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NCERT Solutions for Class 11th: Ch 9 Mechanical properties of Solids Physics

March 13, 2015, 8:39 am
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NCERT Solutions for Class 11th: Ch 9 Mechanical properties of Solids Physics Science

Page No: 242

Excercises

9.1. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10–5m2stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5m2under a given load. What is the ratio of the Young’s modulus of steel to that of copper?

Answer

Length of the steel wire, L1 = 4.7 m
Area of cross-section of the steel wire, A1 = 3.0 × 10–5m2
Length of the copper wire, L2 = 3.5 m
Area of cross-section of the copper wire, A2 = 4.0 × 10–5m2
Change in length = ΔL1= ΔL2 = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire:
Y1 = (F1 / A1) (L1 / ΔL1)
= (F / 3 X 10-5) (4.7 / ΔL)     ....(i)
Young’s modulus of the copper wire:
Y2 = (F2 / A2) (L2 / ΔL2)
= (F / 4 × 10-5) (3.5 / ΔL)     ....(ii)
Dividing (i) by (ii), we get:
Y1 / Y2  =  (4.7 × 4 × 10-5) / (3 × 10-5 × 3.5)
= 1.79 : 1
The ratio of Young’s modulus of steel to that of copper is 1.79 : 1.

9.2. Figure 9.11 shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strength for this material?
Answer

(a) It is clear from the given graph that for stress 150 × 106 N/m2, strain is 0.002.
∴Young’s modulus, Y = Stress / Strain
= 150 × 106 / 0.002  =  7.5 × 1010 Nm-2
Hence, Young’s modulus for the given material is 7.5 ×1010N/m2.

(b) The yield strength of a material is the maximum stress that the material can sustain without crossing the elastic limit.
It is clear from the given graph that the approximate yield strength of this material is 300 × 106 Nm/2 or 3 × 108 N/m2.

Page No: 243

9.3. The stress-strain graphs for materials A and B are shown in Fig. 9.12.


The graphs are drawn to the same scale.
(a) Which of the materials has the greater Young’s modulus?
(b) Which of the two is the stronger material?

Answer

(a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence, Young's modulus (=stress/strain) is greater for A than that of B.

(b) A is stronger than B. Strength of a material is measured by the amount of stress required to cause fracture, corresponding to the point of fracture.

9.4. Read the following two statements below carefully and state, with reasons, if it is true or false.(a) The Young’s modulus of rubber is greater than that of steel;
(b) The stretching of a coil is determined by its shear modulus.

Answer

(a) False, because for given stress there is more strain in rubber than steel and modulus of elasticity is inversely proportional to strain.

(b) True, because the stretching of coil simply changes its shape without any change in the length of the wire used in the coil due to which shear modulus of elsticity is involved.

9.5. Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in Fig. 9.13. The unloaded length of steel wire is 1.5 m and that of brass wire is 1.0 m. Compute the elongations of the steel and the brass wires.
Answer

Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m
Diameter of the wires, d = 0.25 m
Hence, the radius of the wires, r = d/2  =  0.125 cm
Length of the steel wire, L1 = 1.5 m
Length of the brass wire, L2 = 1.0 m
Total force exerted on the steel wire:
F1 = (4 + 6) g = 10 × 9.8 = 98 N
Young’s modulus for steel:
Y1 = (F1/A1) / (ΔL1 / L1)
Where,
ΔL1= Change in the length of the steel wire
A1 = Area of cross-section of the steel wire = πr12
Young’s modulus of steel, Y1 = 2.0 × 1011 Pa
∴ ΔL1 = F1  × L1 / (A1  × Y1)
= (98  × 1.5) /[π(0.125  × 10-2)2  × 2  × 1011]  =   1.49  × 10-4 m

Total force on the brass wire:
F2 = 6 × 9.8 = 58.8 N
Young’s modulus for brass:
Y2 = (F2/A2) / (ΔL2 / L2)
Where,
ΔL2= Change in the length of the brass wire
A1 = Area of cross-section of the brass wire = πr12
∴ ΔL2 = F2  × L2 / (A2  × Y2)
= (58.8 X 1) / [ (π  × (0.125  × 10-2)2  × (0.91  × 1011) ] = 1.3  × 10-4 m
Elongation of the steel wire = 1.49 × 10–4 m
Elongation of the brass wire = 1.3 × 10–4 m.

Page No: 244

9.6. The edge of an aluminium cube is 10 cm long. One face of the cube is firmly fixed to a vertical wall. A mass of 100 kg is then attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?

Answer

Edge of the aluminium cube, L = 10 cm = 0.1 m
The mass attached to the cube, m = 100 kg
Shear modulus (η) of aluminium = 25 GPa = 25 × 109 Pa
Shear modulus, η = Shear stress / Shear strain  =  (F/A) / (L/ΔL)
Where,
F = Applied force = mg = 100 × 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 × 0.1 = 0.01 m2
ΔL = Vertical deflection of the cube
∴ ΔL = FL / Aη
= 980 × 0.1 / [ 10-2 × (25 × 109) ]
= 3.92 × 10–7 m
The vertical deflection of this face of the cube is 3.92 ×10–7m.

9.7. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 cm and 60 cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column.

Answer

Mass of the big structure, M = 50,000 kg
Inner radius of the column, r = 30 cm = 0.3 m
Outer radius of the column, R = 60 cm = 0.6 m
Young’s modulus of steel, Y = 2 × 1011 Pa
Total force exerted, F= Mg = 50000 × 9.8 N
Stress = Force exerted on a single column = 50000 × 9.8 / 4  =  122500 N
Young’s modulus, Y = Stress / Strain
Strain = (F/A) / Y
Where,
Area, A = π (R2– r2) = π ((0.6)2– (0.3)2)
Strain = 122500 / [ π ((0.6)2– (0.3)2) × 2 × 1011 ]  =  7.22 × 10-7
Hence, the compressional strain of each column is 7.22 × 10–7.

9.8. A piece of copper having a rectangular cross-section of 15.2 mm × 19.1 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?

Answer

Length of the piece of copper, l = 19.1 mm = 19.1 × 10–3 m
Breadth of the piece of copper, b = 15.2 mm = 15.2 × 10–3 m
Area of the copper piece:
A = l × b
= 19.1 × 10–3× 15.2 × 10–3
= 2.9 × 10–4 m2
Tension force applied on the piece of copper, F = 44500 N
Modulus of elasticity of copper, η = 42 × 109 N/m2
Modulus of elasticity, η = Stress / Strain
= (F/A) / Strain
∴ Strain = F / Aη
= 44500 / (2.9 × 10-4 × 42 × 109)
= 3.65 × 10–3.

9.9. A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 108 N m–2, what is the maximum load the cable can support?

Answer

Radius of the steel cable, r = 1.5 cm = 0.015 m
Maximum allowable stress = 108 N m–2
Maximum stress = Maximum force / Area of cross-section
∴ Maximum force = Maximum stress × Area of cross-section
= 108× π (0.015)2
= 7.065 × 104 N
Hence, the cable can support the maximum load of 7.065 × 104 N.

9.10. A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each end are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.

Answer

The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.
The relation for Young’s modulus is given as:
Y = Stress / Strain
= (F/A) / Strain  =  (4F/πd2) / Strain     ....(i)
Where,
F = Tension force
A = Area of cross-section
d = Diameter of the wire
It can be inferred from equation (i) that Y ∝ (1/d2)
Young’s modulus for iron, Y1 = 190 × 109 Pa
Diameter of the iron wire = d1
Young’s modulus for copper, Y2 = 120 × 109 Pa
Diameter of the copper wire = d2
Therefore, the ratio of their diameters is given as:


9.11. A 14.5 kg mass, fastened to the end of a steel wire of unstretched length 1.0 m, is whirled in a vertical circle with an angular velocity of 2 rev/s at the bottom of the circle. The cross-sectional area of the wire is 0.065 cm2. Calculate the elongation of the wire when the mass is at the lowest point of its path.

Answer

Mass, m = 14.5 kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev/s = 2 × 2π rad/s = 12.56 rad/s
Cross-sectional area of the wire, a = 0.065 cm2 = 0.065 × 10-4 m2
Let Δl be the elongation of the wire when the mass is at the lowest point of its path.
When the mass is placed at the position of the vertical circle, the total force on the mass is:
F = mg + mlω2
= 14.5 × 9.8 + 14.5 × 1 × (12.56)2
= 2429.53 N
Young's modulus = Strss / Strain
Y = (F/A) / (∆l/l)
∴ ∆l = Fl / AY
Young’s modulus for steel = 2 × 1011 Pa
∆l = 2429.53 × 1 / (0.065 × 10-4 × 2 × 1011)   =   1.87 × 10-3 m
Hence, the elongation of the wire is 1.87 × 10–3 m.

9.12. Compute the bulk modulus of water from the following data: Initial volume = 100.0 litre, Pressure increase = 100.0 atm (1 atm = 1.013 × 105 Pa), Final volume = 100.5 litre. Compare the bulk modulus of water with that of air (at constant temperature). Explain in simple terms why the ratio is so large.

Answer

Initial volume, V1 = 100.0l = 100.0 × 10–3 m3
Final volume, V2 = 100.5 l = 100.5 ×10–3 m3
Increase in volume, ΔV = V2– V1 = 0.5 × 10–3 m3
Increase in pressure, Δp = 100.0 atm = 100 × 1.013 × 105 Pa
Bulk modulus = Δp/ (ΔV/V1)  =  Δp × V1 / ΔV
= 100 × 1.013 × 105 × 100 × 10-3 / (0.5 × 10-3)
= 2.026 × 109 Pa
Bulk modulus of air = 1 × 105 Pa
∴ Bulk modulus of water / Bulk modulus of air  =  2.026 × 109/ (1 × 105)  =  2.026 × 104
This ratio is very high because air is more compressible than water.

9.13. What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 1.03 × 103kg m–3?

Answer

Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 × 1.01 × 105 Pa
Density of water at the surface, ρ1 = 1.03 × 103 kg m–3
Let ρ2 be the density of water at the depth h.
Let V1 be the volume of water of mass m at the surface.
Let V2 be the volume of water of mass m at the depth h.
Let ΔV be the change in volume.
ΔV = V1 - V2
= m [ (1/ρ1) - (1/ρ2) ]
∴ Volumetric strain = ΔV / V1
= m [ (1/ρ1) - (1/ρ2) ] × (ρ1 / m)
ΔV / V1 = 1 - (ρ1/ρ2)     ......(i)
Bulk modulus, B = pV1 / ΔV
ΔV / V1 = p / B
Compressibility of water = (1/B) = 45.8 × 10-11 Pa-1
∴ ΔV / V1 = 80 × 1.013 × 105 × 45.8 × 10-11  =  3.71 × 10-3   ....(ii)
For equations (i) and (ii), we get:
1 - (ρ1/ρ2)   =   3.71 × 10-3
ρ2 = 1.03 × 103/ [ 1 - (3.71 × 10-3) ]
= 1.034 × 103 kg m-3
Therefore, the density of water at the given depth (h) is 1.034 × 103 kg m–3.

9.14. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.

Answer

Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 × 1.013 × 105 Pa
Bulk modulus of glass, B = 37 × 109 Nm–2
Bulk modulus, B = p / (∆V/V)
Where,
∆V/V = Fractional change in volume
∴ ∆V/V = p / B
= 10 × 1.013 × 105 / (37 × 109)
= 2.73 × 10-5
Hence, the fractional change in the volume of the glass slab is 2.73 × 10–5.

9.15. Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0 ×106Pa.

Answer

Length of an edge of the solid copper cube, l = 10 cm = 0.1 m
Hydraulic pressure, p = 7.0 × 106 Pa
Bulk modulus of copper, B = 140 × 109 Pa
Bulk modulus, B = p / (∆V/V)
Where,
∆V/V = Volumetric strain
ΔV = Change in volume
V = Original volume.
ΔV = pV / B
Original volume of the cube, V = l3
∴ ΔV = pl3 / B
= 7 × 106 × (0.1)3 / (140 × 109)
= 5 × 10-8 m3  =  5 × 10-2 cm-3
Therefore, the volume contraction of the solid copper cube is 5 × 10–2 cm–3.

9.16. How much should the pressure on a litre of water be changed to compress it by 0.10%?

Answer

Volume of water, V = 1 L
It is given that water is to be compressed by 0.10%.
∴ Fractional change, ∆V / V = 0.1 / (100 × 1)  =  10-3
Bulk modulus, B = ρ / (∆V/V)
ρ = B × (∆V/V)
Bulk modulus of water, B = 2.2 × 109 Nm-2
ρ = 2.2 × 109 × 10-3  =  2.2 × 106 Nm-2
Therefore, the pressure on water should be 2.2 ×106 Nm–2.

Additional Excercises

9.17. Anvils made of single crystals of diamond, with the shape as shown in Fig. 9.14, are used to investigate behaviour of materials under very high pressures. Flat faces at the narrow end of the anvil have a diameter of 0.50 mm, and the wide ends are subjected to a compressional force of 50,000 N. What is the pressure at the tip of the anvil?
Answer

Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 × 10–3m
Radius, r = d/2  =  0.25 × 10-3 m
Compressional force,F = 50000 N
Pressure at the tip of the anvil:
P = Force / Area  =  50000 /π(0.25 × 10-3)2
= 2.55  × 1011 Pa
Therefore, the pressure at the tip of the anvil is 2.55 × 1011 Pa.

Page No: 245

9.18. A rod of length 1.05 m having negligible mass is supported at its ends by two wires of steel (wire A) and aluminium (wire B) of equal lengths as shown in Fig. 9.15. The cross-sectional areas of wires Aand Bare 1.0 mm2and 2.0 mm2, respectively. At what point along the rod should a mass m be suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.
Answer

Cross-sectional area of wire A, a1 = 1.0 mm2 = 1.0 × 10–6 m2
Cross-sectional area of wire B, a2 = 2.0 mm2 = 2.0 × 10–6 m2
Young’s modulus for steel, Y1 = 2 × 1011 Nm–2
Young’s modulus for aluminium, Y2 = 7.0 ×1010 Nm–2

(a)Let a small mass m be suspended to the rod at a distance y from the end where wire A is attached.
Stress in the wire = Force / Area  =  F / a
If the two wires have equal stresses, then:
F1 / a1  =  F2 / a2
Where,
F1= Force exerted on the steel wire
F2= Force exerted on the aluminum wire
F1 / F2 = a1 / a2  =  1 / 2    ....(i)
The situation is shown in the following figure:

Taking torque about the point of suspension, we have:
F1y = F2 (1.05 - y)
F1 / F2 = (1.05 - y) / y   ......(ii)
Using equations (i) and (ii), we can write:
(1.05 - y) / y  = 1 / 2
2(1.05 - y)  =  y
y = 0.7 m
In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young's modulus = Stress / Strain
Strain = Stress / Young's modulus  =  (F/a) / Y
If the strain in the two wires is equal, then:
(F1/a1) /Y1  =  (F2/a2) /Y2
F1 / F2 = a1Y1 / a2Y2
a1 / a2 = 1/2
F1 / F2 = (1 / 2) (2 × 1011 / 7 × 1010)  =  10 / 7      .......(iii)
Taking torque about the point where mass m, is suspended at a distance y1 from the side where wire Aattached, we get:
F1y1 = F2 (1.05 – y1)
F1 / F2  =  (1.05 - y1) / y1   ....(iii)
Using equations (iii) and (iv), we get:
(1.05 - y1) / y1  =  10 / 7
7(1.05 - y1)  =  10y1
y1 = 0.432 m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire Ais attached.

9.19. A mild steel wire of length 1.0 m and cross-sectional area 0.50 × 10–2 cm2 is stretched, well within its elastic limit, horizontally between two pillars. A mass of 100 g is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

Answer

From the above figure,
Let x be the depression at the mid point i.e. CD = x.
In fig.,
AC= CB = l = 0.5 m ;
m = 100 g = 0.100 Kg
AD= BD = (l2 + x2)1/2
Increase in length, ∆l = AD + DB - AB = 2AD - AB
9.20. Two strips of metal are riveted together at their ends by four rivets, each of diameter 6.0 mm. What is the maximum tension that can be exerted by the riveted strip if the shearing stress on the rivet is not to exceed 6.9 × 107Pa? Assume that each rivet is to carry one quarter of the load.

Answer

Diameter of the metal strip, d = 6.0 mm = 6.0 × 10–3 m
Radius, r = d/2 = 3 × 10-3 m
Maximum shearing stress = 6.9 × 107 Pa
Maximum stress = Manimum load or force / Area
Maximum force = Maximum stress × Area
= 6.9 × 107× π × (r) 2
= 6.9 × 107× π × (3 ×10–3)2
= 1949.94 N
Each rivet carries one quarter of the load.
∴ Maximum tension on each rivet = 4 × 1949.94 = 7799.76 N.

9.21. The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of water. The water pressure at the bottom of the trench is about 1.1 × 108Pa. A steel ball of initial volume 0.32 m3is dropped into the ocean and falls to the bottom of the trench. What is the change in the volume of the ball when it reaches to the bottom?

Answer

Water pressure at the bottom, p = 1.1 × 108 Pa
Initial volume of the steel ball, V = 0.32 m3
Bulk modulus of steel, B = 1.6 × 1011 Nm–2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.
Bulk modulus, B = p / (∆V/V)
∆V  =  B / pV
= 1.1 × 108 × 0.32 / (1.6 × 1011 )  =  2.2 × 10-4 m3
Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 × 10–4 m3.


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NCERT Solutions for Class 11th: Ch 10 Mechanical Properties of Fluids Physics

March 14, 2015, 8:59 am
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NCERT Solutions for Class 11th: Ch 10 Mechanical Properties of Fluids Physics Science

Page No: 268

Excercises

10.1. Explain why
(a) The blood pressure in humans is greater at the feet than at the brain
(b) Atmospheric pressure at a height of about 6 km decreases to nearly half of its value at the sea level, though the height of the atmosphere is more than 100 km
(c) Hydrostatic pressure is a scalar quantity even though pressure is force divided by area.

Answer

(a) The height of the blood column in the human body is more at feet than at the brain. That is why, the blood exerts more pressure at the feet than at the brain.
(Pressure = h ρ g where h = height, ρ = density of liquid and g = accleration due to gravity)

(b) Density of air is the maximum near the sea level. Density of air decreases with increase in height from the surface. At a height of about 6 km, density decreases to nearly half of its value at the sea level. Atmospheric pressure is proportional to density. Hence, at a height of 6 km from the surface, it decreases to nearly half of its value at the sea level.

(c) When force is applied on a liquid, the pressure in the liquid is transmitted in all directions. Hence, hydrostatic pressure does not have a fixed direction and it is a scalar physical quantity.

10.2. Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape

Answer

(a) The angle between the tangent to the liquid surface at the point of contact and the surface inside the liquid is called the angle of contact (θ), as shown in the given figure.
Sla, Ssa, and Ssl are the respective interfacial tensions between the liquid-air, solid-air, and solid-liquid interfaces. At the line of contact, the surface forces between the three media must be in equilibrium, i.e.,
Cosθ = (Ssa - Sla) / Sla
The angle of contact θ , is obtuse if Ssa < Sla (as in the case of mercury on glass). This angle is acute if Ssl < Sla (as in the case of water on glass).

(b) Mercury molecules (which make an obtuse angle with glass) have a strong force of attraction between themselves and a weak force of attraction toward solids. Hence, they tend to form drops.
On the other hand, water molecules make acute angles with glass. They have a weak force of attraction between themselves and a strong force of attraction toward solids. Hence, they tend to spread out.

(c) Surface tension is the force acting per unit length at the interface between the plane of a liquid and any other surface. This force is independent of the area of the liquid surface. Hence, surface tension is also independent of the area of the liquid surface.

(d) Water with detergent dissolved in it has small angles of contact (θ). This is because for a small θ, there is a fast capillary rise of the detergent in the cloth. The capillary rise of a liquid is directly proportional to the cosine of the angle of contact (θ). If θ is small, then cosθ will be large and the rise of the detergent water in the cloth will be fast.

(e) A liquid tends to acquire the minimum surface area because of the presence of surface tension. The surface area of a sphere is the minimum for a given volume. Hence, under no external forces, liquid drops always take spherical shape.

10.3. Fill in the blanks using the word(s) from the list appended with each statement:

(a) Surface tension of liquids generally . . . with temperatures (increases / decreases)
► decreases

(b) Viscosity of gases. .. with temperature, whereas viscosity of liquids . . . with temperature (increases / decreases)
► increases ; decreases

(c) For solids with elastic modulus of rigidity, the shearing force is proportional to . . . , while for fluids it is proportional to . .. (shear strain / rate of shear strain)
► shear strain ; rate of shear strain

(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (conservation of mass / Bernoulli’s principle)
► conservation of mass ; Bernoulli’s principle

(e) For the model of a plane in a wind tunnel, turbulence occurs at a ... speed for turbulence for an actual plane (greater / smaller)
► greater

10.4. Explain why
(a) To keep a piece of paper horizontal, you should blow over, not under, it
(b) When we try to close a water tap with our fingers, fast jets of water gush through the openings between our fingers
(c) The size of the needle of a syringe controls flow rate better than the thumb pressure exerted by a doctor while administering an injection
(d) A fluid flowing out of a small hole in a vessel results in a backward thrust on the vessel
(e) A spinning cricket ball in air does not follow a parabolic trajectory.

Answer

(a)  When we blow over the paper, the velocity of air blow increases and hence pressure of air on it decreases (according to Beroulli's Theorem), where as pressure of air blow the paper is atmospheric. Hence, the paper stays horizontal.

(b) By ddoing so the area of outlet of water jet is reduced, so velocity of water increases according to equation of continuity, Area × Velocity = Constant.

(c) The small opening of a syringe needle controls the velocity of the blood flowing out. This is because of the equation of continuity. At the constriction point of the syringe system, the flow rate suddenly increases to a high value for a constant thumb pressure applied.

(d) When a fluid flows out from a small hole in a vessel, the vessel receives a backward thrust. A fluid flowing out from a small hole has a large velocity according to the equation of continuity:
Area × Velocity = Constant
According to the law of conservation of momentum, the vessel attains a backward velocity because there are no external forces acting on the system.

(e) A spinning cricket ball has two simultaneous motions – rotatory and linear. These two types of motion oppose the effect of each other. This decreases the velocity of air flowing below the ball. Hence, the pressure on the upper side of the ball becomes lesser than that on the lower side. An upward force acts upon the ball. Therefore, the ball takes a curved path. It does not follow a parabolic path.

10.5. A 50 kg girl wearing high heel shoes balances on a single heel. The heel is circular with a diameter 1.0 cm. What is the pressure exerted by the heel on the horizontal floor?

Answer

Mass of the girl, m = 50 kg
Diameter of the heel, d = 1 cm = 0.01 m
Radius of the heel, r = d/2 = 0.005 m
Area of the heel = πr2
= π (0.005)2
= 7.85 × 10–5 m2
Force exerted by the heel on the floor:
F = mg
= 50 × 9.8
= 490 N
Pressure exerted by the heel on the floor:
P = Force / Area
= 490 / (7.85 × 10-5)  =  6.24 × 106 Nm-2
Therefore, the pressure exerted by the heel on the horizontal floor is
6.24 × 106 Nm–2.

Page No: 269

10.6. Toricelli’s barometer used mercury. Pascal duplicated it using French wine of density 984 kg m–3. Determine the height of the wine column for normal atmospheric pressure.

Answer

Density of mercury, ρ1 = 13.6 × 103 kg/m3
Height of the mercury column, h1 = 0.76 m
Density of French wine, ρ2 = 984 kg/m3
Height of the French wine column = h2
Acceleration due to gravity, g = 9.8 m/s2
The pressure in both the columns is equal, i.e.,
Pressure in the mercury column = Pressure in the French wine column
ρ1h1g  =  ρ2h2g
h2 = ρ1h1 / ρ2
= 13.6 × 103 × 0.76 / 984  =  10.5 m
Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.

10.7. A vertical off-shore structure is built to withstand a maximum stress of 109 Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to be roughly 3 km, and ignore ocean currents.

Answer

The maximum allowable stress for the structure, P = 109 Pa
Depth of the ocean, d = 3 km = 3 × 103 m
Density of water, ρ = 103 kg/m3
Acceleration due to gravity, g = 9.8 m/s2
The pressure exerted because of the sea water at depth, d = ρdg
= 3 × 103 × 103 × 9.8
= 2.94 × 107 Pa
The maximum allowable stress for the structure (109 Pa) is greater than the pressure of the sea water (2.94 × 107 Pa). The pressure exerted by the ocean is less than the pressure that the structure can withstand. Hence, the structure is suitable for putting up on top of an oil well in the ocean.

10.8. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425 cm2. What maximum pressure would the smaller piston have to bear?

Answer

The maximum mass of a car that can be lifted, m = 3000 kg
Area of cross-section of the load-carrying piston, A = 425 cm2 = 425 × 10–4 m2
The maximum force exerted by the load, F = mg
= 3000 × 9.8
= 29400 N
The maximum pressure exerted on the load-carrying piston, P = F/A
 29400 / (425 × 10-4)
= 6.917 × 105 Pa
Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 × 105 Pa.

10.9. A U-tube contains water and methylated spirit separated by mercury. The mercury columns in the two arms are in level with 10.0 cm of water in one arm and 12.5 cm of spirit in the other. What is the specific gravity of spirit?

Answer

Height of the spirit column, h1 = 12.5 cm = 0.125 m
Height of the water column, h2 = 10 cm = 0.1 m
P0 = Atmospheric pressure
ρ1 = Density of spirit
ρ2 = Density of water
Pressure at point B = P0 + ρ1h1g
Pressure at point D = P0 + ρ2h2g
Pressure at points B and D is the same.
P0 + ρ1h1g  =  P0 + ρ2h2g
ρ1 / ρ2  =  h2 / h1
= 10 / 12.5  =  0.8
Therefore, the specific gravity of spirit is 0.8.

10.10. In problem 10.9, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6).

Answer

Height of the water column, h1 = 10 + 15 = 25 cm
Height of the spirit column, h2 = 12.5 + 15 = 27.5 cm
Density of water, ρ1 = 1 g cm–3
Density of spirit, ρ2 = 0.8 g cm–3
Density of mercury = 13.6 g cm–3
Let h be the difference between the levels of mercury in the two arms.
Pressure exerted by height h, of the mercury column:
= hρg
= h × 13.6g … (i)
Difference between the pressures exerted by water and spirit:
= ρ1h1g - ρ2h2g
= g(25 × 1 – 27.5 × 0.8)
= 3g … (ii)
Equating equations (i) and (ii), we get:
13.6 hg = 3g
h = 0.220588 ≈ 0.221 cm
Hence, the difference between the levels of mercury in the two arms is 0.221 cm.

10.11. Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.

Answer

Bernoulli’s equation cannot be used to describe the flow of water through a rapid in a river because of the turbulent flow of water. This principle can only be applied to a streamline flow.

10.12. Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.

Answer

No, it does not matter if one uses gauge pressure instead of absolute pressure while applying Bernoulli’s equation. The two points where Bernoulli’s equation is applied should have significantly different atmospheric pressures.

10.13. Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].

Answer

Length of the horizontal tube, l = 1.5 m
Radius of the tube, r = 1 cm = 0.01 m
Diameter of the tube, d = 2r = 0.02 m
Glycerine is flowing at a rate of 4.0 × 10–3 kg s–1.
M = 4.0 × 10–3 kg s–1
Density of glycerine, ρ = 1.3 × 103 kg m–3
Viscosity of glycerine, η = 0.83 Pa s
Volume of glycerine flowing per sec:
V = M / ρ
= 4 × 10-3 / (1.3 × 103)  =  3.08 × 10-6 m3s-1
According to Poiseville’s formula, we have the relation for the rate of flow:
V = πpr4 / 8ηl
Where, p is the pressure difference between the two ends of the tube
∴ p = V8ηl / πr4
= 3.08 × 10-6 × 8 × 0.83 × 1.5 / [ π × (0.01)4 ]
= 9.8 × 102 Pa
Reynolds’ number is given by the relation:
R = 4pV / πdη
= 4 × 1.3 × 103 × 3.08 × 10-6 / ( π × 0.02 × 0.83)
= 0.3
Reynolds’ number is about 0.3. Hence, the flow is laminar.

10.14. In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70 m s–1and 63 m s–1 respectively. What is the lift on the wing if its area is 2.5 m2? Take the density of air to be 1.3 kg m–3.

Answer

Speed of wind on the upper surface of the wing, V1 = 70 m/s
Speed of wind on the lower surface of the wing, V2 = 63 m/s
Area of the wing, A = 2.5 m2
Density of air, ρ = 1.3 kg m–3
According to Bernoulli’s theorem, we have the relation:
P1 + (1/2)ρV12 = P2 + (1/2)ρV22
P2 - P1 = (1/2) ρ (V12 - V22)
Where,
P1 = Pressure on the upper surface of the wing
P2 = Pressure on the lower surface of the wing
The pressure difference between the upper and lower surfaces of the wing provides lift to the aeroplane.
Lift on the wing = (P2 - P1) A
= (1/2) ρ (V12 - V22) A
= (1/2) × 1.3 × [ 702 - 632 ] × 2.5
= 1512.87 N  =  1.51 × 103 N
Therefore, the lift on the wing of the aeroplane is 1.51 × 103 N.

10.15. Figures 10.23 (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?
Answer

Fig. (a) is incorrect. Accoridng to equation of continuity, i.e., av = Constant, where area of cross-section of tube is less, the velcoity of liquid flow is more. So the velocity of liquid flow at a constriction of tube is more than the other portion of tube.
Accroding to Bernoulli's Theorem, P + 1/2ρv2 = Constant, where v is more, P is less and vice versa.

10.16. The cylindrical tube of a spray pump has a cross-section of 8.0 cm2 one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min–1, what is the speed of ejection of the liquid through the holes?

Answer

Area of cross-section of the spray pump, A1 = 8 cm2 = 8 × 10–4 m2
Number of holes, n = 40
Diameter of each hole, d = 1 mm = 1 × 10–3 m
Radius of each hole, r = d/2 = 0.5 × 10–3 m
Area of cross-section of each hole, a = πr2 = π (0.5 × 10–3)2 m2
Total area of 40 holes, A2 = n × a
= 40 × π (0.5 × 10–3)2 m2
= 31.41 × 10–6 m2
Speed of flow of liquid inside the tube, V1 = 1.5 m/min = 0.025 m/s
Speed of ejection of liquid through the holes = V2
According to the law of continuity, we have:
A1V1 = A2V2
V2 = A1V1 / A2
= 8 × 10-4 × 0.025 / (31.61 × 10-6)
= 0.633 m/s
Therefore, the speed of ejection of the liquid through the holes is 0.633 m/s.

10.17. A U-shaped wire is dipped in a soap solution, and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 × 10–2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?

Answer

The weight that the soap film supports, W = 1.5 × 10–2 N
Length of the slider, l = 30 cm = 0.3 m
A soap film has two free surfaces.
∴ Total length = 2l = 2 × 0.3 = 0.6 m
Surface tension, S = Force or Weight / 2l
= 1.5 × 10-2 / 0.6
= 2.5 × 10-2 N/m
Therefore, the surface tension of the film is 2.5 × 10–2 N m–1.

10.18. Figure 10.24 (a) shows a thin liquid film supporting a small weight = 4.5 × 10–2 N. What is the weight supported by a film of the same liquid at the same temperature in Fig. (b) and (c)? Explain your answer physically.
Answer

Take case (a):
The length of the liquid film supported by the weight, l = 40 cm = 0.4 cm
The weight supported by the film, W = 4.5 × 10–2 N
A liquid film has two free surfaces.
∴ Surface tension = W / 2l
= 4.5 × 10-2 / (2 × 0.4)  = 5.625 × 10-2 N/m
In all the three figures, the liquid is the same. Temperature is also the same for each case. Hence, the surface tension in figure (b) and figure (c) is the same as in figure (a), i.e., 5.625 × 10–2 N m–1.
Since the length of the film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10–2 N.

Page No: 270

10.19. What is the pressure inside the drop of mercury of radius 3.00 mm at room temperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10–1 N m–1. The atmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

Answer

Radius of the mercury drop, r = 3.00 mm = 3 × 10–3 m
Surface tension of mercury, S = 4.65 × 10–1 N m–1
Atmospheric pressure, P0 = 1.01 × 105 Pa
Total pressure inside the mercury drop
= Excess pressure inside mercury + Atmospheric pressure
= 2S / r + P0
= [ 2 × 4.65 × 10-1 / (3 × 10-3) ] + 1.01 × 105
= 1.0131 × 105
= 1.01 × 105 Pa
Excess pressure = 2S / r
= [ 2 × 4.65 × 10-1 / (3 × 10-3) ] = 310 Pa.

10.20. What is the excess pressure inside a bubble of soap solution of radius 5.00 mm, given that the surface tension of soap solution at the temperature (20 °C) is 2.50 × 10–2 N m–1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 × 105 Pa).

Answer

Excess pressure inside the soap bubble is 20 Pa;
Pressure inside the air bubble is 1.06 × 105 Pa
Soap bubble is of radius, r = 5.00 mm = 5 × 10–3 m
Surface tension of the soap solution, S = 2.50 × 10–2 Nm–1
Relative density of the soap solution = 1.20
∴ Density of the soap solution, ρ = 1.2 × 103 kg/m3
Air bubble formed at a depth, h = 40 cm = 0.4 m
Radius of the air bubble, r = 5 mm = 5 × 10–3 m
1 atmospheric pressure = 1.01 × 105 Pa
Acceleration due to gravity, g = 9.8 m/s2
Hence, the excess pressure inside the soap bubble is given by the relation:
P = 4S / r
= 4 × 2.5 × 10-2 / 5 × 10-3
= 20 Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation:
P' = 2S / r
= 2 × 2.5 × 10-2 / (5 × 10-3)
= 10 Pa
Therefore, the excess pressure inside the air bubble is 10 Pa.
At a depth of 0.4 m, the total pressure inside the air bubble
= Atmospheric pressure + hρg + P’
= 1.01 × 105 + 0.4 × 1.2 × 103 × 9.8 + 10
= 1.06 × 105 Pa
Therefore, the pressure inside the air bubble is 1.06 × 105 Pa.

Additional Excercises

10.21. A tank with a square base of area 1.0 m2 is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm2. The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.

Answer

Base area of the given tank, A = 1.0 m2
Area of the hinged door, a = 20 cm2 = 20 × 10–4 m2
Density of water, ρ1 = 103 kg/m3
Density of acid, ρ2 = 1.7 × 103 kg/m3
Height of the water column, h1 = 4 m
Height of the acid column, h2 = 4 m
Acceleration due to gravity, g = 9.8
Pressure due to water is given as:
P1 = h1ρ1g
= 4 × 103 × 9.8  =  3.92 × 104 Pa
Pressure due to acid is given as:
P2 = h2ρ2g
= 4 × 1.7 × 103 × 9.8  =  6.664 × 104 Pa
Pressure difference between the water and acid columns:
ΔP = P2 - P1
= 6.664 × 104 - 3.92 × 104
= 2.744 × 104 Pa
Hence, the force exerted on the door = ΔP × a
= 2.744 × 104 × 20 × 10–4
= 54.88 N
Therefore, the force necessary to keep the door closed is 54.88 N.

10.22. A manometer reads the pressure of a gas in an enclosure as shown in Fig. 10.25 (a) When a pump removes some of the gas, the manometer reads as in Fig. 10.25 (b) The liquid used in the manometers is mercury and the atmospheric pressure is 76 cm of mercury.
(a) Give the absolute and gauge pressure of the gas in the enclosure for cases (a) and (b), in units of cm of mercury.
(b) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury) are poured into the right limb of the manometer? (Ignore the small change in the volume of the gas).
Answer

(a) For figure (a)
Atmospheric pressure, P0 = 76 cm of Hg
Difference between the levels of mercury in the two limbs gives gauge pressure
Hence, gauge pressure is 20 cm of Hg.
Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 + 20 = 96 cm of Hg

For figure (b)
Difference between the levels of mercury in the two limbs = –18 cm
Hence, gauge pressure is –18 cm of Hg.
Absolute pressure = Atmospheric pressure + Gauge pressure
= 76 cm – 18 cm = 58 cm

(b) 13.6 cm of water is poured into the right limb of figure (b).
Relative density of mercury = 13.6
Hence, a column of 13.6 cm of water is equivalent to 1 cm of mercury.
Let h be the difference between the levels of mercury in the two limbs.
The pressure in the right limb is given as:
PR = Atmospheric pressure + 1 cm of Hg
= 76 + 1 = 77 cm of Hg … (i)
The mercury column will rise in the left limb.
Hence, pressure in the left limb, PL = 58 + h               .....(ii)
Equating equations (i) and (ii), we get:
77 = 58 + h
∴h = 19 cm
Hence, the difference between the levels of mercury in the two limbs will be 19 cm.

10.23. Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?

Answer

Two vessels having the same base area have identical force and equal pressure acting on their common base area. Since the shapes of the two vessels are different, the force exerted on the sides of the vessels has non-zero vertical components. When these vertical components are added, the total force on one vessel comes out to be greater than that on the other vessel. Hence, when these vessels are filled with water to the same height, they give different readings on a weighing scale.

Page No: 271

10.24. During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000 Pa. At what height must the blood container be placed so that blood may just enter the vein? [Use the density of whole blood from Table 10.1].

Answer

Gauge pressure, P = 2000 Pa
Density of whole blood, ρ = 1.06 × 103 kg m–3
Acceleration due to gravity, g = 9.8 m/s2
Height of the blood container = h
Pressure of the blood container, P = hρg
∴ h = P / ρg
= 2000 / (1.06 × 103 × 9.8)
= 0.1925 m
The blood may enter the vein if the blood container is kept at a height greater than 0.1925 m, i.e., about 0.2 m.

10.25. In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy. (a) What is the largest average velocity of blood flow in an artery of diameter 2 × 10–3 m if the flow must remain laminar? (b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.

Answer

(a) If dissipative forces are present, hen some forces in liquid flow due to pressure difference is spent against dissipative forces. Due to which the pressure drop becomes large.

(b) The dissipative forces become more important with increasing flow velocity, because of turbulence.

10.26. (a) What is the largest average velocity of blood flow in an artery of radius 2 × 10–3 m if the flow must remain laminar? (b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10–3 Pa s).

Answer

(a) Radius of the artery, r = 2 × 10–3 m
Diameter of the artery, d = 2 × 2 × 10–3 m = 4 × 10– 3 m
Viscosity of blood, η = 2.084 X 10-3 Pa s
Density of blood, ρ = 1.06 × 103 kg/m3
Reynolds’ number for laminar flow, NR = 2000
The largest average velocity of blood is given by the relation:
Varg = NRη / ρd
= 2000 × 2.084 × 10-3 / (1.06 × 103 × 4 × 10-3)
= 0.983 m/s
Therefore, the largest average velocity of blood is 0.983 m/s.

(b) Flow rate is given by the relation:
R = π r2Vavg
= 3.14 × (2 × 10-3)2 × 0.983
= 1.235 × 10-5 m3s-1
Therefore, the corresponding flow rate is 1.235 × 10-5 m3s-1.

10.27. A plane is in level flight at constant speed and each of its two wings has an area of 25 m2. If the speed of the air is 180 km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1 kg m–3).

Answer

The area of the wings of the plane, A = 2 × 25 = 50 m2
Speed of air over the lower wing, V1 = 180 km/h = 50 m/s
Speed of air over the upper wing, V2 = 234 km/h = 65 m/s
Density of air, ρ = 1 kg m–3
Pressure of air over the lower wing = P1
Pressure of air over the upper wing= P2
The upward force on the plane can be obtained using Bernoulli’s equation as:
P1 + (1/2) ρ V12  =  P2 + (1/2) ρ V22
P1 - P2 = (1/2) ρ ( V22 - V12)   .....(i)
The upward force (F) on the plane can be calculated as:
(P1 - P2)A  =  (1/2) ρ ( V22 - V12) A
= (1/2) × 1 × (652 - 502) × 50
= 43125 N
Using Newton’s force equation, we can obtain the mass (m) of the plane as:
F = mg
∴ m = 43125 / 9.8 = 4400.51 kg
∼ 4400 kg
Hence, the mass of the plane is about 4400 kg.

10.28. In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 × 10–5 m and density 1.2 × 103 kg m–3? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10–5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

Answer

Terminal speed = 5.8 cm/s
Viscous force = 3.9 × 10–10 N
Radius of the given uncharged drop, r = 2.0 × 10–5 m
Density of the uncharged drop, ρ = 1.2 × 103 kg m–3
Viscosity of air, η = 1.8 × 10-5 Pa s
Density of air (ρ0) can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, g = 9.8 m/s2
Terminal velocity (v) is given by the relation:
v = 2r2 × (ρ - ρ0) g / 9η
= 2 × (2 × 10-5)2 (1.2 × 103 - 0 ) × 9.8 / (9 × 1.8 × 10-5)
= 5.8 × 10-2 m/s
= 5.8 cm s-1
Hence, the terminal speed of the drop is 5.8 cm s–1.
The viscous force on the drop is given by:
F = 6πηrv
∴ F = 6 × 3.14 × 1.8 × 10-5 × 2 × 10-5 × 5.8 × 10-2
= 3.9 × 10-10 N
Hence, the viscous force on the drop is 3.9 × 10–10 N.

10.29. Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 N m–1. Density of mercury = 13.6 × 103 kg m–3.

Answer

Angle of contact between mercury and soda lime glass, θ = 140°
Radius of the narrow tube, r = 1 mm = 1 × 10–3 m
Surface tension of mercury at the given temperature, s = 0.465 N m–1
Density of mercury, ρ =13.6 × 103 kg/m3
Dip in the height of mercury = h
Acceleration due to gravity, g = 9.8 m/s2
Surface tension is related with the angle of contact and the dip in the height as:
s = hρgr / 2 Cosθ
∴ h = 2s Cosθ / ρgr
= 2 × 0.465 × Cos1400 / (1 × 10-3 × 13.6 × 103 × 9.8)
= -0.00534 m
= -5.34 m
Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm.

10.30. Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form a U-tube open at both ends. If the U-tube contains water, what is the difference in its levels in the two limbs of the tube? Surface tension of water at the temperature of the experiment is 7.3 × 10–2 N m–1. Take the angle of contact to be zero and density of water to be 1.0 × 103 kg m–3 (g = 9.8 m s–2).

Answer

Diameter of the first bore, d1 = 3.0 mm = 3 × 10–3 m
Hence, the radius of the first bore, r1 = d1/ 2  =  1.5 × 10-3 m
Diameter of the first bore, d2 = 6.0 mm = 6 × 10–3 mm
Hence, the radius of the first bore, r2 = d2/ 2  =  3 × 10-3 m
Surface tension of water, s = 7.3 × 10–2 N m–1
Angle of contact between the bore surface and water, θ= 0
Density of water, ρ =1.0 × 103 kg/m–3
Acceleration due to gravity, g = 9.8 m/s2
Let h1 and h2 be the heights to which water rises in the first and second tubes respectively. These heights are given by the relations:
h1 = 2s Cosθ / r1ρg      .....(i)
h2 = 2s Cosθ / r2ρg      .....(ii)
The difference between the levels of water in the two limbs of the tube can be calculated as:
= 4.966 × 10-3 m
= 4.97 mm
Hence, the difference between levels of water in the two bores is 4.97 mm.

10.31. (a) It is known that density ρ of air decreases with height y as ρ0e-y/y0
Where ρ0 = 1.25 kg m-3
is the density at sea level, and y0 is a constant. This density variation is called the law of atmospheres. Obtain this law assuming that the temperature of atmosphere remains a constant (isothermal conditions). Also assume that the value of gremains constant.
(b) A large He balloon of volume 1425 m3 is used to lift a payload of 400 kg. Assume that the balloon maintains constant radius as it rises. How high does it rise?
[Take y0= 8000 m and ρHe = 0.18 kg m-3 ]


Answer

(a) Volume of the balloon, V = 1425 m3
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s2
y0 = 8000 m
ρHe = 0.18 kg m-3
ρ0 = 1.25 kg m-3
Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as:
ρ = ρ0e-y/y0
ρ / ρ0 = e-y/y0     ....(i)

This density variation is called the law of atmospherics.
It can be inferred from equation (i) that the rate of decrease of density with height is directly proportional to ρ, i.e.,
-(dρ / dy) ∝ ρ
(dρ / dy) = -kρ
(dρ / ρ) = -k dy
Where, k is the constant of proportionality
Height changes from 0 to y, while density changes from ρ0 to ρ
Integrating the sides between these limits, we get:
Comparing equations (i) and (ii), we get:
y0 = 1/k
k = 1/y0   ....(iii)
From equations (i) and (iii), we get
ρ = ρ0e-y/y0
(b) Density ρ = Mass / Volume
= (Mass of the payload + Mass of helium) / Volume
( m + VρHe) / V
= (400 + 1425 × 0.18) / 1425
= 0.46 kg m-3
From equations (ii) and (iii), we can obtain y as:
ρ = ρ0e-y/y0
loge(ρ / ρ0)  =  -y / y0
∴ y = - 8000 × loge(0.46 / 1.25)
= -8000 × (-1)
= 8000 m  =  8 km
Hence, the balloon will rise to a height of 8 km.


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NCERT Solutions for Class 11th: Ch 11 Thermal Properties of Matter Physics

March 14, 2015, 9:54 pm
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NCERT Solutions for Class 11th: Ch 11 Thermal Properties of Matter Physics Science

Page No: 294

Excercises

11.1. The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Answer

Kelvin and Celsius
scales are related as:
TC = TK– 273.15 … (i)
Celsius and Fahrenheit scales are related as:
TF = (9/5)TC + 32   ....(ii)

For neon:
TK = 24.57 K
∴ TC = 24.57 – 273.15 = –248.58°C
TF = (9/5)TC + 32
= (9/5) × (-248.58) +32
= 415.440 F

For carbon dioxide:
TK = 216.55 K
∴ TC= 216.55 – 273.15 = –56.60°C
TF = (9/5)TC + 32
= (9/5) × (-56.60) +32
= -69.880 C

11.2. Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between TA and TB?

Answer

Triple point of water on absolute scaleA, T1 = 200 A
Triple point of water on absolute scale B, T2 = 350 B
Triple point of water on Kelvin scale, TK = 273.15 K
The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A.
T1 = TK
200 A = 273.15 K
∴ A = 273.15/200
The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.
T2 = TK
350 B = 273.15
∴ B = 273.15/350
TA is triple point of water on scale A.
TB is triple point of water on scale B.
∴ 273.15/200 × TA  = 273.15/350 × TB
Therefore, the ratio TA: TB is given as 4 : 7.

11.3. The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = Ro[1 + α (T – To)]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω?

Answer

It is given that:
R = R0[1 + α (T – T0)] … (i)
where,
R0and T0 are the initial resistance and temperature respectively

Rand T are the final resistance and temperature respectively
α is a constant
At the triple point of water, T0 = 273.15 K
Resistance of lead, R0 = 101.6 Ω
At normal melting point of lead, T = 600.5 K
Resistance of lead, R = 165.5 Ω

Substituting these values in equation (i), we get:
R = Ro[1 + α (T – To)]
165.5 = 101.6 [ 1 + α(600.5 - 273.15) ]
1.629 = 1 + α (327.35)
∴ α = 0.629 / 327.35  =  1.92 × 10-3 K-1

For resistance, R1 = 123.4 Ω
R1= R0[1 + α (T – T0)]
where,
T is the temperature when the resistance of lead is 123.4 Ω
123.4 = 101.6 [ 1 + 1.92 × 10-3( T - 273.15) ]
Solving for T, we get
T = 384.61 K.

11.4. Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?
(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the number 0 °C and 100 °C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?
(c) The absolute temperature (Kelvin scale) T is related to the temperature tcon the Celsius scale by
tc= T – 273.15
Why do we have 273.15 in this relation, and not 273.16?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Answer

(a) The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.

(b) The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.

(c) The temperature 273.16 K is the triple point of water. It is not the melting point of ice. The temperature 0°C on Celsius scale is the melting point of ice. Its corresponding value on Kelvin scale is 273.15 K.
Hence, absolute temperature (Kelvin scale) T, is related to temperature tc, on Celsius scale as:
tc = T– 273.15

(d) Let TF be the temperature on Fahrenheit scale and TK be the temperature on absolute scale. Both the temperatures can be related as:
(TF - 32) / 180  =  (TK - 273.15) / 100    ....(i)
Let TF1 be the temperature on Fahrenheit scale and TK1 be the temperature on absolute scale. Both the temperatures can be related as:
(TF1 - 32) / 180  =  (TK1 - 273.15) / 100    ....(ii)
It is given that:
TK1– TK = 1 K
Subtracting equation (i) from equation (ii), we get:
(TF1 - TF) / 180  =  (TK1 - TK) / 100  =  1 / 100
TF1 - TF = (1 ×180) / 100  =  9/5
Triple point of water = 273.16 K
∴ Triple point of water on absolute scale = 273.16 × (9/5)  =  491.69

Page No: 295

11.5. Two ideal gas thermometers Aand Buse oxygen and hydrogen respectively. The following observations are made:

TemperaturePressure thermometer APressure thermometer B
Triple-point of water1.250 × 105 Pa0.200 × 105 Pa
Normal melting point of sulphur1.797 × 105 Pa0.287 × 105 Pa

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers Aand B?
(b) What do you think is the reason behind the slight difference in answers of thermometers Aand B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Answer

(a) Triple point of water, T = 273.16 K.
At this temperature, pressure in thermometer A, PA= 1.250 × 105 Pa
Let T1 be the normal melting point of sulphur.
At this temperature, pressure in thermometer A, P1= 1.797 × 105 Pa
According to Charles’ law, we have the relation:
PA / T  =  P1 / T1
∴ T1 = (P1T) /PA  =  (1.797 × 10 × 273.16) / (1.250 × 105)
= 392.69 K
Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.
At triple point 273.16 K, the pressure in thermometer B, PB= 0.200 × 105 Pa
At temperature T1, the pressure in thermometer B, P2 = 0.287 × 105 Pa
According to Charles’ law, we can write the relation:
PB / T  =  P1 / T1
(0.200 × 105) / 273.16  =  (0.287 × 105) / T1
∴ T1 = [ (0.287 × 105) / (0.200 × 105) ] × 273.16  =  391.98 K
Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.

(b) The oxygen and hydrogen gas present in thermometers A and B respectively are not perfect ideal gases. Hence, there is a slight difference between the readings of thermometers A and B.
To reduce the discrepancy between the two readings, the experiment should be carried under low pressure conditions. At low pressure, these gases behave as perfect ideal gases.

11.6. A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0 °C? Coefficient of linear expansion of steel = 1.20 × 10–5K–1.

Answer

Length of the steel tape at temperature T = 27°C, l = 1 m = 100 cm
At temperature T1= 45°C, the length of the steel rod, l1 = 63 cm
Coefficient of linear expansion of steel, α= 1.20 × 10–5 K–1
Let l2be the actual length of the steel rod and l' be the length of the steel tape at 45°C.
l' = l + αl(T1 - T)
∴ l' = 100 + 1.20 × 10-5 × 100(45 - 27)
= 100.0216 cm
Hence, the actual length of the steel rod measured by the steel tape at 45°C can be calculated as:
l2 = (100.0216 / 100) × 63  =  63.0136 cm
Therefore, the actual length of the rod at 45.0°C is 63.0136 cm. Its length at 27.0°C is 63.0 cm.

11.7. A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: αsteel = 1.20 × 10–5 K–1.

Answer

The given temperature, T = 27°C can be written in Kelvin as:
27 + 273 = 300 K
Outer diameter of the steel shaft at T, d1 = 8.70 cm
Diameter of the central hole in the wheel at T, d2 = 8.69 cm
Coefficient of linear expansion of steel, αsteel = 1.20 × 10–5 K–1
After the shaft is cooled using ‘dry ice’, its temperature becomes T1.
The wheel will slip on the shaft, if the change in diameter, Δd = 8.69 – 8.70
= – 0.01 cm
Temperature T1, can be calculated from the relation:
Δd = d1αsteel (T1– T)
0.01 = 8.70 × 1.20 × 10–5 (T1– 300)
(T1– 300) = 95.78
∴ T1= 204.21 K
= 204.21 – 273.16
= –68.95°C
Therefore, the wheel will slip on the shaft when the temperature of the shaft is –69°C.

11.8. A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in the diameter of the hole when the sheet is heated to 227 °C? Coefficient of linear expansion of copper = 1.70 × 10–5K–1.

Answer

Initial temperature, T1 = 27.0°C
Diameter of the hole at T1, d1 = 4.24 cm
Final temperature, T2 = 227°C
Diameter of the hole at T2 = d2
Co-efficient of linear expansion of copper, αCu= 1.70 × 10–5 K–1
For co-efficient of superficial expansion β,and change in temperature ΔT, we have the relation:
Change in area (∆)  /  Original area (A)  =  β∆T
[ (πd22/ 4) - (πd12 / 4) ] / (πd11 / 4)  =  ∆A / A
∴ ∆A / A = (d22 - d12) / d12
But β = 2α
∴ (d22 - d12) /d12 = 2α∆T
(d22 / d12) - 1  =  2α(T2 - T1)
d22 / 4.242 = 2 × 1.7 × 10-5 (227 - 27) +1
d22 = 17.98 × 1.0068  =  18.1
∴ d2 = 4.2544 cm
Change in diameter = d2– d1 = 4.2544 – 4.24 = 0.0144 cm
Hence, the diameter increases by 1.44 × 10–2 cm.

11.9. A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm? Co-efficient of linear expansion of brass = 2.0 × 10–5K–1; Young’s modulus of brass = 0.91 × 1011Pa.

Answer

Initial temperature, T1 = 27°C
Length of the brass wire at T1, l = 1.8 m
Final temperature, T2 = –39°C
Diameter of the wire, d = 2.0 mm = 2 × 10–3 m
Tension developed in the wire = F
Coefficient of linear expansion of brass, α = 2.0 × 10–5 K–1
Young’s modulus of brass, Y = 0.91 × 1011 Pa
Young’s modulus is given by the relation:
γ = Stress / Strain  =  (F/A) / (∆L/L)
∆L = F X L/ (A X Y)      ......(i)

Where,
F = Tension developed in the wire
A = Area of cross-section of the wire.
ΔL = Change in the length, given by the relation:
ΔL = αL(T2– T1) … (ii)
Equating equations (i) and (ii), we get:
αL(T2 - T1) = FL / [ π(d/2)2X Y ]
F = α(T2 - T1)πY(d/2)2
F = 2 × 10-5 × (-39-27) × 3.14 × 0.91 × 1011 × (2 × 10-3 / 2 )2
= -3.8 × 102 N
(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is 3.8 ×102 N.

11.10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1).

Answer

Initial temperature, T1 = 40°C
Final temperature, T2 = 250°C
Change in temperature, ΔT = T2– T1 = 210°C
Length of the brass rod at T1, l1 = 50 cm
Diameter of the brass rod at T1, d1 = 3.0 mm
Length of the steel rod at T2, l2 = 50 cm
Diameter of the steel rod at T2, d2 = 3.0 mm
Coefficient of linear expansion of brass, α1= 2.0 × 10–5K–1
Coefficient of linear expansion of steel, α2= 1.2 × 10–5K–1
For the expansion in the brass rod, we have:
Change in length (∆l1) / Original length (l1)  =  α1ΔT
∴ ∆l1 = 50 × (2.1 × 10-5) × 210
= 0.2205 cm
For the expansion in the steel rod, we have:
Change in length (∆l2) / Original length (l2)  =  α2ΔT
∴ ∆l1 = 50 × (1.2 × 10-5) × 210
= 0.126 cm
Total change in the lengths of brass and steel,
Δl= Δl1 + Δl2
= 0.2205 + 0.126
= 0.346 cm
Total change in the length of the combined rod = 0.346 cm
Since the rod expands freely from both ends, no thermal stress is developed at the junction.

11.11. The coefficient of volume expansion of glycerin is 49 × 10–5K–1. What is the fractional change in its density for a 30 °C rise in temperature?

Answer

Coefficient of volume expansion of glycerin, αV = 49 × 10–5 K–1
Rise in temperature, ΔT = 30°C
Fractional change in its volume = ΔV/V
This change is related with the change in temperature as:
ΔV/V = αVΔT
VT2 - VT1  =  VT1αVΔT
(m /ρT2)- (m /ρT1) = (m /ρT1)αVΔT
Where,
m = Mass of glycerine
ρT1 = Initial density at T1
ρT2 = Initial density at T2
(ρT1 - ρT2 ) /ρT2  =  Fractional change in density
∴ Fractional change in the density of glycerin = 49 ×10–5 × 30 = 1.47 × 10–2.

11.12. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings. Specific heat of aluminium = 0.91 J g–1K–1.

Answer

Power of the drilling machine, P= 10 kW = 10 × 103 W
Mass of the aluminum block, m= 8.0 kg = 8 × 103g
Time for which the machine is used, t= 2.5 min = 2.5 × 60 = 150 s
Specific heat of aluminium, c= 0.91 J g–1K–1
Rise in the temperature of the block after drilling = δT
Total energy of the drilling machine = Pt
= 10 × 103 × 150
= 1.5 × 106J
It is given that only 50% of the power is useful.
Useful energy, ∆Q = (50/100) × 1.5 × 106  =  7.5 × 105 J
But ∆Q = mc∆T
∴ ∆T = ∆Q / mc
= (7.5 × 105) / (8 × 103 × 0.91)
= 103o C
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

11.13. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 J g–1 K–1; heat of fusion of water = 335 J g–1).

Answer

Mass of the copper block, m = 2.5 kg = 2500 g
Rise in the temperature of the copper block, Δθ = 500°C
Specific heat of copper, C = 0.39 J g–1 °C–1
Heat of fusion of water, L = 335 J g–1
The maximum heat the copper block can lose, Q = mCΔθ
= 2500 × 0.39 × 500
= 487500 J
Let m1 g be the amount of ice that melts when the copper block is placed on the ice block.
The heat gained by the melted ice, Q = m1L
∴ m1 = Q / L  =  487500 / 335  =  1455.22 g
Hence, the maximum amount of ice that can melt is 1.45 kg.

11.14. In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 cm3of water at 27 °C. The final temperature is 40 °C. Compute the specific heat of the metal. If heat losses to the surroundings are not negligible, is your answer greater or smaller than the actual value for specific heat of the metal?

Answer

Mass of the metal, m = 0.20 kg = 200 g
Initial temperature of the metal, T1 = 150°C
Final temperature of the metal, T2 = 40°C
Calorimeter has water equivalent of mass, m’ = 0.025 kg = 25 g
Volume of water, V = 150 cm3
Mass (M) of water at temperature T = 27°C:
150 × 1 = 150 g
Fall in the temperature of the metal:
ΔT = T1– T2 = 150 – 40 = 110°C
Specific heat of water, Cw= 4.186 J/g/°K
Specific heat of the metal = C
Heat lost by the metal, θ = mCΔT… (i)
Rise in the temperature of the water and calorimeter system:
ΔT′’ = 40 – 27 = 13°C
Heat gained by the water and calorimeter system:
Δθ′′ = m1CwΔT’
= (M + m′) CwΔT’… (ii)
Heat lost by the metal = Heat gained by the water and colorimeter system
mCΔT = (M + m’) CwΔT’
200 × C × 110 = (150 + 25) × 4.186 × 13
∴ C = (175 × 4.186 × 13) / (110 × 200)  =  0.43 Jg-1K-1
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

Page No: 296

11.15. Given below are observations on molar specific heats at room temperature of some common gases.
Observations
The measured molar specific heats of these gases are markedly different from those for monatomic gases. Typically, molar specific heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you infer from the somewhat larger (than the rest) value for chlorine?

Answer

The gases listed in the given table are diatomic. Besides the translational degree of freedom, they have other degrees of freedom (modes of motion).
Heat must be supplied to increase the temperature of these gases. This increases the average energy of all the modes of motion. Hence, the molar specific heat of diatomic gases is more than that of monatomic gases.
If only rotational mode of motion is considered, then the molar specific heat of a diatomic gas = (5/2)R
= (5/2) × 1.98 = 4.95 cal mol-1 K-1
With the exception of chlorine, all the observations in the given table agree with (5/2)R.
This is because at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.

11.16. Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in equilibrium?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO2?
(c) What are the critical temperature and pressure for CO2? What is their significance?
(d) Is CO2 solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60° C under 10 atm, (c) 15 °C under 56 atm?

Answer

The P-T phase diagram for CO2 is shown in the following figure:
P-T phase diagram for CO2

(a) The solid, liquid and vapour phase of carbon dioxide exist in equilibrium at the triple point, i.e., temprature = - 56.6° C and pressure = 5.11 atm.

(b) With the decrease in pressure, both the fusion and boiling point of carbon dioxide will decrease.

(c) For carbon dioxide, the critical temperature is 31.1° C and critical pressure is 73.0 atm. If the temprature of carbon dioxide is more than 31.1° C, it can not be liquified, however large pressure we may apply.

(d) Carbon dioxide will be (a) a vapour, at =70° C under 1atm. (b) a solid, at -6° C under 10 atm (c) a liquid, at 15° C under 56 atm.

11.17. Answer the following questions based on the P–T phase diagram of CO2:
(a) CO2 at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid phase?
(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at constant pressure?
(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm pressure and temperature –65 °C as it is heated up to room temperature at constant pressure.
(d) CO2 is heated to a temperature 70° C and compressed isothermally. What changes in its properties do you expect to observe?

Answer

The P-T phase diagram for CO2 is shown in the following figure:
P-T phase diagram for CO2


(a) Since the temprature -60° C lies to the left of 56.6° C on the curve i.e. lies in the region vapour and solid phase, so carbon dioxide will condense directly into the solid without becoming liquid.

(b) Since the pressure 4 atm is less than 5.11 atm the carbon dioxide will condense directly into solid without becoming liquid.

(c) When a solid CO2 at 10 atm pressure and -65° C temprature is heated, it is first converted into liquid. A further increase in temprature brings it into the vapour phase. At P = 10 atm, if a horizontal line is drawn parallel to the T-axis, then the points of intersection of this line with the fusion and vaporization curve will give the fusion and boiling points of CO2 at 10 atm.

(d) Since 70° C is higher than the critical temprature of CO2, so the CO2 gas can not be converted into liquid state on being compressed isothermally at 70° C. It will remain in the vapour state. However, the gas will depart more and more from its perfect gas behaviour with the increase in pressure.

11.18. A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g–1.

Answer

Initial temperature of the body of the child, T1 = 101°F
Final temperature of the body of the child, T2 = 98°F
Change in temperature, ΔT = [ (101 - 98) × (5/9) ] o C
Time taken to reduce the temperature, t = 20 min
Mass of the child, m = 30 kg = 30 × 103 g
Specific heat of the human body = Specific heat of water = c
= 1000 cal/kg/ °C
Latent heat of evaporation of water, L = 580 cal g–1
The heat lost by the child is given as:
∆θ = mc∆T
= 30 × 1000 × (101 - 98) × (5/9)
= 50000 cal
Let m1 be the mass of the water evaporated from the child’s body in 20 min.
Loss of heat through water is given by:
∆θ = m1L
∴ m1 = ∆θ / L
= (50000 / 580)  =  86.2 g
∴ Average rate of extra evaporation caused by the drug = m1 / t
= 86.2 / 200
= 4.3 g/min.

11.19. A ‘thermacole’ icebox is a cheap and efficient method for storing small quantities of cooked food in summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and co-efficient of thermal conductivity of thermacole is 0.01 J s–1m–1K–1. [Heat of fusion of water = 335 × 103J kg–1]

Answer

Side of the given cubical ice box, s= 30 cm = 0.3 m
Thickness of the ice box, l= 5.0 cm = 0.05 m
Mass of ice kept in the ice box, m= 4 kg
Time gap, t = 6 h = 6 × 60 × 60 s
Outside temperature, T= 45°C
Coefficient of thermal conductivity of thermacole, K= 0.01 J s–1m–1K–1
Heat of fusion of water, L= 335 × 103J kg–1
Let m’be the total amount of ice that melts in 6 h.
The amount of heat lost by the food:
θ = KA(T - 0)t/ l
Where,
A= Surface area of the box = 6s2= 6 × (0.3)2= 0.54 m3
θ = 0.01 × 0.54 × 45 × 6 × 60 × 60 / 0.05  =  104976 J
But θ = m'L
∴ m' = θ/L
= 104976/(335 × 103)  =  0.313 kg
Mass of ice left = 4 – 0.313 = 3.687 kg
Hence, the amount of ice remaining after 6 h is 3.687 kg.

11.20. A brass boiler has a base area of 0.15 m2and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the part of the flame in contact with the boiler. Thermal conductivity of brass = 109 J s –1m–1 K–1; Heat of vaporisation of water = 2256 × 103J kg–1.

Answer

Base area of the boiler, A = 0.15 m2
Thickness of the boiler, l = 1.0 cm = 0.01 m
Boiling rate of water, R= 6.0 kg/min
Mass,m = 6 kg
Time, t = 1 min = 60 s
Thermal conductivity of brass, K= 109 J s –1m–1 K–1
Heat of vaporisation, L= 2256 × 103J kg–1
The amount of heat flowing into water through the brass base of the boiler is given by:
θ = KA(T1 - T2) t/ l    ....(i)

where,
T1= Temperature of the flame in contact with the boiler
T2= Boiling point of water = 100°C
Heat required for boiling the water:
θ = mL … (ii)
Equating equations (i)and (ii), we get:
∴ mL = KA(T1 - T2) t/l
T1 - T2 = mLl/KAt
= 6 × 2256 × 103 × 0.01 / (109 × 0.15 × 60)
= 137.98 o C
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

Page No: 297

11.21. Explain why:
(a) a body with large reflectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more efficient in warming a building than those based on circulation of hot water

Answer

(a) A body with a large reflectivity is a poor absorber of light radiations. A poor absorber will in turn be a poor emitter of radiations. Hence, a body with a large reflectivity is a poor emitter.

(b) Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and one feels cooler.
Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is conducted from the body to the wooden tray. Hence, there is only a negligible drop in the temperature of the body and one does not feel cool.
Thus, a brass tumbler feels colder than a wooden tray on a chilly day.

(c) An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature of a red hot iron piece kept in the open.
Black body radiation equation is given by:
E = σ (T4 - T04)
Where,
E = Energy radiation
T = Temperature of optical pyrometer
To = Temperature of open space
σ = Constant
Hence, an increase in the temperature of open space reduces the radiation energy.
When the same piece of iron is placed in a furnace, the radiation energy, E = σT4

(d) Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra heat will be trapped. All the heat would be radiated back from earth’s surface.

(e) A heating system based on the circulation of steam is more efficient in warming a building than that based on the circulation of hot water. This is because steam contains surplus heat in the form of latent heat (540 cal/g).

11.22. A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

Answer

According to Newton’s law of cooling, we have:
(-dT/T) = K(T - T0)
dT / K(T - T0)  =  -Kdt            ....(i)
Where,
Temperature of the body = T
Temperature of the surroundings = T0 = 20°C
K is a constant
Temperature of the body falls from 80°C to 50°C in time, t = 5 min = 300 s
Integrating equation (i), we get:


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NCERT Solutions for Class 11th: Ch 12 Thermodynamics Physics

March 15, 2015, 9:06 am
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NCERT Solutions for Class 11th: Ch 12 Thermodynamics Physics Science

Page No: 316

Excercises

12.1. A geyser heats water flowing at the rate of 3.0 litres per minute from 27 °C to 77 °C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if its heat of combustion is 4.0 × 104 J/g?

Answer

Water is flowing at a rate of 3.0 litre/min.
The geyser heats the water, raising the temperature from 27°C to 77°C.
Initial temperature, T1 = 27°C
Final temperature, T2 = 77°C
∴Rise in temperature, ΔT = T2– T1
= 77 – 27 = 50°C
Heat of combustion = 4 × 104 J/g
Specific heat of water, c = 4.2 J g–1 °C–1
Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, ΔQ = mcΔT
= 3000 × 4.2 × 50
= 6.3 × 105 J/min
  ∴ Rate of consumption = 6.3 × 105 / (4 × 104)  =  15.75 g/min.

12.2. What amount of heat must be supplied to 2.0 × 10–2 kg of nitrogen (at room temperature) to raise its temperature by 45 °C at constant pressure? (Molecular mass of N2 = 28; R = 8.3 J mol–1 K–1.)

Answer

Mass of nitrogen, m = 2.0 × 10–2 kg = 20 g
Rise in temperature, ΔT = 45°C
Molecular mass of N2, M = 28
Universal gas constant, R = 8.3 J mol–1 K–1
Number of moles, n = m/M
= (2 × 10-2 × 103) / 28
= 0.714
Molar specific heat at constant pressure for nitrogen, Cp = (7/2)R
= (7/2) × 8.3
= 29.05 J mol-1 K-1
The total amount of heat to be supplied is given by the relation:
ΔQ = nCP ΔT
= 0.714 × 29.05 × 45
= 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J.

12.3. Explain why
(a) Two bodies at different temperatures T1and T2if brought in thermal contact do not necessarily settle to the mean temperature (T1+ T2)/2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.

Answer

(a) In thermal contact, heat flows from the body at higher temprature to the body at lower temprature till tempratures becomes equal. The final temprature can be the mean temprature (T1 + T2)/2only when thermal capicities of the two bodies are equal.

(b) This is bcause heat absorbed by a substance is directly proportional to the specific heat of the substance.

(c) During driving, the temprature of air inside the tyre increases due to moion. Accordingto Charle's law, P ∝ T. Therefore, air pressure inside the tyre increases.

(d) This is because in a harbour town, the relative humidity is more than in a desert town. hence, the climate of a harbour town is without extremes of hot and cold.

12.4. A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?

Answer

The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder = P1
Final pressure inside the cylinder = P2
Initial volume inside the cylinder = V1
Final volume inside the cylinder = V2
Ratio of specific heats, γ = 1.4
For an adiabatic process, we have:
P1V1γ =P2V2γ
The final volume is compressed to half of its initial volume.
∴ V2 = V1/2
P1V1γ = P2(V1/2)γ
P2/P1 =V1γ / (V1/2)γ
= 2γ = 21.4 = 2.639
Hence, the pressure increases by a factor of 2.639.

12.5. In changing the state of a gas adiabatically from an equilibrium state A to another equilibrium state B, an amount of work equal to 22.3 J is done on the system. If the gas is taken from state A to B via a process in which the net heat absorbed by the system is 9.35 cal, how much is the net work done by the system in the latter case? (Take 1 cal = 4.19 J)

Answer

The work done (W) on the system while the gas changes from state A to state B is 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.
∴ ΔQ = 0
ΔW = –22.3 J (Since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = ΔU + ΔW
Where,
ΔU = Change in the internal energy of the gas
∴ ΔU = ΔQ– ΔW = – (– 22.3 J)
ΔU = + 22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is:
ΔQ = 9.35 cal = 9.35 × 4.19 = 39.1765 J
Heat absorbed, ΔQ = ΔU + ΔQ
∴ΔW = ΔQ– ΔU
= 39.1765 – 22.3
= 16.8765 J
Therefore, 16.88 J of work is done by the system.

12.6. Two cylinders A and B of equal capacity are connected to each other via a stopcock. A contains a gas at standard temperature and pressure. B is completely evacuated. The entire system is thermally insulated. The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P-V-T surface?

Answer

(a) When the stopcock is suddenly opened, the volume available to the gas at 1 atmospheric pressure will become two times. Therefore, pressure will decrease to one-half, i.e., 0.5 atmosphere.

(b) There will be no change in the internal energy of the gas as no work is done on/by the gas.

(c) Since no work is being done by the gas during the expansion of the gas, the temperature of the gas will not change at all.
(d) No, because the process called free expansion is rapid and cannot be controlled. the intermediate states are non-equilibrium states and do not satisfy the gas equation. In due course, the gas does return to an equilibrium state.

12.7. A steam engine delivers 5.4×108 J of work per minute and services 3.6 × 109 J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?

Answer

Work done by the steam engine per minute, W= 5.4 × 108J
Heat supplied from the boiler, H= 3.6 × 109J
Efficiency of the engine = Output energy / Input energy
∴ η = W / H  =  5.4 × 108 / (3.6 × 109)  =  0.15
Hence, the percentage efficiency of the engine is 15 %.
Amount of heat wasted = 3.6 × 109– 5.4 × 108
= 30.6 × 108= 3.06 × 109J
Therefore, the amount of heat wasted per minute is 3.06 × 109J.

12.8.An electric heater supplies heat to a system at a rate of 100W. If system performs work at a rate of 75 Joules per second. At what rate is the internal energy increasing?

Answer

Heat is supplied to the system at a rate of 100 W.
∴Heat supplied, Q= 100 J/s
The system performs at a rate of 75 J/s.
∴Work done, W = 75 J/s
From the first law of thermodynamics, we have:
Q = U + W
Where,
U = Internal energy
∴U= Q – W
= 100 – 75
= 25 J/s
= 25 W
Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

12.9. A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in Fig. (12.13).
Its volume is then reduced to the original value from E to F by an isobaric process. Calculate the total work done by the gas from D to E to F.

Answer

Total work done by the gas from D to E to F = Area of ΔDEF
Area of ΔDEF = (1/2) DE × EF
Where,
DF = Change in pressure
= 600 N/m2– 300 N/m2
= 300 N/m2
FE = Change in volume
= 5.0 m3– 2.0 m3
= 3.0 m3
Area of ΔDEF = (1/2) × 300 × 3 = 450 J
Therefore, the total work done by the gas from D to E to F is 450 J.

12.10. A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36° C, calculate the coefficient of performance.

Answer

Temperature inside the refrigerator, T1 = 9°C = 282 K
Room temperature, T2 = 36°C = 309 K
Coefficient of performance = T1 / (T2 - T1)
= 282 / (309 - 282)
= 10.44
Therefore, the coefficient of performance of the given refrigerator is 10.44.


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Study Material and Summary of The ant and the cricket NCERT Class 8th

March 27, 2015, 8:36 am
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Study Material of The ant and the cricket (Summary, Extra Questions/Answer and Word Meanings)

Summary of Poem
Study Material and Summary of The ant and the cricket

It is a fable in poetic form. A fable is a story, often with animals as characters,that conveys a moral.

This is a poem about a silly young cricket and an ant. Cricket was only singing all day long and used to enjoy his good times during summer season. He didn’t plan anything for the future.

When winter arrives, cricket couldn’t find a small amount of food to eat. So, the cricket thinks of going to miser ant to borrow food and to get shelter. Then, the cricket knocked on the ant´s door asking for help.

The ant gives a very important lesson of life during its conversation with the cricket. Ant says that ants neither borrow from somebody nor lend to somebody. Ants are hardworking creatures and save for the future. The ant asks the cricket what it was doing during happier times. On hearing the cricket indulged in dancing and singing and making merry, the ant asks the cricket to try dancing and singing once again during rough times.

Moral: We should happy moments but should not ignore our future. Careful planning is essential for everyone and it ensures a secure future.

Terms and meanings from the Chapter:

• Accustomed to sing: used to singing
• Lay nothing by: save nothing
• Quoth: said
• Hastily: Hurriedly

Extra Questions and Answer

What was the young cricket accustomed to do?
► The young cricket accustomed to sang all day long and enjoyed his good times.

When was the cricket more happy?
► The cricket was more happy through the warm,sunny months of gay summar and spring.

Why did he complain?
► He complained because he found his cupboard was empty,and winter was come.

Give the opposite of : empty, warm.
► Empty – Full
Warm – cold

What made the cricket bold ?
► Starvation and famine made the cricket bold.

Why cricket go to the ant?
► The cricket went to the ant for shlter and grains to eat.

What did the ant tell the cricket?
► The ant told the cricket that thay neither borrow from somebody nor lend to somebody.

What did the ant ask the cricket?
► The ant asked the cricket that what he was doing in summer times.

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Study Material and Summary of Geography lesson NCERT Class 8th

March 27, 2015, 8:35 am
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Study Material of Geography lesson (Summary, Extra Questions/Answer and Word Meanings)

Summary of Poem

The poem is about the way earth looks from different altitudes. It is divided into three parts.

When the jet takes off and starts to climb up in the sky, you can have full height view of the city. The city grew as per its necessity and did not grow as per proper planning. It does not have any particular style. It even looked six inches from a certain height revealing its true structure.

When the jet climbs higher up to about ten thousand feet, he could apprehend the fact that cities grew water resources. Water fulfilled the necessities like agriculture, transport, business and others. He found that valleys were populated which met the necessities of the people.

When the jet went above six miles, there is more water than land on the earth. While appreciating the geographical niceties of our planet, the poet is unable to understand the tendency to build borders, to erect walls, to create fences. Then he reflected that people on earth selfish and narrow for they hate each other .He found the earth to be one but not the people living on this earth are divided for shallow reasons.

Terms and Meanings from Chapter

• Sprang – Rise
• Inevitable – unavoidable
• Haphazard - without plan or order
• Delineated – Shown

Extra Questions and Answer

1. What geographic lessons did the poet in the jet learn when the jet just took off? 
► The city that human have devloped have not been well planned, it grew as per necessity.

2. What geographic lessons did the poet in the jet learn when the jet reached ten thousand feet?
► From a height of ten thousand feet above the ground, the earth’s cities were found located on the banks of rivers and a little above the river level, on valleys.

3. What was once most necessary for the emergence of a city in the past?
► The easy availability of water for drinking to irrigation was the prime reason why civilizations flocked around rivers.

4. What was difficult to understand about the earth when the jet was six miles high?
► It was diffcult to understood why human hate each other, build borders, walls and create fences.

5. What is the poet trying to bring out about human beings, the earth and learning?
► The poet is trying to bring out the fact that human beings have achieved a lot of knowledge about the earth and beyond. We have learnt much about the earth, its shape, the emergence of cities and nations, but we have failed to learn how to love each other live peacefully on this wonderful planet.

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Study Material and Summary of Macavity: The Mystery Cat NCERT Class 8th

March 27, 2015, 8:35 am
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Study Material of Macavity: The Mystery Cat (Summary, Extra Questions/Answer and Word Meanings)

Summary of Poem
Study Material and Summary of Macavity: The Mystery Cat

Macavity is a mystery cat because no crime agency was ever able to arrest this criminal mastermind during or after his committing a crime. He is too clever to leave any evidence of his guilt. He is a puzzle for every detective agency in the world including Scotland Yard and Flying Squad who are specialized investigating crime. Whenever they reach the scene of crime-Macavity’s not there. Macavity is too clever to be caught and he is nowhere near at the crime spot.

Not only does he breaks the human law but also breaks the law of gravity. His brows are deeply lining as a result of continuous planning of crime. Macavity has sunken eyes and “his head is highly domed”. He never combs his whiskers. His movements resemble that of a snake. When you think he is sleeping, he is wide awake in fact.

Macavity is a devil in cat’s shape. He is morally corrupt. You may meet and see him everywhere but whenever his crime is discovered you will not find him there.

Terms and Meanings from Chapter

• Defy – Disobey
• Bafflement – To confuse
• Scotland Yard - the headquarters of the London police force
• Flying Squad - a group of police or soldiers ready to move into action quickly
• Levitation - floating in the air without support
• Fiend – Devil
• Feline – a cat
• Depravity – moral corruption

Extra Questions and Answer

1. How does Macavity outwits the world’s top investigation agencies?
► Whenever investigation agencies reached the spot of crime, Macavity is not presnet there. He also didn’t leave any clue of him. These agencies never able to found any sign of Macavity. Therefore, he outwitted the world’s top investigation agencies.

2. What is the most remarkable thing about Macavity?
► Macavity, known as the Mystery Cat has defied all laws. He even defies the law of gravity. He possesses supernatural powers which allow him to levitate up in the air. He is so confident in his manner that whenever the crime is discovered, Macavity disappears without leaving a single trace.

3. Describe Macavity.
► Macavity is a tall and thin ginger cat whose eyebrows are deep with lines. He has sunken eyes which gives him a devil look. His coat is untidy and his whiskers are uncombed. He moves his head from side to side and his body movement is like that of a snake.

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Study Material and Summary of Bepin Choudhury’s lapse of memory NCERT Class 8th

March 27, 2015, 8:34 am
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Study Material of Bepin Choudhury’s lapse of memory (Summary, Character Sketch and Word Meanings)

Character Sketch

• Bepin Choudhury: is an old man who had no children. He was living a lonely life after his wife’s demise. He was working in reputed at responsible post and have few friends.
Personality: He had keen interest in reading books. This is shown in starting of story, he stop his car every Monday in New market to buy books. He never wanted to waste his time in idle chatting.

• Parimal Ghose: He is round faced, meek looking man who intrude Bepin Choudhury in book shop. He tried to make belive Bepin babu that he had gone to Ranchi. He also provided many evidence that make Bepin rely on his statement.

• Chunni Lal: He is school friend of Bepin Choudhary. He previously worked in travel agency but now in search of job. He kept visitng Bepins house about a job but Bepin always disregards him. Bepin babu had also not helped him financially at time of need so he tried to took revenge by frustating Bepin’s mind through his imaginative.

• Dr. Paresh Chandra: He is a young physician with pair of bright eyes and a sharp nose. He is called by Bepin choudhury to treat memory loss problem.

Terms and Meanings from the Chapter

• Idle Chat - Unnecessary conversation
• Meek – Quiet
• Utter disbelief – Complete surprise
• Regretted – felt sad
• Intimate – very personal and private
• Deliberate – intentionally
• Insanity – mental illness
• Bracing – Stimulating
• Going nuts – going mad
• Sarcasm – satire
• Getting into people’s hair – interfering with and annoying people.
• Snuggling – comfortable position
• Sleuthing – investigating (an event)
• Conscientious – careful and correct
• Head was in a whirl - confused and unable to think clearly.
• Gather his wits together - make an effort to become calm and think clearly.
• Having a rough time – Having a lot of problems
• Turning up like a bad penny - appearing at a place where one is not welcome.
• Didn’t beat about the bush - came straight to the point
• Off and on – now and then
• Throw your mind back - think back and recall a past event
• Must see about consulting – may have to consult
• Tranquilliser - a medicine to reduce stress and anxiety
• Boulder – a large rock
• Came round – regained consciousness

Summary of the Chapter

It was habit of Bepin Choudhury to stop on a book shop present in New market to purchase the books of his interest.
But one day, he encountered with Parimal Ghose,who had recognised Mr. Bepin Babu but Bepin refused to identify him. As proof, Mr.Ghose narrate him Ranchi incident but Bepin replied that he had never been to Ranchi and told that he also don’t know Ghose. Later, Ghose described him his personal details as family-wife and kids, Dinesh Mukerji,his bunglow etc but Bepin Babu was not ready to believe.

Bepin think over it. He was confused whether he Ghose was telling lie or he was losing his memory then he decided to ring up Dinesh Mukerji so that he could confirm the matter. He talked to Dinesh but he also replied in positive and recollected all other events also, hearing all his head began to spin.
Next day when he reached home from office,his servant told him that Mr. Chunni Lal wanted to meet him.Chunni asked him for money but Bepin Babu refused, besides he began to confirms about the incident of Ranchi and his last job just to verify if he was losing his memory or not .But Chunni Lal also replied positively. Now he called for Dr. Paresh Chanda, who gave him remedy.

At the end of the lesson, Bepin himself went to Hudroo but couldn’t retrieve his memory. When he reached Hudroo, he felt unconsciousness there. He returned calcutta and feeling that he will lose everything, there is no hope left in him. Few minutes later Bepin's servant brought a letter which was describing Chunni’s revenge for not helping him in at the time of need. Now Bepin understood all the matter. When Dr. Chanda again came to treat him, he was totally surprised when Bepin told him that he was better now. He just needed a pain killer as he has fallen in Ranchi.

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Study Material and Summary of The Last Bargain NCERT Class 8th

March 27, 2015, 8:34 am
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Study Material of The Last Bargain (Summary, Extra Questions and Word Meanings)

Summary of Poem

One morning, speaker walking on the stone paved road and asking out to hire him.
Firstly, a king came near him and told that he would hire him with his power but his power counted for nothing so he disliked his agreement.
During mid-day, an old man came with gold coins and said that he would hire with money but the poetspeaker rejects all his money because money will soon be spent and money cannot buy us everlasting happiness.
In the evening, the garden fences were full of flowers. A fair lady came out from garden and said that he would hire him with a smile but her smile faded away and she melted into tears and returns into the dark leaving the speaker alone.
At last a child playing with shells and said that he would hire him with nothing an poet accepted his agreement but speaker was attracted by the flawless character of the child so he accepted the agreement of child. The bargain of the child makes the speaker a free man.

Terms and Meanings from the poem

• Naught – nothing
• Crooked – bent
• Pondered – consider
• Hedge – fence
• Glistened – shine

Extra Questions and Answers

1. Why speaker deny the king proposal?
► The speaker stated the king’s power as nothing so he denied the proposal of king.

2. What did the old man want? Was he successful in his bargain?
► The old man wanted to hire the speaker with his money. No, he was not successful in his bargain.

3. Explain, “I hire you with nothing”?
► The child has no material thing so he used the word nothing here. He has only goodwill and cheer to hire the speaker.

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Study Material and Summary of Summit Within NCERT Class 8th

March 27, 2015, 8:33 am
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Study Material of Summit Within (Summary, Extra Questions and Word Meanings)

Summary of Chapter

This story is experiences of Major H.P.S. Ahluwalia, a member of the first successful Indian expedition to Mount Everest. Ahluwalia was full of humility when he stood on the summit of Everest. He thanked God for his physical success. It was the highest of the goals for him. While getting down from the Summit he asked himself why he climbed Everest and what made him to do so.
The people climb mountains because they present great difficulties. Man takes delight in overcoming obstacles. A climb to a summit means the demonstration of climber’s endurance, Persistence and will power. Since childhood, the narrator has been attracted by mountains. For him mountains are nature at its best. They have a challenging beauty and majesty. They are a means of communion with God. Everest is the highest and the mightiest and has defied many previous attempts. On climbing its summit, one has the sense of victory and of happiness. Its view brings a spiritual change in his mind. It poses a challenge before him which was difficult to resist. One feels connected with the supernatural element on reaching the Summit. One becomes conscious of his own smallness in this large Universe. It provides physical, emotional and spiritual fulfilment. 
Climbing a mountain is a highly risky job and needs others’ help also. The fellow climbers prove to be a source of inspiration. They remember their Gods to feel confident. It is far more difficult to climb the summit within oneself than to climb the summit of a mountain. One can get a fuller knowledge of oneself merely by climbing one’s personal and internal mountain peak. It is fearful and unscalable like the climb to the summit of a mountain. Both the climbs teach one much about the world and about oneself. The internal summits are much Higher than Everest. The climber gets the inspiration to face life’s ordeals with determination.

Terms and Meanings from the Chapter

• Surged – arose suddenly and intensely
• Panorama – view of a wide area
• Humility – modesty
• Jubliant – very happy due to success
• Tinge – shade
• Exhaustion – tiredness
• Endurance – toleration
• Persistence – determination
• Exhilarating – Very exciting
• Communion - sharing or exchanging of intimate thoughts and feelings.
• Defied – frustrated
• Mystical – spiritual
• Aloofness –state of being distant
• Firm in – make yourselff firm
• Belays – fixes a rope
• Ascent – climb
• Ennobling – giving a noble rank or title
• Akin – similar
• Ordeals – painful experiences
• Resolutely – with determination

Extra Question and Answer

1. Why author climb the summit of everest?
► In author’s view, climbing a mountain presents great difficulties. A climber have to cross many obstacles, it means endurance, persistence and will power. These qualities attract author to climb summit of everest. Also, from my childhood author is attracted by mountains.

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Study Material and Summary of The School Boy NCERT Class 8th

March 27, 2015, 8:33 am
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Study Material of The School Boy (Summary, Extra Question and answers and Word Meanings)

Summary of Poem
Study Material and Summary of The School Boy

The speaker of poem is a school boy who love to rise in summer morning, when birds are singing on the trees. The boy gets entertained by the company of the hunter who blows his clarion from a distance field and sweet lullabies of skylark.
But the thing he don’t like is going to school which pulls all his happiness and joy. He is tired and even puzzled under the strict supervision of his teacher. Instead of enjoying the pleasures of summer, the child has to spend many tensed hours in his school nor in the garden where he can learn many things in interesting way with the nature.
A bird can never sing sweet song when he if caged. Similarly, a child if remained under the umbrella of annoying fear and tension, the skepticism of his teacher can never enjoy the natural instincts of joy and playfulness.
In last stanzas, he tried to make undertsand his parents that if a budding child is picked and swept of in the early stage of life, where there is no one to care for then how he could grow in mature plant.

Terms and Meanings from the poem

• Outworn – out of date
• Sighing – breathe out
• Dismay – shock
• Drooping - To hang down as from exhaustion
• Anoxious – worried
• Bower - climbing plants in a garden
• Dreary – Uninteresting
• Nip’d (nipped) - destroy it at an early stage of its development
• Strip’d (stripped) – remove all coverings

Extra Question and Answers

1. Why child hates going to school?
► Child hated school because there he is under strict control of his teacher like he has been caged.

2. How child happiness turn into sorrowness?
► The child rose in the fresh and delightful summer morning. He is very happy but his parents force him to go to the school where he spend his time in sorrowness.

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Study Material and Summary of This is Jody’s Fawn NCERT Class 8th

March 27, 2015, 8:32 am
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Study Material of This is Jody’s Fawn (Summary, Character Sketch and Word Meanings)

Character Sketch

• Jody: He is a small young boy. He felt bad for fawn which is alone in the forest without his mother.

• Ezra Baxter (Penny): Father of Jody. He was bitten by a rattle snake in his hand. He killed a doe and used its heart and liver to cure himself. He allowed his son Jody to bring the fawn back home and raise it.

• Ora baxter: Mother of Jody. She also agreed to bring the fawn at house and raise it.

• Mill – Wheel and Dec wilson: Forester, who helped Jody in finding fawn in the forest.

Terms and Meanings from the Chapter

• Drift back to – go back to
• Close shave – narrow escape
• Kept your head – stayed calm in your situation
• Hemmed in - caught in a situation where one can’t say ’no’.
• Acorns – small brown nuts
• Sidled back – walked back quietly
• Gasped – breathed heavily
• Mounted – riding an animal
• Endure – suffer
• Makes a bearing - acts as a compass and helps to identify directions.
• Buzzard - a large bird like the vulture that eats the flesh of dead animals.
• Carrcass – dead body of animal
• Adjacent – tree
• Carrion – the decaying flesh of dead animals
• Parted – moved
• Quivering – shivering
• Delirious – extermely excited
• Convulsion – shiver
• Sleek – smooth and shiny
• Hoist – pull up higher
• Light-headed – unable to think clearly
• Romp – play
• Balked – was unwilling (to do something)
• Wobbled – move
• Gurgle - make a hollow bubbling sound

Summary of the Chapter

Jody's father, Mr. Baxter, had been bitten by a rattle snake. He killed a doe and used its heart and liver to suck out the venom from his wound. The next morning he felt much better but his son, Jody, felt bad to have left the doe's fawn all alone in the forest. He went to his father and reasoned with him the need to bring the fawn to the safety of their home. His father allowed him to go on a search for the fawn. Jody's mother though gave her consent, feared for her son's safety in the forest. Doc Wilson and Mill-wheel also approved considering that they had already left the fawn motherless.
Jody went with Mill-wheel on his horse assuring his mother to reach home by dinner. After sometime they reached closer to the place where his father was bitten. Jody wanted go on further all alone because he didn't want Mill-wheel to see his disappointment if they failed in finding the fawn. Contrarily, if he found the fawn, he wanted to experience the joy of it all alone as he felt that their meeting would be intense, full of emotion and thus, personal. Thus, assuring Mill-wheel of his knowledge of directions and his ability to take care of himself, he move on. 
When Jody reached the spot where his father was bitten, he found buzzards hovering over the carcass of the dead doe. He also found footprints of cats and for a moment he feared for the life of the fawn. After an intense search he finally found the fawn behind a bush. Jody noticed that the fawn was shivering and was distrustful. He tried to calm the fawn and tried to establish some kind of understanding with him. 
However, though the fawn allowed the proximity of Jody, he did not move. Jody then decided to carry the fawn all the way home. He first patted the fawn and then lightly lifted him. He went around the area where his father was bitten and where lay the carcass of the fawn's mother, fearing that the scent of his mother would make the fawn restless. He had to stop often for breaks as himself being little he was severely struggling with the weight of the fawn and the vines and bushes that lay on his path. Though, Jody's arms had started hurting, he carried on his journey. He even managed to win the fawn's trust, who gradually grew willing to follow him. 
After a point, Jody felt such a grand connection with the fawn that all the struggle and pain he was going through no longer mattered. By and by they reached their destination. The fawn refused to go upstairs, probably sensing the presence of his mother's killer, but Jody carried him to his father who expressed joy at seeing the fawn. Later, Jody lovingly fed him milk in the kitchen and enjoyed the fawn's trust and love for him.

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Study Material and Summary of The Duck and the Kangaroo NCERT Class 8th

March 27, 2015, 8:32 am
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Study Material of The Duck and the Kangaroo (Summary, Extra Question and answers and Word Meanings)

Summary of Poem


Study Material and Summary of The Duck and the Kangaroo

The Duck and the Kangaroo, both were very good friends. As the duck lived in a pond and does not get a variegated life of visiting world, he wanted to have a pleasure tour all around the world. So he requested the kangaroo to allow him to sit on the top of his tail and have a pleasure tour. The kangaroo accepted the wish, but at the same time put some conditions too. According to him the duck’s feet were unpleasantly wet and cold. This may cause with rheumatism. At this the Duck assured him with the remedies he thought about. According to him he has already bought four pairs of woolen socks to put on. Besides he has bought a cloak to cover himself and he will smoke cigar too. In this way both the duck and the kangaroo started their tour and continued their happy journey.

Terms and Meanings from the Poem

• Gracious – kind
• Nasty – unpleasent
• Roo-matiz (Rheumatism) – any disease marked by inflammation and pain in the joints, muscles or tissue.
• Cloak – robe
• Worsted socks – woollen socks
• Pale – light in colour

Extra Questions and Answers

1. Why duck wants to go out for tour?
► The duck wanted to saw world beyond his pond so he wanted went out for tour.

2. What objection Kangaroo have?
► The kangaroo's objection is that the duck's webbed feet are cold and damp and it would give him the roo-matiz.

3. What are the conditions put by Kangaroo for the duck?
► The conditions put by Kangaroo for the duck is that he sit steady at the end of his tail so he can balance himself.

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Study Material and Summary of When I sat out for Lyonnesse NCERT Class 8th

March 27, 2015, 8:32 am
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Study Material of When I sat out for Lyonnesse (Summary, Extra Question and answers and Word Meanings)

Summary of Poem

The poet is a young architect who went to Lyonnesse to supervise the renovation work of a dilapidated church. It was 100 miles away. It was winter season and snowfall was at the peak. He was in solitary state and during his journey he witnessed starlight. He becomes conscious about what would happen at Lyonnesse when he would stay there. According to him no prophet can declare this and even a wizard would not be able to say what would happen at Lyonnesse. When he returned from Lyonnesse, he had magic in his eyes. All could understand that he was filled with a rare and immeasurable radiance.

Terms and Meanings from the poem

• Lyonnesse – an imaginary place
• Rime – frost
• Spray – leaves andd branches of tree
• Sojourn - stay
• Durst – dared
• Bechance – chance to happen
• Radiance – glow
• Surmise - guess
• Fathomless - so deep that the depth can’t be measured.

Extra Questions and Answers

1. How poet is feeling while travelling to lyonnesse?
► The poet is feeling lonely while travelling as no one accompanied him.

2. What changes occur in poet after he returned from Lyonnesse?
► When the poet returned from Lyonnesse he have strange glows in his eyes. It seemed as if he had magical eyes that glowed from inner radiance.

3. Which season is there when poet is travelling?
► The winter season is in the poem. This can be understood by the line: The rime was on the spray.

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Study Material and Summary of On the grasshooper and the cricket NCERT Class 8th

March 27, 2015, 8:31 am
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Study Material of On the grasshooper and the cricket (Summary, Extra Question and answers and Word Meanings)

Summary of Poem

This is fine sonnet and symbolic poem in which the grasshopper is a symbol of hot summer and the cricket is of very cold winter.
Every poet has been attracted towards the beauty of nature and so has Keats been. He finds nature beautiful in all seasons not excluding the hot summer and cold winter.
He says that the earth is always singing. When the birds stop singing in the hot summer finding cool place, we find the grasshopper singing and flying from hedge. He sings tirelessly and when tired, rests beneath some weed.
When the birds are silent in very cold the earth never stops and expresses its pleasure through different beings like the cricket. It sings the beautiful songs from the stones. It seems to be increasing in warmth every moment and half-asleep human being feels it to be a grasshopper’s song coming from grassy hills.

Terms and Meanings from the poem

• Faint – ill-defined
• Hedge – fence or boundary
• Wrought – brought about
• Shrills – comes through loud and clear

Extra Questions and Answers

1. What is the meanings of the line: ‘The poetry of earth is never dead’?
► The poet wanted to say that poetry of nature is never going to end. No matter what the season is, whether it is the sweltering summer or the harsh cold winter, the music and the poetry of the nature is never dead.

2. What is main theme of the poem?
► The main theme of poem is that poetry and music in nature do not perish.

3. Where do birds take rest in hot summer day?
► The birds took rest under shady trees to secure themselves from the scorching heat of the sun.

4. Where do grasshooper take rest when he tired?
► When grashooper became tired, he rests ease beneath some pleasant weeds.

5. In which season cricket sing?
► The cricket sung in winter. When it is very cold and quiet, the winter silence is broken by a shrill sound.

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Study Material and Summary of The Great Stone Face I NCERT Class 8th

March 27, 2015, 8:31 am
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Study Material of The Great Stone Face -I (Summary, Character Sketch and Word Meanings)

Character Sketch

• Ernest: He is upright, hard-working, and a benevolent presence to his neighbors. He spends his leisure hours gazing at a mountain rock formation called the Great Stone Face.

• Ernest's Mother: Loving woman who tells her son about an old story predicting that a child born in the valley below the Great Stone Face will become the greatest person of his time. The story says his face will resemble the one on themountain.

• Mr. Gathergold: A wealthy merchant who is a native of the valley. He arrives in the valley during his end days.

• Old Blood and Thunder: Great general who is a native of the valley returned in valley after becoming old and tired.

• Statesman: Great orator who is a native of the valley.

• Poet: Great writer who is a native of the valley who declared that ernest resembles as Great Stone Face.

Terms and Meanings from the Chapter

• Immense – Huge
• Enormous – very big
• Resembled – similar appearance
• Benign – kind
• Prophecy – statement that tells what will happen in the future.
• Pensive – thoughtful
• Gaze – stare
• Stirred – moved
• Labour - work
• Folly – foolishness
• Sympathises – feeling of sorrow, approval and understanding.
• Beheld – saw
• Renowned – famous
• Banquet – feast
• Proclaimed – announced

Summary of Chapter

One afternoon , a mother with her son, Ernest , was at the door of their cottage. They were talking about the great stone face. It was clearly visible in the bright sunshine. Thousands of people lived there. Everybody there, was familiar with the great stone face. It was the work of the nature. One day the great stone face smiled on Ernest and looked kindly . He wished to hear its pleasant voice. He longed to see a man with such a face in order to love him dearly. There was a man called gather gold he had left his native many years ago he had became quite rich man there. He decided to return to his native valley. There was a rumour that gathergold looked like the great stone face. Gathergold had the face of an old man with yellow skin. The people considered him the image of the great stone face Ernest gazed up the valley. The great stone face seemed to reject gathergold as its likeness.
Ernest had grown up to be a young man. Everyday, he would go off by himself and gaze upon the great stone face. He wondered why its likeness was delaying its appearance. By his time gathergold had become poor and died without establishing his likeness with the stone face. After son of the valley had joined the army as a soldier many years before. He had be come a general by then and came to be known as blood and thunder on the battled. In his old age, he desired to return to his valley. The people considered him as the likeness of the great stone face. On his return, he was welcomed warmly. The gathering mistook him as the greatest man of all time. Ernest failed to recognise any likeness between Blood and thunder and the great stone face. His heart assured him that the real face had still to come.

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Study Material and Summary of The Great Stone Face II NCERT Class 8th

March 27, 2015, 8:41 am
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 Study Material of The Great Stone Face -I (Summary and Word Meanings)

Summary of the Chapter

Ernest still lived in his native valley. He was simple hearted man. He always worked for the betterment of the world. Though he was considered to be an ordinary man yet he was humble and rich in thoughts. With the passage of time ,Ernest became old. He had wise thoughts in his mind. He had become famous all over the world. Men came from distant places to see and speak to him. A new poet had appeared on earth while Ernest had been growing old. The poet was a native of Ernests valley but had stayed in distant cities for a long period. The poet had heared of Ernests character. One day, he came to his door. Ernest was reading a book and glancing lovingly at the mountain from time to time. Ernest gave him shelter for the night. The great stone face looked kindly at the poet. The poet found Ernest ,wise ,gentle and kind.
Ernest used to speak to his neighbours every evening. The poet also accompanied him the poet listened to his talk. He felt that Ernests life and character were a far nobler kind of poetry than his own poems just then ,the poet saw the great stone face . He declared ernest as the likeness of the great stone face every body agreed with him. Ernest still hoped that someone wiser and better than himself would appear sometime, bearing a likeness to the great stone face.

Terms and Meanings from the Chapter

• Furrows – deep lines
• Obscure – not well known
• Unawares – unknowingly
• Customary toil – usual work
• Hospitably – kindly
• Corresponded – been in harmony with
• Custom – habit
• Harmonised with – agreed with
• Sage – wise man
• Diffused – spread all around.

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Study Material and Summary of How I taught my Grandmother to read NCERT Class 9th

March 27, 2015, 9:02 pm
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Study Material of How I taught my Grandmother to read (Summary, Character Sketch and Word Meanings)

Character Sketch

• Sudha Murty: The authoress, was a twelve year old child. She had been living with her grandmother. As her grandmother was illiterate, she used to read a novel serial published weekly to her grandmother. Later, she taught her grandmother to read Kannada alphabet.

• Krishtakka: The authoress’s grandmother, was an illiterate person. He could not read own so she used to listen the novel serial and after she discussed the story with her friends. She was pure religious person and solely dedicated herself to god. Later, she learnt to read Kannada alphabet.

Summary of the Chapter



The authoress was twelve years old and she used to stay with her grandparents in north Karnataka. Since there were very few diversions, the entire family would eagerly wait for a weekly magazine called Karamveera. This magazine was publishing famous writer Triveni’s novel, ‘Kashi Yatre’ as a serial. The novel dealt with the protagonist’s struggle to visit Kashi. Since grandmother believed in Kashi-Yatra as a pilgrimage, she could identify with the trials and tribulations of the main characters. Every Wednesday, the authoress would read the next episode of the story to her. The grandmother would not only listen with great attention but also memorize it by heart. Later she would discuss it with friends in the temple courtyard.
One day the authoress had to go to the neighbouring village for a cousin’s wedding and stayed back there for a week. When she returned, she was surprised to see her grandmother in tears. At night, the grandmother narrated how her past life, feeding and bringing up children, kept her busy, sparing no time for education of any kind. Moreover, education for girls was not considered as important in those days. Without the granddaughter, she had a tough time trying to decipher and guess the next episode of ‘Kashi Yatre’. She had felt very desperate and helpless, mutely gazing at the pictures, unable to read a single word.
This incident fanned the grandmother’s desire to learn the Kannada alphabet and she vowed to learn it by Dassera. The granddaughter’s pleas, her mockery of the grandmother’s age, all fell on deaf ears. No wonder, the grandmother proved to be an ideal student. In no time, she could read, repeat and write Kannada. The authoress presented her with a copy of the novel ‘Kashi Yatre’ as a token of appreciation and grandmother touched the feet of her granddaughter, her teacher, as a befitting tribute.

Terms and Meanings from the Chapter

• Psychological – relating to mental and emotional state of a person
• Ardent – showing strong feelings
• Savouring – enjoying an experience slowl in order to appreciate it as much as possible.
• Scriptures – holy writings

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Study Material and Summary of A Dog named Duke NCERT Class 9th

March 28, 2015, 12:26 am
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Study Material of A Dog named Duke (Summary, Character Sketch and Word Meanings)

Character Sketch

• Charles Hooper: He was a six-foot well built man of highly competitive nature. He was a hard-charging zone sales manager for a chemical company. One day he met with accident which paralysed his left side. Slowly he recover with the help of his dog, Duke.

• Duke: He was a a huge Doberman, weighing 23 kilos, with a red coat and a fawn vest. He helped his master, Chuck in recovering. Later, he was killed in accident.

• Marcy: She was wife of Chrles hooper. He was not a dog-lover intially and not wannted to purchase Duke. He supported his husband chuck in times of need. Later, he also affectionate towards Duke.

Terms and Meanings from the Chapter

• Haemorrhage – heavy bleading
• Headway – progress
• Nudged – poked
• Snorted – made explosive sound
• Anticipation – expectation
• Reproachful - a look to show that you are criticising someone.
• Taut – stretched

Summary of Chapter

Study Material and Summary of A Dog named Duke NCERT Class 9th

Duke was a rough playing Doberman Pinscher, four year old and of 23 kilos, with a red coat and a fawn vest. Chuck Hooper had doubts at first about buying him because his wife, Marcy, was not really a dog lover but finally he purchased the dog because he was very impressed with the energy and agility of the dog. Hooper himself had an athletic built and professionally also he was fortunate to be the Zonal Sales Manager of a chemical company. Life was going well but one day, Hooper met with an accident. He had to be admitted to the hospital, due to a haemorrhage in the motor section of his brain, completely paralysing his left side. His wife Marcy was also compelled to look after her huband in the hospital. They had no option but to leave the dog in kennel. After he was discharged from the hospital , the doctor advised him to do regular excercises.
In his first meeting with Duke, after his accident, Duke shouted, charged at Hooper and hit him above the belt, making him struggle to keep his balance. Perhaps the dog sensed his master’s need and from that moment, he never left Hooper’s side. But Hooper remained grim and didn’t reciprocate at all. One fine day, Duke appeared to be in no mood to tolerate his master’s indifference. He nudged, needled, poked and snorted. By chance, Chuck’s right hand hooked onto Duke’s collar to hold him still Duke pranced, Chuck asked Marcy to make him stand, and in a few seconds as he straightened his left foot to drag forward, he had taken a step. Chuck was exhausted but a beginning was made. Next day, Duke walked to the end of the leash and tugged. Chuck managed to take four steps. In two weeks, both of them managed to reach the front porch and by next month, they were along the sidewalk. Seeing this progress, the doctor prescribed a course of physiotherapy with weights, pulleys and whirlpool baths. Soon Duke began walking with Hooper, without pausing and relaxing at each step.
On January 4, Hooper managed to walk to his district office without Duke. But Hooper was not mentally fit to undertake a full day’s work. There was no time for physiotherapy sessions so Chuck again took Duke’s help who pulled him along the street faster and faster. Chuck regained his stability, endurance and next year, he was promoted as the Regional Manager. Chuck, Marcy and Duke moved to a new locality. But it was not to be. One day on October 12, 1957, after being run over by a car, Duke succumbed to his injuries in a nearby hospital. People who had seen Chuck and Duke walk together, sympathised with Chuck, who walked alone now. A few weeks later, the chemical company promoted Charles Hooper to the post of Assistant National Sales Manager. As a special tribute to Duke, they called this promotion as advancing ‘towards our objective, step by step.’

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Study Material and Summary of The man who knew too much NCERT Class 9th

March 28, 2015, 6:02 am
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Study Material of A Dog named Duke (Summary and Word Meanings)

Summary of the Chapter

Private Quelch was a serious looking, lanky, stooping and bespectacled man whom the writer met at the training centre. He was very fond of showing off his knowledge and was in the habit of sermonising. So he became an object of fun, sarcasm and was nicknamed ‘Professor’. During lessons of musketry, he would interrupt the Sergeant during the lecture and try to correct him. Private Quelch could answer a lot of questions and when others marvelled, his standard answer was, “It's all a matter of intelligent reading.”
In course of time, the writer and his teammates discovered more about Private Quelch. He was very competitive, wanted quick promotion and always tried to appear better than others. He read a lot, attacked his instructors with questions and on every occasion tried to belittle and overshadow others. Private Quelch’s behaviour was very condescending and he would try to appear exceptional before his seniors. Whether it was to identify an aircraft or handling of a rifle, he would tower over others and irritate them. He was unstoppable in his pretentious behaviour.
Corporal Turnbull, who was highly renowned for his toughness, was not a person to be taken lightly. When Quelch tried to correct the corporal, he got offended and asked Quelch to change places with him. Though Quelch delivered a brilliant lecture, it was for sure that he had stirred up trouble. Corporal Turnbull took his revenge by delegating Quelch for permanent cook house duties. But who could stop Quelch, his sermonising continued even within the four walls of the kitchen. However, the writer and his friends had a surge of immense relief that they were temporarily rid of a great nuisance like Private Quelch.

Terms and Meanings from the Chapter

• Private – Soldier without rank
• Musketry - art of using the infantry soldier's handgun
• N.W. Frontier ribbons - decorations showing service in the N.W. province in British India, today a part of modern Pakistan.
• Commission: become an army officer
• Stripe – V-shaped band to indicate the rank of a soldier.
• Route Marches - training marches of battalions
• Orderly Officer – Officer of the day
• Condescending – to look down
• Sprawling - lying with arms and legs outstretched.
• Trifled with - to play with or fool around with, talk or act frivolously with.
• Unabashed – unashamed
• Cowed – subdued

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