख थे। सुभाष चन्द्र बोस के जुलुस का भार पूर्णोदास पर था।

स्त्री समाज भी अपना जुलुस निकालने और ठीक स्थान पर पहुँचनें की कोशिश कर रहीं थीं। तीन बजे से ही मैदान में भीड़ जमा होने लगी और लोग टोलियां बनाकर घूमने लगे। इतनी बड़ी सभा कभी नही की गयी थी पुलिस कमिश्नर के नोटिस के आधार पर अमुक-अमुक धारा के अनुसार कोई सभा नहीं हो सकती थी और भाग लेने वाले व्यक्तियों को दोषी समझा जाएगा। कौंसिल के नोटिस के अनुसार चार बजकर चौबीस मिनट पर झंडा फहराया जाना था और स्वतंत्रता की प्रतिज्ञा पढ़ी जानी थी।

ठीक चार बजे सुभाष चन्द्र बोस जुलुस के साथ आए। भीड़ ज्यादा होने की वजह से पुलिस उन्हें रोक नही पायी। पुलिस ने लाठियां चलायीं, कई लोग घायल हुए और सुभाष बाबू पर भी लाठियां पड़ीं। वे जोर से 'वन्दे मातरम्' बोल रहे थे और आगे बढ़ते रहे। पुलिस भयानक रूप से लाठियां चला रहीं थी जिससे क्षितीश चटर्जी का सिर फैट गया था। उधर स्त्रियां मोनुमेंट की सीढियाँ चढ़कर झंडा फहरा रही थीं। सुभाष बाबू को पकड़ लिया गया और गाडी में बैठाकर लॉकअप भेज दिया गया।

कुछ देर बाद वहाँ से स्त्रियां जुलुस बनाकर चलीं और साथ में बहुत बड़ी भीड़ इकट्ठी हो गयी। पुलिस ने डंडे बरसाने शुरू कर दिए जिससे बहुत आदमी घायल हो गए। धर्मतल्ले के मोड़ के पास आकर जुलुस टूट गया और करीब 50-60 महिलाएँ वहीँ बैठ गयीं जिसे पुलिस से पकड़कर लालबाजार भेज दिया। स्त्रियों का एक भाग आगे विमला देवी के नेतृत्व में आगे बढ़ा जिसे बहू बाजार के मोड़ पर रोक गया और वे वहीँ बैठ गयीं। डेढ़ घंटे बाद एक लारी में बैठाकर लालबाजार ले जाया गया।

वृजलाल गोयनका को पकड़ा गया और मदालसा भी पकड़ी गयीं।  सब मिलाकर 105 स्त्रियां पकड़ी गयीं थीं जिसे बाद में रात 9 बजे छोड़ दिया गया। कलकत्ता में आज तक एक साथ इतनी ज्यादा गिरफ्तारी कभी नहीं हुई थी। करीब दो सौ लोग घायल हुए थे। पकड़े गए आदमियों की संख्या का पता नही चला पर लालबाजार के लॉकअप में स्त्रियों की संख्या 105 थी। आज का दिन कलकत्तावासियों के लिए अभूतपूर्व था। आज वो कलंक धुल गया की कलकत्तावासियों की यहाँ काम नही हो सकता।

लेखक परिचय

सीताराम सेकसरिया

इनका जन्म 1892 में राजस्थान के नवलगढ़ में हुआ परन्तु अधिकांश जीवन कलकत्ता में बिता। ये व्यापार से जुड़े होने के साथ अनेक साहित्यिक, सांस्कृतिक और नारी शिक्षण संस्थाओं के प्रेरक, संस्थापक और संचालक रहे। महात्मा गांधी के आह्वाहन पर ये स्वतंत्रता संग्राम से जुड़े। कुछ साल तक ये आजाद हिंद फ़ौज के मंत्री भी रहे। इन्होने स्वाध्याय से पढ़ना-लिखना सीखा।

प्रमुख कार्य

कृतियाँ - स्मृतिकण, मन की बात, बिता युग, नयी याद और दो भागों में एक कार्यकर्ता की डायरी।
पुरस्कार - पद्मश्री पुरस्कार।

कठिन शब्दों के अर्थ

• पुनरावृति - फिर से आना
• गश्त - पुलिस कर्मचारी का पहरे के लिए घूमना
• सार्जेंट - सेना में एक पद
• मोनुमेंट - स्मारक
• कौंसिल - परिषद
• चौरंगी - कलकत्ता के एक शहर का नाम
• वालेंटियर - स्वयंसेवक
• संगीन - गंभीर
• मदालसा - जानकी देवी और जमना लाल बजाज की पुत्री का नाम

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NCERT Solutions for Class 7th: Ch 15 Light Science

August 28, 2015, 1:12 am

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NCERT Solutions for Class 7th: Ch 15 Light Science

Page No: 189

Exercises

1. Fill in the blanks:
(a) An image that cannot be obtained on a screen is called ____________.
(b) Image formed by a convex __________ is always virtual and smaller in size.
(c) An image formed by a __________ mirror is always of the same size as that of the object.
(d) An image which can be obtained on a screen is called a _________ image.
(e) An image formed by a concave ___________ cannot be obtained on a screen.

Answer

(a) An image that cannot be obtained on a screen is called virtual image.

(b) Image formed by a convex mirror is always virtual and smaller in size.

(c) An image formed by a plane mirror is always of the same size as that of the object.

(d) An image which can be obtained on a screen is called a real image.

(e) An image formed by a concave lens cannot be obtained on a screen.

2. Mark T if the statement is true and F if it is false:
(a) We can obtain an enlarged and erect image by a convex mirror. (T/F)
(b) A concave lens always form a virtual image. (T/F)
(c) We can obtain a real, enlarged and inverted image by a concave mirror. (T/F)
(d) A real image cannot be obtained on a screen. (T/F)
(e) A concave mirror always form a real image. (T/F)

Answer

(a) F
(b) T
(c) T
(d) F
(e) F

3. Match the items given in Column I with one or more items of Column II.

Column I	Column II
(a) A plane mirror	(i) Used as a magnifying glass.
(b) A convex mirror	(ii) Can form image of objects spread over a large area.
(c) A convex lens	(iii) Used by dentists to see enlarged image of teeth.
(d) A concave mirror	(iv) The image is always inverted and magnified.
(e) A concave lens	(v) The image is erect and of the same size as the object.
-	(vi) The image is erect and smaller in size than the object.

Answer

Column I	Column II
(a) A plane mirror	(v) The image is erect and of the same size as the object.
(b) A convex mirror	(ii) Can form image of objects spread over a large area.
(c) A convex lens	(i) Used as a magnifying glass.
(d) A concave mirror	(iii) Used by dentists to see enlarged image of teeth.
(e) A concave lens	(vi) The image is erect and smaller in size than the object.

4. State the characteristics of the image formed by a plane mirror.

Answer

Characteristics of the image formed by a plane mirror:
(i) The image formed is virtual
(ii) The image is laterally inverted.
(iii) It is of the same size as the object.
(iv) The image is situated at the same distance from the mirror as the object.
(v) The image is erected.

5. Find out the letters of English alphabet or any other language known to you in which the image formed in a plane mirror appears exactly like the letter itself. Discuss your findings.

Answer

A, H, I, M, O, T, U, V, W, X, Y are the letters of English alphabet in which the image formed in a plane mirror appears exactly like the letter itself.
Discuss with your classmates to find the same types of words from other languages.

Page No: 190

6. What is a virtual image? Give one situation where a virtual image is formed.

Answer

The image which cannot be formed or obtained on the screen is called virtual image.
When we stand in front of our dressing table mirror, we use to see our virtual image. The virtual image is formed in case of plane and convex mirror.

7. State two differences between a convex and a concave lens.

Answer

Convex lens	Concave lens
(i) Convex lens converges the light falling on it.	(i) Concave lens diverges the light falling on it.
(ii) Convex lens is thicker in the middle.	(ii) Concave lens is thinner in the middle.

8. Give one use each of a concave and a convex mirror.

Answer

Concave mirror forms large images therefore it is used by dentists to see enlarged image of teeth.
Convex mirror forms diminished image therefore it used as rear view mirror in vehicles.

9. Which type of mirror can form a real image?

Answer

Concave mirror can form a real image.

10. Which type of lens forms always a virtual image?

Answer

Concave lens forms always a virtual image.

Choose the correct option in questions 11–13

11. A virtual image larger than the object can be produced by a
     (i) concave lens                 (ii) concave mirror
    (iii) convex mirror              (iv) plane mirror

Answer

(ii) concave mirror can produce a virtual image larger than the object.

12. David is observing his image in a plane mirror. The distance between the mirror and his image is 4 m. If he moves 1 m towards the mirror, then the distance between David and his image will be
          (i) 3 m            (ii) 5 m              (iii) 6 m            (iv) 8 m

Answer

(iii) 6 m
In case of plane mirror, the image is situated at the same distance from the mirror as the object.
Initial distance between David and the mirror = 4 m
Final distance between David and the mirror = 4-1 = 3 m
Therefore, distance between David and his image = 3+3 = 6m

13. The rear view mirror of a car is a plane mirror. A driver is reversing his car at a speed of 2 m/s. The driver sees in his rear view mirror the image of a truck parked behind his car. The speed at which the image of the truck appears to approach the driver will be
      (i) 1 m/s                    (ii) 2 m/s
     (iii) 4 m/s                   (iv) 8 m/s

Answer

(iii) 4 m/s
In case of plane mirror, the distance is always doubled therefore the speed is in case between the image and the object. So, when driver is reversing his car at a speed of 2 m/s, then the image is also coming closer to the mirror by 2m/s. Thus, it seems that the image of the truck appears to approach the driver at 4 m/s.

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NCERT Solutions for Class 7th: Ch 16 Water: A Precious Resource Science

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NCERT Solutions for Class 7th: Ch 16 Water: A Precious Resource Science

Page No: 203

Exercises

1. Mark ‘T’ if the statement is true and ‘F’ if it is false:
(a) The freshwater stored in the ground is much more than that present in the rivers and lakes of the world. (T/F)
(b) Water shortage is a problem faced only by people living in rural areas. (T/F)
(c) Water from rivers is the only source for irrigation in the fields. (T/F)
(d) Rain is the ultimate source of water. (T/F)

Answer

(a) T
(b) F
(c) F
(d) T

2. Explain how groundwater is recharged?

Answer

The rainwater and water from other sources such as rivers and ponds seeps through the soil and fills the empty spaces and cracks deep below the ground. The process of seeping of water into the ground is called infiltration. The groundwater thus gets recharged by this process.

3. There are ten tubewells in a lane of fifty houses. What could be the long term impact on the water table?

Answer

This will lead to exploitation of water resource present underground if people of all the fifty houses in the lane use the ten tubewells regularly. In long term, the water level will go below and thus the depletion of tubewells.

4. You have been asked to maintain a garden. How will you minimise the use of water?

Answer

For minimising the use of water in maintenance of garden, i will use the technique of drip irrigation. In this method, required amount water is supplied directly to the roots of the plants using narrow pipes thus preventing the loss of water.

5. Explain the factors responsible for the depletion of water table.

Answer

Factors responsible for the depletion of water level:
(i) Increase in the demand of water due to overpopulation.
(ii) Deforestation leads to less rainfall which affect the recharge of river and pounds.
(iii) Increase the number of industries as they use a lot of water for their purpose.
(iv) Insignificant uses in agriculture activities as in many parts of the world people uses old techniques which waste a lot of water.
(v) Lack awareness in people is also a main reason behind the depletion of water level as they waste water without any uses.

6. Fill in the blanks with the appropriate answers:
(a) People obtain groundwater through________ and ________ .
(b) Three forms of water are ________, ________ and ________.
(c) The water bearing layer of the earth is ________.
(d) The process of water seepage into the ground is called ________.

Answer

(a) People obtain groundwater through tube wells and hand pumps.

(b) Three forms of water are ice, water and vapour.

(c) The water bearing layer of the earth is aquifer.

(d) The process of water seepage into the ground is called infiltration.

Page No: 204

7. Which one of the following is not responsbile for water shortage?
(i) Rapid growth of industries
(ii) Increasing population
(iii) Heavy rainfall
(iv) Mismanagement of water resources

Answer

(iii) Heavy rainfall

8. Choose the correct option. The total water
(i) in the lakes and rivers of the world remains constant.
(ii) under the ground remains constant.
(iii) in the seas and oceans of the world remains constant.
(iv) of the world remains constant.

Answer

(iv) of the world remains constant.

9. Make a sketch showing groundwater and water table. Label it.

Answer

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NCERT Solutions for Class 10th: पाठ - 15 नीलकंठ (रेखाचित्र) हिंदी वसंत भाग-II

- महादेवी वर्मा

पृष्ठ संख्या - 116

प्रश्न अभ्यास

निबंध से

1. मोर-मोरनी के नाम किस आधार पर रखे गए?

उत्तर

नीली गर्दन होने के कारण मोर का नाम नीलकंठ रखा गया और मोरनी सदा मोर की छाया के समान उसके साथ रहती इसलिए उसका नाम राधा रखा गया।

2. जाली के बड़े घर में पहुँचने पर मोर के बच्चों का किस प्रकार स्वागत हुआ?

उत्तर

जाली के बड़े घर में पहुँचने पर मोर के बच्चों का उसी तरह स्वागत हुआ जैसा नववधू के आगमन पर परिवार में होता है। लक्का कबूतर नाचना छोड़ उनके चारों ओर घूम-घूम कर गुटरगूं-गुटरगूं की रागिनी अलापने लगे, बड़े खरगोश सभ्य सभासदों के समान क्रम से बैठकर उनका निरीक्षण करने लगे, छोटे खरगोश उनके चारों ओर उछलकूद मचाने लगे और तोते एक आँख बंद करके उनका परीक्षण करने लगे।

3. लेखिका को नीलकंठ की कौन-कौन सी चेष्टाएँ बहुत भाती थीं?

उत्तर

नीलकंठ देखने में बहुत सुंदर था और लेखिका को उसकी हर चेष्टाएँ आकर्षक लगती थीं परन्तु कुछ चेष्टाएँ उन्हें बहुत भाती थीं जैसे -
• मेघों की गर्जन ताल पर उसका इंद्रधनुष के गुच्छे जैसे पंखों को मंडलाकार बनाकर तन्मय नृत्य करना।
• लेखिका के हाथों से हौले-हौले चने उठाकर खाते समय उसकी चेष्टाएँ हँसी और विस्मय उत्पन्न करती थी।
• नीलकंठ का दयालु स्वभाव और सबकी रक्षा करने की चेष्टा करना।

4. 'इस आनंदोंत्सव की रागिनी में बेमेल स्वर कैसे बज उठा' - वाक्य किस घटना की ओर संकेत कर रहा है?

उत्तर

यह वाक्य लेखिका द्वारा कुब्जा मोरनी को लाने की ओर संकेत कर रहा है। कुब्जा मोरनी के आने से पहले नीलकंठ, राधा और अन्य पशु-पक्षी बाड़े में आराम से रह रहे थे जिसे लेखिका ने आनंदोंत्सव की रागिनी कहा है। परन्तु कुब्जा मोरनी के आ जाने से वहाँ अशांति फ़ैल गयी। वह स्वभाव से मेल-मिलाप वाली न थी। ईर्ष्यालु प्रकृति की होने के कारण वह नीलकंठ और राधा को साथ न देख पाती थी। उसने राधा के अंडे भी तोड़ डाले थे। नीलकंठ अप्रसन्न रहने लगा था और अंत में यह उसकी मृत्यु का कारण बना।

5. वसंत ऋतु में नीलकंठ के लिए जालीघर में बंद रहना असहनीय क्यों हो जाता था?

उत्तर

वसंत में आम के वृक्ष मंजरियों से लदे जाते और अशोक लाल पत्तों से ढक जाता जिसे देखकर नीलकंठ के लिए जालीघर में रहना असहनीय हो जाता। उसे फलों के वृक्षों से भी अधिक सुगन्धित व खिले पत्तों वाले वृक्ष अच्छे लगते थे।

6. जालीघर में रहनेवाले सभी जीव एक-दूसरे के मित्र बन गए थे, पर कुब्जा के साथ ऐसा संभव क्यों नहीं हो पाया?

उत्तर

कुब्जा का स्वभाव मेल-मिलाप वाला न था। ईर्ष्यालु होने के कारण वह सबसे झगड़ा करती रहती थी और अपनी चोंच से नीलकंठ के पास जाने वाले हर-एक पक्षी को नोंच डालती थी। वह किसी को भी नीलकंठ के पास आने नहीं देती थी यहाँ तक की उसने इसी ईर्ष्यावश राधा के अंडें भी तोड़ दिए थे। इसी कारण वह किसी की मित्र न बन सकी।

पृष्ठ संख्या: 117

7. नीलकंठ ने खरगोश के बच्चे को साँप से किस तरह बचाया? इस घटना के आधार पर नीलकंठ के स्वभाव की विशेषताओं का उल्लेख कीजिए।

उत्तर

एक बार एक साँप पशुओं के जाली के भीतर पहुँच गया। सब जीव-जंतु इधर-उधर भागकर छिप गए, केवल एक शिशु खरगोश साँप की पकड़ में आ गया। निगलने के प्रयास में साँप ने उसका आधा पिछला शरीर मुँह में दबा लिया। नन्हा खरगोश धीरे-धीरे चीं-चीं कर रहा था परन्तु आवाज़ इतना तीव्र नही था की किसी को स्पष्ट सुनाई दे। सोये हुए नीलकंठ ने जब यह मंद स्वर सुना तो वह झट से अपने पंखों को समेटता हुआ झूले से नीचे आ गया। उसने सावधानी से साँप के फन के पास पंजों से दबाया और फिर अपनी चोंच से इतने प्रहार उस पर किए कि वह अधमरा हो गया और फन की पकड़ ढीली होते ही खरगोश का बच्चा मुख से निकल आया। इस प्रकार नीलकंठ ने खरगोश के बच्चे को साँप से बचाया।
इस घटना के आधार पर नीलकंठ के स्वभाव की विशेषताओं निम्नलिखित हैं -
• सतर्कता - जालीघर के ऊँचे झूले पर सोते हुए भी उसे  खरगोश की मंद पुकार सुनकर यह शक हो गया कोई प्राणी कष्ट में है और वह झट से झूले से नीचे उतरा।
• साहसी और वीर - अकेले ही उसने साँप से खरगोश के बच्चों को बचाया और साँप के दो खंड कर दिया जिससे उसके साहस और वीरता का पता चलता है।• रक्षक - खरगोश को मौत के मुँह से बचाकर नीलकंठ ने यह सिद्ध कर दिया कि वह रक्षक है।
• दयालु - वह खरगोश के बच्चे को सारी रात अपने पंखों में छिपाकर ऊष्मा देता रहा जिससे उसके दयालु होने का पता चलता है।
भाषा की बात

1. 'रूप' शब्द से 'कुरूप', 'स्वरूप', 'बहुरूप' आदि शब्द बनते हैं। इसी प्रकार नीचे लिखे शब्दों से अन्य शब्द बनाओ -
गंध, रंग, फल, ज्ञान

उत्तर

गंध - सुगंध, दुर्गन्ध, गंधक, गंधहीन।
रंग - बदरंग, बेरंग, रंगबिरंगा।
फल - सफल, निष्फल, असफल, विफल।
ज्ञान - विज्ञान, अज्ञान, सद्ज्ञान।

2. नीचे दिए गए शब्दों के संधि विग्रह कीजिए

संधि	विग्रह
नील + आभ =	सिंहासन =
नव + आगंतुक =	मेघाच्छन्न =

उत्तर

संधि	विग्रह
नील + आभ = नीलाभ	सिंहासन = सिंह + आसन
नव + आगंतुक = नवागंतुक	मेघाच्छन्न = मेघ + आच्छन्न

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NCERT Solutions for Class 7th: Ch 17 Forests: Our Lifeline Science

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NCERT Solutions for Class 7th: Ch 17 Forests: Our Lifeline Science

Page No: 217

Exercises

1. Explain how animals dwelling in the forest help it grow and regenerate.

Answer

Animals dwelling in the forest help in various way for growth and regenerate. Some of them are:
(i) Herbivores animals clear the land by eating grass for the new growth of vegetation.
(ii) Animals also help in the seed dispersal and pollination.
(iii) Dead and decaying bodies of animals convert to humus after decomposition which increase the fertility of the soil of the forest.
(iv) Dungs of animals provide nutrient to various types of seedling to grow.

2. Explain how forests prevent floods.

Answer

Forest  acts as a natural absorber of rainwater and allows it to seep. It helps in controlling the flow of water and slows it down which helps in preventing flood. Also, the trees present in the forest prevents the rain from directly hitting the ground and bind the soil together which helps in absorption of the rain water and thus prevent flood.

3. What are decomposers? Name any two of them. What do they do in the forest?

Answer

The micro-organisms which convert the dead plants and animals to humus are known as decomposers.
Bacteria and fungi are the two examples of decomposers.
They play an important role in the decomposition of dead remains of the plants and animals
 and converting them to humus which add nutrient to the soil.

4. Explain the role of forest in maintaining the balance between oxygen and carbon dioxide in the atmosphere.

Answer

Forest play an important role in maintaining the balance between oxygen and carbon dioxide in the atmosphere. The green plants and trees present in the forest take in carbon dioxide which is released by the animals and release oxygen through the process of photosynthesis which are breathe in by the animals and thus maintains the balance between oxygen and carbon dioxide in the atmosphere.

5. Explain why there is no waste in a forest.

Answer

There is no waste in a forest because whatever produced here are utilized by the other organisms or plants for sustainability. Even the waste materials and dead remains which are produced are bodegradable and converted into humus which returns back the nutrient back to the soil.

6. List five products we get from forests?

Answer

Five products we get from forests are:
(i) Wood
(ii) Resins
(iii) Gum
(iv) Honey
(v) Medicines

Page No: 218

7. Fill in the blank:
(a) The insects, butterflies, honeybees and birds help flowering plants in ___________ .
(b) A forest is a purifier of _____________ and _______________ .
(c) Herbs form the ______________ layer in the forest.
(d) The decaying leaves and animal droppings in a forest enrich the ______________ .

Answer

(a) The insects, butterflies, honeybees and birds help flowering plants in pollination .

(b) A forest is a purifier of water and air .

(c) Herbs form the lowest layer in the forest.

(d) The decaying leaves and animal droppings in a forest enrich the soil .

8. Why should we worry about the conditions and issues related to forests far from us?

Answer

We should worry about the conditions and issues related to forests far from us because more or less we are dependent on it because:
(i) The amount of carbon dioxide in air will increase if forest will disappear and thus less oxygen in atmosphere o breathe in.
(ii) There will be more soil erosion and thus increase in floods as soil will not able to hold water.
(iii) Increase in the percentage of carbon dioxide will also leads to global warming.
(iv) Deforestation will endanger our life and environment and also there will be no shelter for animals.
(v) There is also imbalance in nature and thus causing climate changes and less rainfall.

9. Explain why there is a need of variety of animals and plants in a forest.

Answer

There is a need of variety of animals and plants in a forest as each of them contribute their for maintaining the ecosystem of the forest. It also helps forest in regeneration and growth. Herbivores are needed to eat green plants and to provide food for the carnivores and Carnivores are needed to eat herbivores and check their population. Also, decomposers are needed to maintain the supply of nutrients to the soil and to the growing plants.

10. In Fig. 17.15, the artist has forgotten to put the labels and directions on the arrows. Mark the directions on the arrows and label the diagram using the following labels:
         clouds, rain, atmosphere, carbon dioxide, oxygen, plants, animals, soil, roots, water table.

Answer

11. Which of the following is not a forest product?
(i) Gum
(ii) Plywood
(iii) Sealing wax
(iv) Kerosene
► (iv) Kerosene

12. Which of the following statements is not correct?
(i) Forests protect the soil from erosion.
(ii) Plants and animals in a forest are not dependent on one another.
(iii) Forests influence the climate and water cycle.
(iv) Soil helps forests to grow and regenerate.
► (ii) Plants and animals in a forest are not dependent on one another.

Page No: 219

13. Micro-organisms act upon the dead plants to produce
(i) sand 
(ii) mushrooms
(iii) humus
(iv) wood
► (iii) humus

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NCERT Solutions for Class 7th: Ch 18 Wastewater Story Science

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NCERT Solutions for Class 7th: Ch 18 Wastewater Story Science

Page No: 228

Exercises

1. Fill in the blanks:
(a) Cleaning of water is a process of removing ____________.
(b) Wastewater released by houses is called ____________.
(c) Dried ____________ is used as manure.
(d) Drains get blocked by ____________ and ____________.

Answer

(a) Cleaning of water is a process of removing pollutants.

(b) Wastewater released by houses is called sewage.

(c) Dried sludge is used as manure.

(d) Drains get blocked by cooking oil and fats.

2. What is sewage? Explain why it is harmful to discharge untreated sewage into rivers or seas.

Answer

Sewage is a liquid waste released by homes, industries, hospitals, offices and other users which also includes rainwater that has run down the street during a storm or heavy rain.
Sewage water contains harmful substances with it which washes off roads and rooftops. It is a complex
mixture containing suspended solids, organic and inorganic impurities, nutrients, saprotrophic and disease causing bacteria and other microbes. So, when sewage discharge untreated into rivers or seas, it will be dangerous for aquatic plants and animals.

3. Why should oils and fats be not released in the drain? Explain.

Answer

Oil and fats should not be released in the drain because they can harden and block the pipes. In an
open drain the fats block the soil pores reducing its effectiveness in filtering water.

4. Describe the steps involved in getting clarified water from wastewater.

Answer

Steps involved in getting clarified water from wastewater:
(i) Wastewater is shaken well and left to stand in the sun for some days.
(ii) After that, it is stir several times using an aerator or mechanical stirrer or a mixer which remove bad odour from water.
(iii) The stirred product is then filtered and sediment to remove insoluble waste.
(iv) Chlorination is used to kill the germs in filtered water. Now, the wastewater is fit for drinking purpose.

5. What is sludge? Explain how it is treated.

Answer

Solids waste like faeces settle at the bottom of tank during sedimentation is called sludge.
Sludge is removed with a scraper and then transferred to a separate tank where it is decomposed by the anaerobic bacteria to produce biogas.

6. Untreated human excreta is a health hazard. Explain.

Answer

Untreated human excreta is a health hazard as it can cause water pollution and soil pollution. Both the surface water and groundwater get polluted. Thus, it becomes the most common route for water borne diseases which include cholera, typhoid, polio, meningitis, hepatitis and dysentery. It is also a big threat to sanitation.

7. Name two chemicals used to disinfect water.

Answer

Two chemicals used to disinfect water are chlorine and ozone.

8. Explain the function of bar screens in a wastewater treatment plant.

Answer

Bar screen in a wastewater treatment plant is used to remove the large objects like rags, sticks, cans, plastic packets and napkins.

9. Explain the relationship between sanitation and disease.

Answer

Sanitation and disease are closely related as lack of sanitation leads to unhygienic condition and ultimately disease. Proper sanitation is necessary to avoid certain kinds of disease. Where there is sanitation, there is no disease and vice versa.

10. Outline your role as an active citizen in relation to sanitation.

Answer

Role as an active citizen in relation to sanitation:
(i) Contribute in maintaining sanitation at public places and should not scatter litter anywhere.
(ii) Throwing garbage or waste at its desired place.
(iii) Ensure that my surrounding and neighbourhood is clean and help the municipality in maintaining proper sanity.
(iv) Preventing clogging of drains by not releasing oil and fat and solid waste into the drain.
(v) Immediately inform the municipality about any sewage leakage and for the maintenance of dustbins.

11. Here is a crossword puzzle: Good luck!

Across
3. Liquid waste products
4. Solid waste extracted in sewage treatment
6. A word related to hygiene
8. Waste matter discharged from human body
Down
1. Used water
2. A pipe carrying sewage
5. Micro-organisms which causes cholera
7. A chemical to disinfect water

Answer

Across
3. Liquid waste products → Sewage
4. Solid waste extracted in sewage treatment → Sludge
6. A word related to hygiene → Sanitation
8. Waste matter discharged from human body → Excreta
Down
1. Used water → Wastewater
2. A pipe carrying sewage → Sewer
5. Micro-organisms which causes cholera → Bacteria
7. A chemical to disinfect water → Ozone

Page No: 229

12. Study the following statements about ozone:
     (a) It is essential for breathing of living organisms.
     (b) It is used to disinfect water.
     (c) It absorbs ultraviolet rays.
     (d) Its proportion in air is about 3%.
Which of these statements are correct?
    (i) (a), (b) and (c)
    (ii) (b) and (c)
    (iii) (a) and (d)
    (iv) All four
► (ii) (b) and (c)

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Our Environment- Class 7th NCERT Solutions Geography

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Our Environment Class 7th NCERT Solutions of Geography

Get answers of your textbook. If you have any problem in finding the correct answers of Geography Textbookthen you can find here. This page will help in finding those NCERT Solutions of books.Here you find complete chapter detailed questions and answers of Class 7 Geography. The answer of each chapter is provided in the list so that you can easily browse throughout different chapters and select needy one. Also you can read NCERT book online in this section.

Chapter 1- Enviroment

Chapter 2- Inside Our Earth

Chapter 3- Our Changing Earth

Chapter 4- Air

Chapter 5- Water

Chapter 6- Natural Vegetation and Wildlife

Chapter 7- Human Environment–Settlement, Transport and Communication

Chapter 8- Human Environment Interactions (The Tropical and the Subtropical Region)

Chapter 9- Life in the Temperate Grasslands

Chapter10- Life in the Deserts

NCERT Solutions for Class 7th: Ch 1 Environment Geography

August 30, 2015, 8:34 pm

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NCERT Solutions for Class 7th: Ch 1 Environment Geography Social Studies Our Enviroment

Page No: 6

Exercises

1. Answer the following questions.
(i) What is an ecosystem?

Answer

An ecosystem is a community in a given area where living organisms interact with each other and their immediate surroundings including non livings.

(ii) What do you mean by natural environment?

Answer

The environment which is created by nature comprises of land, water, air, plants and animals is known as natural environment.
 
(iii) Which are the major components of the environment?

Answer

The major components of the environment are:
(i) Natural environment
(ii) Human environment
(iii) Human made environment

(iv) Give four examples of human made environment.

Answer

Four examples of human made environment are:
(i) Vehicles
(ii) Buildings
(iii) Roads
(iv) Parks

(v) What is lithosphere?

Answer

Lithosphere is the solid crust or the hard top layer of the earth which is made up of rocks and minerals and covered by a thin layer of soil.

(vi) Which are the two major components of biotic environment?

Answer

Two major components of biotic environment are:
(i) Plants
(ii) Animals

(vii) What is biosphere?

Answer

Biosphere is a narrow zone of the earth where land, water and air interact with each other to support life. It consists of plant and animal kingdom together.

2. Tick the correct answer.
(i) Which is not a natural ecosystem?
    (a) Desert          (b) Aquarium          (c) Forest
✓(b) Aquarium

(ii) Which is not a component of human environment?
     (a) Land           (b) Religion            (c) Community
✓ (a) Land

(iii) Which is a human made environment?
      (a) Mountain           (b) Sea            (c) Road
✓(c) Road

(iv) Which is a threat to environment?
      (a) Growing plant             (b) Growing population            (c) Growing crops
✓ (b) Growing population

3. Match the following.

(i) Biosphere	(a) blanket of air which surrounds the earth
(ii) Atmosphere	(b) domain of water
(iii) Hydrosphere	(c) gravitational force of the earth
(iv) Environment	(d) our surroundings
-	(e) narrow zone where land water and air interact
-	(f) relation between the organisms and their surroundings

Answer

(i) Biosphere	(e) narrow zone where land water and air interact
(ii) Atmosphere	(a) blanket of air which surrounds the earth
(iii) Hydrosphere	(b) domain of water
(iv) Environment	(d) our surroundings

4. Give reasons.
(i) Man modifies his environment
(ii) Plants and animals depend on each other

Answer

(i) Man modifies his environment because the are capable of modifying it according to their need and live comfortable life. Humans learn new ways to use and change environment and invented many things. Industrial revolution enabled large scale production. Transportation became faster. Information revolution made communication easier and speedy across the world.

(ii) Plants and animals depend on each other for their sustainability. Animals consume plants as they are the only producers and also takes oxygen from them. Plants are dependent on animals as they give out carbon dioxide which is important for photosynthesis. Also, dead remains of animals provide nutrients to the plants.

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NCERT Solutions for Class 10th: पाठ - 16 भोर और बरखा (कविता) हिंदी

August 31, 2015, 1:04 am

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NCERT Solutions for Class 10th: पाठ - 16 भोर और बरखा (कविता) हिंदी वसंत भाग-II

- मीराबाई

पृष्ठ संख्या: 120

प्रश्न अभ्यास

कविता से

1. 'बंसीवारे ललना', 'मोरे प्यार', 'लाल जी', कहते हुए यशोदा किसे जगाने का प्रयास करती हैं और वे कौन-कौन सी बातें कहती हैं?

उत्तर

'बंसीवारे ललना', 'मोरे प्यार', 'लाल जी', कहते हुए यशोदा अपने पुत्र श्रीकृष्ण को जगाने का प्रयास करती हैं। वे कहतीं हैं कि रात बीत गयी है, सुबह हो गयी है, सभी के दरवाजें खुल चुके हैं। गोपियाँ दही से मक्खन निकाल रही हैं जिससे उनके कंगन बज रहे हैं, उन्हें सुनो। दरवाजे पर देव और मानव सभी तुम्हारी प्रतीक्षा में खड़े हैं, ग्वाल-बाल भी शोर मचा रहे हैं और जय-जयकार कर रहें हैं, उनके हाथ में माखन रोटी लेकर गाएँ चराने के लिए तुम्हारी प्रतीक्षा कर रहें हैं।

2. नीचे दी गई पंक्ति का आशय अपने शब्दों में लिखिए - 'माखन-रोटी हाथ मँह लीनी, गउवन के रखवारे।'

उत्तर

प्रस्तुत पंक्ति का आशय यह है कि गायों के रखवाले ग्वाल-बालों के हाथ में माखन और रोटी है।

3. पढ़े हुए पद के आधार पर ब्रज की भोर का वर्णन कीजिए।

उत्तर

पढ़े हुए पद के आधार पर ब्रज में भोर होते ही सभी घरों के किवाड़ खुल जाते हैं। गोपियाँ दही मथना शुरू कर देती हैं जिससे उनके कंगन खनकने की आवाज़ होती है। ग्वाल-बाल गायें चराने के लिए तैयार होने लगते हैं।

4. मीरा को सावन मनभावन क्यों लगने लगा?

उत्तर

मीरा को सावन मनभावन इसलिए लगा क्योंकि यह मौसम मीरा को श्रीकृष्ण के आने का अहसास कराता है।इसमें प्रकृति बड़ी सुहावनी होती है इसलिए मन में उमंग भर जाती है।

5. पाठ के आधार पर सावन की विशेषताएँ लिखिए।

उत्तर

सावन में प्रकृति मनोहारी दृश्य प्रस्तुत करती है। चारों तरफ बादल फैल जाते हैं, गरजते हैं और बिजली चमकती है। इस मौसम में मनभावन वर्षा होती है जिससे सभी प्रसन्न हो जाते हैं। गर्मी में कमी आती है और ठंडी हवाएँ बहती हैं।

पाठ से आगे

1. मीरा भक्तिकाल की प्रसिद्द कवयित्री थीं। इस काल के दूसरे कवियों के नामों की सूची बनाइए तथा उनकी एक-एक रचना का नाम लिखिए।

उत्तर

कवि - उनकी रचना
सूरदास - सूरसागर
रसखान - प्रेम वाटिका
परमानंद - परमानंदसागर
तुलसीदास - रामचरितमानस

2. सावन वर्षा ऋतु का महीना है, वर्षा ऋतु से संबंधित दो अन्य महीनों के नाम लिखिए।

उत्तर

आषाढ़ और भादो

पृष्ठ संख्या: 121

भाषा की बात

1. कृष्ण को 'गउवन के रखवारे' कहा गया है। जिसका अर्थ है गौओं का पालन करनेवाला। इसके लिए एक शब्द दें।

उत्तर

गोपाला

2. नीचे दो पंक्तियाँ दी गई हैं। इनमें से पहली पंक्ति में रेखांकित शब्द दो बार आए हैं, और दूसरी पंक्ति में भी दो बार। इन्हें पुनरुक्ति (पुन:उक्ति) कहते हैं। पहली पंक्ति में रेखांकित शब्द विशेषण हैं और दूसरी पंक्ति में संज्ञा।
'नन्हीं-नन्हींबूँदन मेहा बरसे'
'घर-घरखुले किंवारे'
इस प्रकार के दो-दो उदाहरण खोजकर वाक्य में प्रयोग कीजिए और देखिए कि विशेषण तथा संज्ञा की पुनरुक्ति के अर्थ में क्या अंतर हैं?
जैसे - मीठी-मीठी बातें, फूल-फूल महके।

उत्तर

विशेषण पुनरुक्ति
नए-नए - कल मैंने नए-नए कपडे पहने थे।
ठंडे-ठंडे - समोसे बड़े ठंडे-ठंडे हैं।

संज्ञा पुनरुक्ति
गली-गली - मैंने उसे गली-गली ढूंढा।
नगर-नगर - आजकल नगर-नगर छापेमारी चल रही है।

पाठ में वापिस जाएँ

NCERT Solutions for Class 7th: Ch 2 Inside Our Earth Geography

August 31, 2015, 8:35 pm

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NCERT Solutions for Class 7th: Ch 2 Inside Our Earth Geography Social Studies Our Enviroment

Page No: 10

Exercises

1. Answer the following questions.
(i) What are the three layers of the earth?

Answer

The three layers of the Earth are the crust, the mantle and the core.

(ii) What is a rock?

Answer

Any natural mass of mineral matter that makes up the earth’s crust is called a rock.

(iii) Name three types of rocks.

Answer

The three types of rocks are igneous rocks, sedimentary rocks and metamorphic rocks.

(iv) How are extrusive and intrusive rocks formed?

Answer

Extrusive rocks are formed by the molten lava which comes on the earth’s surface and rapidly cools down
to becomes solid.
When the molten magma cools down deep inside the earth’s crust then the solid rocks so formed are called intrusive rocks.

(v) What do you mean by a rock cycle?

Answer

When one type of rock changes to another type under certain conditions in a cyclic manner then this process of transformation of the rock from one to another is known as the rock cycle.

(vi) What are the uses of rocks?

Answer

Uses of rocks:
(i) Hard rocks are used in construction of buildings and roads.
(ii) Some rocks are shiny and precious therefore used for making jewellery.
(iii) Rocks are made up of different minerals and are very important to humankind.
(iv) Some are used as fuels. For example, coal, natural gas and petroleum.
(v) Soft rocks are used for making talcum powder, chalks etc.

(vii) What are metamorphic rocks?

Answer

The rocks which are formed due to conversion of igneous and sedimentary rocks under great heat and
pressure is called metamorphic rocks.

2. Tick the correct answer.
(i) The rock which is made up of molten magma is
    (a) Igneous            (b) Sedimentary         (c) Metamorphic
✓ (a) Igneous

(ii) The innermost layer of the earth is
     (a) Crust          (b) Core        (c) Mantle
✓ (b) Core

(iii) Gold, petroleum and coal are examples of
     (a) Rocks           (b) Minerals        (c) Fossils
✓ (b) Minerals

(iv) Rocks which contain fossils are
    (a) Sedimentary rocks
    (b) Metamorphic rocks
    (c) Igneous rocks
✓ (a) Sedimentary rocks

(v) The thinnest layer of the earth is
     (a) Crust           (b) Mantle         (c) Core
✓ (a) Crust

3. Match the following.

(i) Core	(a) Earth’s surface
(ii) Minerals	(b) Used for roads and buildings
(iii) Rocks	(c) Made of silicon and alumina
(iv) Clay	(d) Has definite chemical composition
(v) Sial	(e) Innermost layer
-	(f) Changes into slate
-	(g) Process of transformation of the rock

Answer

(i) Core	(e) Innermost layer
(ii) Minerals	(d) Has definite chemical composition
(iii) Rocks	(b) Used for roads and buildings
(iv) Clay	(f) Changes into slate
(v) Sial	(c) Made of silicon and alumina

Page No: 11

4. Give reasons.
(i) We cannot go to the centre of the earth.
(ii) Sedimentary rocks are formed from sediments.
(iii) Limestone is changed into marble.

Answer

(i) We cannot go to the centre of the earth because the it has very high temperature and pressure and lies 6000 km below the ocean floor. We will not able to survive there because there is no oxygen or favourable conditions.

(ii) Sedimentary rocks are formed from sediments because of extreme compression and hardening of the particles of sediment which are transported and deposited by wind, water etc.

(iii) Limestone is changed into marble because of extreme heat and pressure as it is a sedimentary rock.

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