ncert11.1

9.MECHANICAL PROPERTIES OF SOLIDS

Q. A steel wire of length 4.7 m and cross-sectional area 3.0 × 10–5 m2 stretches
by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 × 10–5 m2

under 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 × 10–5 m2

Length of the copper wire, L2 = 3.5 m A2 = 4.0 × 10–5 m2 Change in length = L1 = L2 = L
Area of cross-section of the copper wire,

Force applied in both the cases = F Young’s modulus of the steel wire:

Y1 = (F1 / A1) (L1 / L1)
= (F / 3 X 10-5) (4.7 / L) ....(i) Young’s modulus of the copper wire: Y2 = (F2 / A2) (L2 / L 2)
= (F / 4 × 10-5) (3.5 / L) ....(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.

Q.. 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?

Ans
(a) It is clear from the given graph that for stress 150 × 106N/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 ×1010 N/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.
Q.. 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?

Ans.
(a) From the two graphs we note that for a given strain, stress for A is more than that of B. Hence,

(b) Young's modulus

(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.

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

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

Q. 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 × 1100––44 m Diameter of the wires, d = 0.25 m
Elongation of the brass wire = 1.3 × 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) / ( L1 / L1) Where,
L1 = 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 ∴ L1 = 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) / ( L2 / L2) Where,
L2 = Change in the length of the brass wire
A1 = Area of cross-section of the brass wire = πr12 ∴ L2 = 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.

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?

Ans.
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/ L) 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
L = Vertical deflection of the cube
∴ L = 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 ×10–7 m.

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.

Ans.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?

Ans.
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.

Q. 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?

Ans.
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 c
= 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 th
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.

Ans.
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 l 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.

Ans.
Initial volume, V1 = 100.0l = 100.0 × 10 –3 m3
Final volume, V2 = 100.5 l = 100.5 ×10 –3 m3
Increase in volume,V = V2 – V1 = 0.5 × 10–3 m3
Increase in pressure, p = 100.0 atm = 100 × 1.013 × 105 Pa Bulk modulus = p / ( V/V1) = p × V 1 / V
= 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 × 103 kg m–3?

Ans.

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 V be the change in volume.

V = V[ 1(1-/ρV12) - (1/ρ2) ]
= m

∴ Volumetric strain = V / V1

= m [ (1/ρ1) - (1/ρ2) ] × (ρ1 / m) V / V1 = 1 - (ρ1/ρ2) ......(i)

Bulk modulus, B = pV1 / V V / V1 = p / B 10-11 Pa-1
Compressibility of water = (1/B) = 45.8 ×

∴ V / 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.

Q. Compute the fractional change in volume of a glass slab, when subjected to a hydraulic

pressure of 10 atm.

Ans.

Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 ×

1.013 × 105 Pa glass, B = 37 × 109 Nm–2 Bulk modulus, B = p / (∆V/V)
Bulk modulus of

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.

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

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 V = Change in volume

V = Original volume.
V = pV / B
Original volume of the cube, V = l3
∴ V = 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%?

Ans.
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.

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 × 10–3 m
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.

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 A and B are 1.0 mm2 and 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

(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

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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
A attached, 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
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.
Ans
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 × 107 Pa? Assume that each rivet is to carry one quarter
of the load.

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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
= 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.

Q. 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 ×108 Pa. A steel ball of initial volume 0.32 m3 is 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?

Ans.
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 V.
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.

10.MECHANICAL PROPERTIES OF FLUIDS

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)

(e) 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.

(f) 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
= The angle of contact of mercury with glass is obtuse, while that of water with
glass is acute.
= 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.)
= Surface tension of a liquid is independent of the area of the surface
= Water with detergent dissolved in it should have small angles of contact.
= 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).
= 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.

= 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.

= 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.

= 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:

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

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

= 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

= 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 doing 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)2–5 2
= 7.85 × 10 m
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

-5 6 -2
= 490 / (7.85 × 10 ) = 6.24 × 10 Nm
Therefore, the pressure exerted by the heel on the horizontal floor is

6 –2
6.24 × 10 Nm .

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 × 3 3 Height of the mercury column, h1 = 0.76 m Density of French wine, ρ2 = 984 3
10 kg/m kg/m

Height of the French wine column = h2

2
Acceleration due to gravity, g = 9.8 m/s
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

3
= 13.6 × 10 × 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 ocean to be roughly 3 km, and ignore ocean

currents.

Ans.

93
The maximum allowable stress for the structure, P = 10 Pa Depth of the ocean, d = 3 km = 3 × 10 m

Density of water, ρ = 3 3 2
10 kg/m

Acceleration due to gravity, g = 9.8 m/s

The pressure exerted because of the sea water at depth, d = ρdg

33
= 3 × 10 × 10 × 9.8

7
= 2.94 × 10 Pa

9
The maximum allowable stress for the structure (10 Pa) is

7
greater than the pressure of the sea water (2.94 × 10 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

2
piston carrying the load is 425 cm . What maximum pressure would the smaller piston have to bear?

Answer

The maximum mass of a car that can be lifted, m = 3000 kg

2 –4 2
Area of cross-section of the load-carrying piston, A = 425 cm = 425 × 10 m
The maximum force exerted by the load, F = mg
= 3000 × 9.8
= 29400 N
The maximum pressure exerted on the load-carrying

-4
piston, P = F/A 29400 / (425 × 10 )

5
= 6.917 × 10 Pa

Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston

5
would have to bear is 6.917 × 10 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.

Height of the water column, h1 = 10 + 15 = 25 cm Height of the spirit column, h2 = 12.5 + 15 = 27.5 cm Density
–3
of water, ρ1 = 1 g cm

Density of spirit, ρ2 = 0.8 g –3
cm

–3
Density of mercury = 13.6 g cm

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.

Q.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.

Q.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
–3 –1

collected per second at one end is 4.0 × 10 kg s , what is the pressure difference between the two ends of the tube?

(Density

3 –3
of glycerine = 1.3 × 10 kg m 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
–3 –1 –3 –1
Glycerine is flowing at a rate of 4.0 × 10 kg s . M = 4.0 × 10 kg s
–3
Density of glycerine, ρ = 1.3 × 3 kg m Viscosity of glycerine, η = 0.83 Pa s Volume of glycerine flowing per sec:
10

V=M/ρ

-3 3 -6 3 -1
= 4 × 10 / (1.3 × 10 ) = 3.08 × 10 m s

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 × -6 × 8 × 0.83 × 1.5 / [ π × (0.01)4 ]
10
2
= 9.8 × 10 Pa

Reynolds’ number is given by the relation:

R = 4pV / πdη

= 4 × 1.3 × 3 × 3.08 × -6 / ( π × 0.02 × 0.83)
10 10

= 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

–1 –1
wing are 70 m s and 63 m s respectively. What is the lift on

2 –3
the wing if its area is 2.5 m ? Take the density of air to be 1.3 kg m .

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

2
= 63 m/s Area of the wing, A = 2.5 m

Density of air, ρ = 1.3 kg –3
m

According to Bernoulli’s theorem, we have the relation: P1 + (1/2)ρV12 = P2 + (1/2)ρV22
2 2
P2 - P1 = (1/2) ρ (V1 - V2 ) 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

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= (1/2) ρ 2 - 2 A
(V1 V2 )

22
= (1/2) × 1.3 × [ 70 - 633 ] × 2.5
= 1512.87 N = 1.51 × 10 N

3
Therefore, the lift on the wing of the aeroplane is 1.51 × 10 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.

2
10.16. The cylindrical tube of a spray pump has a cross-section of 8.0 cm one end of which has 40 fine holes
each of diameter

–1
1.0 mm. If the liquid flow inside the tube is 1.5 m min , what is the speed of ejection of the liquid through
the holes?

Answer

2–

Area of cross-section of the spray pump, A1 = 8 cm = 8 × 10 4 m2

Number of holes, n = 40

–3 –3
Diameter of each hole, d = 1 mm = 1 × 10 m Radius (o0f.e5a×ch1h0o–le3, )r2=md/22 = 0.5 × 10 m

Area of cross-section of each hole, a = πr2 = π Total area of 40 holes, A2 = n × a
10–3)2
= 40 × π (0.5 × 2 2
–6 m

= 31.41 × 10 m

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

-4 -6
= 8 × 10 × 0.025 / (31.61 × 10 )
= 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

–2
light slider supports a weight of 1.5 × 10 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?

Ans. –2

The weight that the soap film supports, W = 1.5 × 10 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

-2
= 1.5 × 10 / 0.6

-2
= 2.5 × 10 N/m

–2 –1
Therefore, the surface tension of the film is 2.5 × 10 N m .

10.18. Figure 10.24 (a) shows a thin liquid film supporting a
–2

small weight = 4.5 × 10 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

–2
The weight supported by the film, W = 4.5 × 10 N A liquid film has two free surfaces.
∴ Surface tension = W / 2l

-2 -2
= 4.5 × 10 / (2 × 0.4) = 5.625 × 10 N/m
In all the three figures, the liquid is the same. Temperature is also the same for each case. Hence, the surface

–2 –1
tension in figure (b) and figure (c) is the same as in figure (a), i.e., 5.625 × 10 N m .Since the length of the

–2
film in all the cases is 40 cm, the weight supported in each case is 4.5 × 10 N.

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.

Ans.

–3 –1 –1
Radius of the mercury drop, r = 3.00 mm = 3 × 10 m Surface tension of mercury, S = 4.65 × 10 N m Atmospheric

5
pressure, P0 = 1.01 × 10 Pa
Total pressure inside the mercury drop

= Excess pressure inside mercury + Atmospheric pressure

= 2S / r + P0

-1 -3 5
= [ 2 × 4.65 × 10 / (3 × 10 ) ] + 1.01 × 10

= 1.0131 × 510 5

= 1.01 × 10 Pa

Excess pressure = 2S / r

-1 -3
= [ 2 × 4.65 × 10 / (3 × 10 ) ] = 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

–2 –1
soap solution at the temperature (20 °C) is 2.50 × 10 N m ? 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

5
bubble? (1 atmospheric pressure is 1.01 × 10 Pa).

Answer

Excess pressure inside the soap bubble is 20 Pa;

5 –3

Pressure inside the air bubble is 1.06 × 10 Pa Soap bubb–le2is of r–ad1ius, r = 5.00 mm = 5 × 10 m
Surface tension of the soap solution, S = 2.50 × 10 Nm Relative density of the soap solution = 1.20

∴ Density of the soap solution, ρ = 1.2 × 3 3 Air bubble formed at a depth, h = 40 cm = 0.4 m
10 kg/m

–3 5
Radius of the air bubble, r = 5 mm = 5 × 10 m 1 atmospheric pressure = 1.01 × 10 Pa

2
Acceleration due to gravity, g = 9.8 m/s

Hence, the excess pressure inside the soap bubble is given by the relation:

P = 4S / r
-2 -3

= 4 × 2.5 × 10 / 5 × 10

= 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 -3

= 2 × 2.5 × 10 / (5 × 10 )

= 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’

53
= 1.01 × 10 + 0.4 × 1.2 × 10 × 9.8 + 10

5
= 1.06 × 10 Pa

5
Therefore, the pressure inside the air bubble is 1.06 × 10 P

2
10.21. A tank with a square base of area 1.0 m is divided by a vertical partition in the middle. The bottom of the

partition has

2
a small-hinged door of area 20 cm . 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 m 2

Area of the hinged door, a = 20 2 = 20 × –4 2 Density of water, ρ1 = 3 3
cm 10 m 10 kg/m

Density of acid, ρ2 = 1.7 × 3 3 Height of the water column, h1 = 4 m Height of the acid column, h2 = 4 m
10 kg/m

Acceleration due to gravity, g = 9.8 Pressure due to water is given as:

P1 = h1ρ1g

34
= 4 × 10 × 9.8 = 3.92 × 10 Pa Pressure due to acid is given as:

P2 = h2ρ2g
34

= 4 × 1.7 × 10 × 9.8 = 6.664 × 10 Pa
Pressure difference between the water and acid columns:

P = P2 - P1
44

= 6.664 × 10 - 3.92 × 10
4

= 2.744 × 10 Pa
Hence, the force exerted on the door = P × a

4 –4
= 2.744 × 10 × 20 × 10

= 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?

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.

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 –3
m
Density of whole blood, ρ = 1.06 × 3 2kg
10

Acceleration due to gravity, g = 9.8 m/s Height of the blood container = h Pressure of the blood container, P =

hρg ∴ h = P / ρg
3

= 2000 / (1.06 × 10 × 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

–3
flow in an artery of diameter 2 × 10 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
–3

an artery of radius 2 × 10 m if the flow must remain laminar?

–3
(b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 × 10 Pa s).

Answer

–3
(a) Radius of the artery, r = 2 × 10 m
–3 –
Diameter of the artery, d = 2 × 2 × 10 m = 4 × 10 3 m Viscosity of blood, η = 2.084 X 10-3 Pa s

Density of blood, ρ = 1.06 × 3 3 Reynolds’ number for laminar flow, NR = 2000
10 kg/m

The largest average velocity of blood is given by the relation:

Varg = NRη / ρd
-3 3 -3

= 2000 × 2.084 × 10 / (1.06 × 10 × 4 × 10 )

= 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 2
r Vavg
-3 2
= 3.14 × (2 × 10 ) × 0.983
-5 3 -1
= 1.235 × 10 m s

-5 3 -1
Therefore, the corresponding flow rate is 1.235 × 10 m s .

10.27. A plane is in level flight at constant speed and each of its

2
two wings has an area of 25 m . 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 1 12
–3 22
P
to be 1 kg m ). 1
Answer

2
The area of the wings of the plane, A = 2 × 25 = 50 m 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
–3
Density of air, ρ = 1 kg m

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)………(1)

The upward force (F) on the plane can be calculated as: (P1 - P2)A = (1/2) ρ ( 2 - 2 A
V2 V1 )
22
= (1/2) × 1 × (65 - 50 ) × 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 k .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

–5
the experiment to be 1.8 × 10 Pa s. How much is the viscous force on the drop at that speed? Neglect

buoyancy of the drop due to air.

AnS.

–10
Terminal speed = 5.8 cm/s Viscous force = 3.9 × 10 N

Radius of the given uncharged drop, r = 2.0 × –5 m Density of the uncharged drop, ρ = 1.2 × 3 kg –3
-5 10 10 m
10
Viscosity of air, η = 1.8 × 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 2 Terminal velocity (v) is given by the relation: v = 2 × (ρ - ρ0) g / 9η
m/s 2r

-5 2 3 -5
= 2 × (2 × 10 ) (1.2 × 10 - 0 ) × 9.8 / (9 × 1.8 × 10 )
-2
= 5.8 × 10 m/s
-1
= 5.8 cm s
–1
Hence, the terminal speed of the drop is 5.8 cm s . The viscous force on the drop is given by:

F = 6πηrv

-5 -5 -2 -10
∴ F = 6 × 3.14 × 1.8 × 10 × 2 × 10 × 5.8 × 10 = 3.9 × 10 N

–10
Hence, the viscous force on the drop is 3.9 × 10 N.

Q. 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°
–3
Radius of the narrow tube, r = 1 mm = 1 × 10 m
–1
Surface tension of mercury at the given temperature, s = 0.465 N m
3 3
Density of mercury, ρ =13.6 × 10 kg/m Dip in the height of mercury = h
2
Acceleration due to gravity, g = 9.8 m/s

Surface tension is related with the angle of contact and the dip in the height as:

s = hρgr / 2 Cosθ

∴ h = 2s Cosθ / ρgr

0 -3 3
= 2 × 0.465 × Cos140 / (1 × 10 × 13.6 × 10 × 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

–3 -3
Diameter of the first bore, d1 = 3.0 mm = 3 × 10 m Hence, the radius of the first bore, r1 = d1/ 2 = 1.5 × 10 m Diameter

–3 -3
of the first bore, d2 = 6.0 mm = 6 × 10 mm Hence, the radius of the first bore, r2 = d2/ 2 = 3 × 10 m
–2 –1
Surface tension of water, s = 7.3 × 10 N m
–3
Angle of contact between the bore surface and water, θ= 0 Density of water, ρ =1.0 × 3 kg/m
2 10
Acceleration due to gravity, g = 9.8 m/s

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:

-3
= 4.966 × 10 m
= 4.97 manually or automatically with PDFmyURL

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 g remains
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 ]

3
Ans.(a) Volume of the balloon, V = 1425 m Mass of the payload, m = 400 kg

2
Acceleration due to gravity, g = 9.8 m/s

y0 = 8000 m
-3
ρHe = 0.18 kg m ....(i)

ρ0 = 1.25 kg -3
m

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

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

-3
= 0.46 kg m
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.

11.THERMAL PROPERTIES OF MATTER

Q. 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
(c) (9/5) × (-248.58) +32

0
(d) 415.44 F

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

0
(g) (9/5) × (-56.6 ) +32

0
(g) -69.88 C

Q. 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,
R0 and T0 are the initial resistance and temperature respectively

R and 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 × -3 -1
10 K

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

-3
123.4 = 101.6 [ 1 + 1.92 × 10 ( T - 273.15) ] Solving for T, we get
T = 384.61 K.
11.4. Answer the following:

= 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)?
= 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?

= The absolute temperature (Kelvin scale) T is related to the temperature tc on 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

= 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.

= The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.

= 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 - T K) / 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

11.5. Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following
observations are made:

Temperature Pressure Pressure

Triple-point of thermometer A thermometer B
water

1.250 ×105 Pa 0.200 ×105 Pa

Normal melting

1.797 ×105 Pa 0.287 ×105 Pa

point of sulphur

= What is the absolute temperature of normal melting point of sulphur as read by thermometers
Aand 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.
5

At this temperature, pressure in thermometer A, PA = 1.250 × 10

Pa

Let T1 be the normal melting point of sulphur.

5
At this temperature, pressure in thermometer A, P1 = 1.797 × 10

Pa
According to Charles’ law, we have the relation:

PA / T = P1 / T1 5
10 )
∴ T1 = (P1T) / PA = (1.797 × 10 × 273.16) / (1.250 ×

= 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 =
5

0.200 × 10 Pa

At temperature T1, the pressure in thermometer B, P2 = 0.287 ×

5
10 Pa
According to Charles’ law, we can write the relation:

PB / T = P1 / T1
55

(0.200 × 10 ) / 273.16 = (0.287 × 10 ) / T1

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55
∴ T1 = [ (0.287 × 10 ) / (0.200 × 10 ) ] × 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

–5 –1
expansion of steel = 1.20 × 10 K .

Ans.
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 × –5 –1
10 K

Let l2 be the actual length of the steel rod and l' be the length of

the steel tape at 45°C. l' =l+ αl(T1 - T)

-5
∴ l' = 100 + 1.20 × 10 × 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:automatically!

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: α steel = 1.20 × –5 –1
10 K.

Ans.
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 = –5 –1

1.20 × 10 K

After the shaft is cooled using ‘dry ice’, its temperature becomes T1.
The wheel will slip on the shaft, if the change in diameter, d = 8.69 – 8.70

= – 0.01 cm

Temperature T1, can be calculated from the relation:

d= d1αsteel (T1 – T) (T1 – 300) (T1 – 300) = 95.78
0.01 = 8.70 × 1.20 × –5

10

∴ 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

–5 –1
linear expansion of copper = 1.70 × 10 K .

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 –5 –1
10 K
Co-efficient of linear expansion of copper, αCu= 1.70 ×

For co-efficient of superficial expansion β,and change in temperature T, we have the relation:
Change in area (∆) / Original area (A) = β∆T [ (πd22/ 4) - (πd12 / 4) ] / (πd11 / 4) = ∆A / A

= ∆A / A = 2 - 2 / 2 β = 2α
(d2 d1 ) d1 But

= 2 2 / 2 = 2α∆T
(d2 - d1 ) d1

22
(d2 / d1 ) - 1 = 2α(T2 - T1)

22 -5 2
d2 / 4.24 = 2 × 1.7 × 10 (227 - 27) +1 d2 = 17.98 × 1.0068 = 18.1

∴ d2 = 4.2544 cm d2 – d1 = 4.2544 – 4.24 = 0.0144 cm Hence, the diameter increases by 1.44 × –2 cm.nvert a
in diameter = 10
Change

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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
–5 K–1;
of brass = 2.0 × 10 Young’s modulus of brass = 0.91 × 1011 Pa.

Answer

Initial temperature, T1 = 27°C

Length of the brass wire at T1, l = 1.8 m Final temperature, T2 = –39°C
–3
Diameter of the wire, d = 2.0 mm = 2 × 10 m Tension developed in the wire = F
–5 –1
Coefficient of linear expansion of brass, α = 2.0 × 10 K

Young’s modulus of brass, Y = 0.91 × 11 Pa Young’s modulus is given by the relation: γ = Stress / Strain = (F/A) /
10

(∆L/L)

∆L = F X L / (A X Y) ......(i)

Where,

F = Tension developed in the wire

= = Area of cross-section of the wire.

L = Change in the length, given by the relation:

L = αL(T2 – T1) … (ii)

Equating equations (i) and (ii), we get: αL(T2 - T1) = FL / [ π(d/2)2 X Y ]

= = α(T2 - T1)πY(d/2)2 2
-5 11 -3 2
== 2 × 10 × (-39-27) × 3.14 × 0.91 × 10 × (2 × 10 / 2 ) = -3.8 × 10 N

(The negative sign indicates that the tension is directed
inward.)

2
Hence, the tension developed in the wire is 3.8 ×10 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

–5 – 1 –5 –1
expand (Co-efficient of linear expansion of brass = 2.0 × 10 K , steel = 1.2 × 10 K ).

Answer

Initial temperature, T1 = 40°C Final temperature, T2 = 250°C

Change in temperature, T = T2 – T1 = 210°C Length of the brass rod at T1, l1 = 50 cm Diameter of

the brass rod at T1, d 1 = 3.0 mm Length of the steel rod at T2, l2 = 50 cm Diameter of the steel rod

at T2, d 2 = 3.0 mm –5 –1
10 K
Coefficient of linear expansion of brass, α1 = 2.0 ×

Coefficient of linear expansion of steel, α2 = 1.2 × –5 –1
10 K

For the expansion in the brass rod, we have:

Change in length (∆l1) / Original length (l1) = α1 T
-5
∴ ∆l1 = 50 × (2.1 × 10 ) × 210

= 0.2205 cm

For the expansion in the steel rod, we have:

Change in length (∆l2) / Original length (l2) = α2 T
-5
∴ ∆l1 = 50 × (1.2 × 10 ) × 210

= 0.126 cm

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Total change in the lengths of brass and steel, l = l1 + l2
= 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.

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

Answer

Coefficient of volume expansion of glycerin, αV = 49 × –5 –1
10 K

Rise in temperature, T = 30°C Fractional change in its volume = V/V

This change is related with the change in temperature as:

V/V = αV T α T
V -V =V

T2 T1 T1 V

(m /ρT2)- (m /ρT1) = (m /ρT1)αV T Where,

m = Mass of glycerine ρT1 = Initial density at T1

ρT2 = Initial density at T2

(ρT1 - ρT2 ) / ρT2 = Fractional change in density

–5 –2
∴ Fractional change in the density of glycerin = 49 ×10 × 30 = 1.47 × 10 .

Q.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–1 K–1.

Ans.
3

Power of the drilling machine, P = 10 kW = 10 × 10 W
3

Mass of the aluminum block, m = 8.0 kg = 8 × 10 g

Time for which the machine is used, t = 2.5 min = 2.5 × 60 = 150 s

–1 –1
Specific heat of aluminium, c = 0.91 J g K

Rise in the temperature of the block after drilling = δT Total energy of the drilling machine = Pt
3

= 10 × 10 × 150
6

= 1.5 × 10 J

It is given that only 50% of the power is useful. 5
6 10
Useful energy, ∆Q = (50/100) × 1.5 × 10 = 7.5 × J But ∆Q = mc∆T

∴ ∆T = ∆Q / mc

53
= (7.5 × 10 ) / (8 × 10 × 0.91)

o
= 103 C

Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

Q.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).

Ans. θ
Mass of the copper block, m = 2.5 kg = 2500 g
Rise in the temperature of the copper block, θ = 500°C

–1 –1
Specific heat of copper, C = 0.39 J –g1 °C
Heat of fusion of water, L = 335 J g
The maximum heat the copper block can lose, Q = mC

= 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.

Q. 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 coppercalorimeter (of water equivalent 0.025 kg)
containing 150 cm3 of 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?

Ans.

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

3
Volume of water, V = 150 cm
Mass (M) of water at temperature T = 27°C: 150 × 1 = 150 g

Fall in the temperature of the metal: T = T1 – T2 = 150 – 40 = 110°C

Specific heat of water, Cw = 4.186 J/g/°K Specific heat of the metal = C
aHeat lost by the metal, θ = mC T … (i)

Rise in the temperature of the water and calorimeter system:

’
T′ = 40 – 27 = 13°C

’
Heat gained by the water and calorimeter system: Δθ′′ = m1 Cw T

= (M + m′) Cw ’
T … (ii)

Heat lost by the metal = Heat gained by the water and colorimeter system
’’

mC T = (M + m ) Cw T

200 × C × 110 = (150 + 25) × 4.186 × 13

∴ C = (175 × 4.186 × 13) / (110 × 200) = 0.43 -1 -1
Jg K

If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

Q.Given below are observations on molar specific heats at room temperature of some common

gases.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?

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

-1 -1
= (5/2) × 1.98 = 4.95 cal mol K
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.
Q. Answer the following questions based on the P-T phase diagram of carbon dioxide:

= At what temperature and pressure can the solid, liquid and vapour phases of CO2 co-exist in
equilibrium?

= What is the effect of decrease of pressure on the fusion and boiling point of CO2?

= 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?
Ans.

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

= 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.

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

= 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.

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

Q.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?

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

= 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.

= 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

–1
580 cal g .

Ans.

Initial temperature of the body of the child, T1 = 101°F Final temperature of the body of the child, T2 = 98°F
o

Change in temperature, T = [ (101 - 98) × (5/9) ] C Time taken to reduce the temperature, t = 20 min

3
Mass of the child, m = 30 kg = 30 × 10 g

Specific heat of the human body = Specific heat of water = c
= 1000 cal/kg/ °C

–1
Latent heat of evaporation of water, L = 580 cal g 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 thermocole is 0.01 J –1 –1 –1 [Heat of fusion of
s m K.

3 –1
water = 335 × 10 J kg ]

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
–1

Coefficient of thermal conductivity of thermacole, K = 0.01 J s

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m–1 K–1

3 –1
Heat of fusion of water, L = 335 × 10 J kg

’
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 = 2 = 6 × 2 = 0.54 3 θ = 0.01 × 0.54 × 45 × 6 × 60 × 60 / 0.05 = 104976
6s (0.3) m

J But θ = m'L

∴ m' = θ/L

= 104976/(335 × 3 = 0.313 kg Mass of ice left = 4 – 0.313 = 3.687 kg
10 )

Hence, the amount of ice remaining after 6 h is 3.687 kg.

2
11.20. A brass boiler has a base area of 0.15 m and 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

–1 –1 –1 3 –1
m K ; Heat of vaporisation of water = 2256 × 10 J kg .

Answer

2
Base area of the boiler, A = 0.15 m 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 –1 –1 –1

Thermal conductivity of brass, K = 109 J s–1 m K
3
Heat of vaporisation, L = 2256 × 10 J kg

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
3

= 6 × 2256 × 10 × 0.01 / (109 × 0.15 × 60)
o

= 137.98 C
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

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. 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:
(T4 4
= = σ - T0 ) Where,

= = 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 = σ 4
T

(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).

Q. 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.

Ans.
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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Q. 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, T = T2 – T1 = 77 – 27 = 50°C
Heat of combustion = 4 × 104
Specific heat of water, c = 4.2 JJ/gg–1 °C–1

Mass of flowing water, m = 3.0 litre/min = 3000 g/min Total heat used, Q = mc T

(e) 3000 × 4.2 × 50

(f) 6.3 × 105 J/min
∴ Rate of consumption = 6.3 × 105 / (4 × 104) = 15.75 g/min.

Q. 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.)

Ans.
Mass of nitrogen, m = 2.0 × 10–2 kg = 20 g

Rise in temperature, T = 45°C

Molecular mass of N2, M = 28 J mol–1 K–1
Universal gas constant, R = 8.3

Number of moles, n = m/M
(h) (2 × 10-2 × 103) / 28

(i) 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:

Q = nCP T
= 0.714 × 29.05 × 45

= 933.38 J

Therefore, the amount of heat to be supplied is 933.38 J.

Q. Explain why
(a) Two bodies at different temperatures T1 and T2 if brought in
thermal contact do not necessarily settle to the mean temperature (T1 + T 2)/2.
A.The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts
of a
B. 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.

Ans.
(a) In thermal contact, heat flows from the body at higher temprature to the body at lower temprature
(b) till tempratures becomes equal. The final temprature can be the mean temprature (T1 + T2)/2only
(c) when thermal capicities of the two
bodies are equal.
This is because heat absorbed by a substance is directly proportional to the specific heat of the

substance.

D.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.

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.
Q. 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?
Ans.
The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between
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.
∴Q = 0

W = –22.3 J (Since the work is done on the system)
From the first law of thermodynamics, we have:

Q= U+ W
Where,

U = Change in the internal energy of the gas
∴U = Q – W = – (– 22.3 J)

U = + 22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is:

Q = 9.35 cal = 9.35 × 4.19 = 39.1765 J Heat absorbed, Q = U + Q
∴ W= Q– U
= 39.1765 – 22.3
= 16.8765 J
Therefore, 16.88 J of work is done by the system.

Q. 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
Aa. 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?
Ans.
Work done by the steam engine per minute, W = 5.4 × 108 J
Heat supplied from the boiler, H = 3.6 × 109 J 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 × 109 J
Therefore, the amount of heat wasted per minute is 3.06 × 109 J.
Q. 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?
Ans.
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.

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 DEF Area of DEF = (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 DEF = (1/2) × 300 × 3 = 450 J
Therefore, the total work done by the gas from D to E to F is 450 J.PDFmyURL converts any url to pdf

Q. A refrigerator is to maintain eatables kept inside at 9°C. If room temperature

is 36° C, calculate the coefficient of performance.

Ans.
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.

13.KINETIC THEORY

Q.Estimate the fraction of molecular volume to the actual volume occupied

by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3Å.

Answer

–8
Diameter of an oxygen molecule, d = 3Å Radius, r = d/2 = 3/2 = 1.5 Å = 1.5 × 10 cm

3
Actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm
Molecular volume of oxygen gas, V = (4/3)πr3N

Where, N is Avogadro’s number = 6.023 × 1023 molecules/mole

∴ V = (4/3) × 3.14 × (1.5 × -8 3 × 6.023 × 23 = 8.51 3
10 ) 10 cm

Ratio of the molecular volume to the actual volume of oxygen

(g) 8.51 / 22400
-4

(h) 3.8 × 10 .

Q. Molar volume is the volume occupied by 1 mol of any (ideal) gas at
standard temperature and pressure (STP: 1 atmospheric pressure, 0 °C).
Show that it is 22.4 litres.

Ans.
The ideal gas equation relating pressure (P), volume (V), and absolute
temperature (T) is given as:
PV = nRT
Where,

–1 –1
R is the universal gas constant = 8.314 J mol K n = Number of moles = 1
T = Standard temperature = 273 K

5 –2
P = Standard pressure = 1 atm = 1.013 × 10 Nm
∴ V = nRT / P

5

(j) 1 × 8.314 × 273 / (1.013 × 10 )

3
(k) 0.0224 m
(l) 22.4 litres
Hence, the molar volume of a gas at STP is 22.4 litres.

Q. Figure 13.8 shows plot of PV/T versus P for 1.00×10–3 kg of
oxygen gas at two different temperatures.

A. What does the dotted plot signify?
B.Which is true: T1 > T2 or T1 < T2?
C.What is the value of PV/T where the curves meet on the y-axis?
D.If we obtained similar plots for 1.00 ×10–3 kg of hydrogen, would we get
the same value of PV/T at the point where the curves meet on the y-axis?
If not, what mass of hydrogen yields the same value of PV/T (for low
pressure high temperature region of the plot)? (Molecular mass of H2 =
2.02μ, of O2 =32.0μ, R = 8.31 J mo1–1 K–1.)

Answer

=The dotted plot in the graph signifies the ideal behaviour of the gas, i.e., the ratio PV/T is
equal.
μR (μ is the number of moles and R is the universal gas constant) is a constant quality. It is
not dependent on the pressure of the gas.

=The dotted plot in the given graph represents an ideal gas. The curve of the gas at temperature T1
is closer to the dotted
plot than the curve of the gas at temperature T2. A real gas
approaches the behaviour of an ideal gas when its temperature increases. Therefore, T1 > T2 is true
for the given plot.

(c) The value of the ratio PV/T, where the two curves meet, is μR. This is because the ideal gas
equation is given as:

PV = μRT

PV/T = μR Where,

P is the pressure
T is the temperature V is the volume
μ is the number of moles R is the universal constant

–3 –1 –1
Molecular mass of oxygen = 32.0 g Mass of oxygen = 1 × 10 kg = 1 g R = 8.314 J mole K

-1
∴ PV/T = (1/32) × 8.314 = 0.26 J K
Therefore, the value of the ratio PV/T, where the curves meet on the y-axis, is

–1
0.26 J K .

–3
(d) If we obtain similar plots for 1.00 × 10 kg of hydrogen, then we will not get the same
value of PV/T at the point where the curves meet the y-axis. This is because the molecular
mass of hydrogen (2.02 u) is different from that of oxygen (32.0 u). We have:

-1
PV/T = 0.26 J K –1 –1
R = 8.314 J mole K
Molecular mass (M) of H2 = 2.02 u
PV/T = μR at constant temperature Where, μ = m/M

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m = Mass of H2
∴ m = (PV/T) × (M/R)

= 0.26 × 2.02 / 8.31
–2 –5

= 6.3 × 10 g = 6.3 × 10 kg
–5

Hence, 6.3 × 10 kg of H2 will yield the same value of PV/T.

13.4. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm and a

temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressure

drops to 11 atm and its temperature drops to 17 °C. Estimate

–1 –1
the mass of oxygen taken out of the cylinder (R = 8.31 J mol K , molecular mass

of O2 = 32μ).

Answer

–3 3
Volume of oxygen, V1 = 30 litres = 30 × 10 m

Gauge 5
pressure, P1 = 15 atm = 15 × 1.013 × 10 Pa Temperature, T1 = 27°C = 300 K
–1 –1
Universal gas constant, R = 8.314 J mole K

Let the initial number of moles of oxygen gas in the cylinder be n1.
The gas equation is given as:

P1V1 = n1RT1
∴ n1 = P1V1/ RT1

5 -3

= (15.195 × 10 × 30 × 10 ) / (8.314 × 300) = 18.276 But n1 = m1 / M

Where,

m1 = Initial mass of oxygen

M = Molecular mass of oxygen = 32 g ∴m1 = n1M = 18.276 × 32 = 584.84 g

After some oxygen is withdrawn from the cylinder, the pressure and temperature reduces.
–3 3

Volume, V2 = 30 litres = 30 × 10 m
5

Gauge pressure, P2 = 11 atm = 11 × 1.013 × 10 Pa Temperature, T2 = 17°C = 290 K

Let n2 be the number of moles of oxygen left in the cylinder. The gas equation is given as:

P2V2 = n2RT2
∴ n2 = P2V2/ RT2

5 -3

= (11.143 × 10 × 30 × 10 ) / (8.314 × 290) = 13.86 But n2 = m2 / M

Where,

m2 is the mass of oxygen remaining in the cylinder
∴ m2 = n2M = 13.86 × 32 = 453.1 g
The mass of oxygen taken out of the cylinder is given by the relation:
Initial mass of oxygen in the cylinder – Final mass of oxygen in the cylinder

= m1 – m2
= 584.84 g – 453.1 g
= 131.74 g
= 0.131 kg
Therefore, 0.131 kg of oxygen is taken out of the cylinder.

3
Q. An air bubble of volume 1.0 cm rises from the bottom of a lake 40 m deep at a

temperature of 12 °C. To what volume does it grow when it reaches the surface,
which is at a temperature of 35 °C?
Answer

3 –6 3
Volume of the air bubble, V1 = 1.0 cm = 1.0 × 10 m

Bubble rises to height, d = 40 m

Temperature at a depth of 40 m, T1 = 12°C = 285 K Temperature at the surface of the lake, T2

= 35°C = 308 K The pressure on the surface of the lake:

5
P2 = 1 atm = 1 ×1.013 × 10 Pa

The pressure at the depth of 40 m: P1 = 1 atm + dρg

Where,

ρ is the density of water = 3 3
10 kg/m

g is the acceleration due to gravity = 9.8 2 ∴ P1 = 1.013 × 5 + 40 × 3 × 9.8 = 493300 Pa
m/s 10 10

We have P1V1 / T1 = P2V2 / T2

Where, V2 is the volume of the air bubble when it reaches the surface

V2 = P1V1T2 / T1P2
-6 5

= 493300 × 1 × 10 × 308 / (285 × 1.013 × 10 )

–6 3 3
= 5.263 × 10 m or 5.263 cm

3
Therefore, when the air bubble reaches the surface, its volume becomes 5.263 cm .

Q.Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour

and other constituents) in a

3
room of capacity 25.0 m at a temperature of 27 °C and 1 atm pressure.

3
Ans. Volume of the room, V = 25.0 m Temperature of the room, T = 27°C = 300 K

5
Pressure in the room, P = 1 atm = 1 × 1.013 × 10 Pa
The ideal gas equation relating pressure (P), Volume (V), and absolute temperature (T) can
be written as:

PV = kBNT
Where,

–23 2 –2 –1
KB is Boltzmann constant = 1.38 × 10 m kg s K N is the number of air molecules in the
room
∴ N = PV / kBT

5 -23
= 1.013 × 10 × 25 / (1.38 × 10 × 300)

26
= 6.11 × 10 molecules

26
Therefore, the total number of air molecules in the given room is 6.11 × 10 .

13.7. Estimate the average thermal energy of a helium atom at
(i) room temperature (27 °C), (ii) the temperature on the surface of the Sun (6000 K), (iii) the

temperature of 10 million Kelvin (the typical core temperature in the case of a star).

Answer

(i) At room temperature, T = 27°C = 300 K Average thermal energy = (3/2)kT

–23 2 –2 –1
Where k is Boltzmann constant = 1.38 × 10 m kg s K ∴ (3/2)kT = (3/2) × 1.38 ×

-38
10 × 300–21
= 6.21 × 10 J
Hence, the average thermal energy of a helium atom at room temperature (27°C) is 6.21 ×

–21
10 J.

(ii) On the surface of the sun, T = 6000 K Average thermal energy = (3/2)

-38
=(3/2) × 1.38 × 10 × 6000

-19
= 1.241 × 10 J
Hence, the average thermal energy of a helium atom on the surface of the sun is 1.241 ×

–19
10 J.

7
(iii) At temperature, T = 10 K Average thermal energy = (3/2)kT

-23 7
= (3/2) × 1.38 × 10 × 10

-16
= 2.07 × 10 J

–16
Hence, the average thermal energy of a helium atom at the core of a star is 2.07 × 10 J.

Q. Three vessels of equal capacity have gases at the same temperature and pressure.
The first vessel contains neon (monatomic), the second contains chlorine (diatomic),
and the third contains uranium hexafluoride (polyatomic). Do the vessels contain equal
number of respective molecules? Is the root mean square speed of molecules the same

in the three cases? If not, in which case is vrms the largest?

Ans.All the three vessels have the same capacity, they have the same volume.

Hence, each gas has the same pressure, volume, and temperature.
According to Avogadro’s law, the three vessels will contain an equal number of the respective

molecules. This number is

equal to Avogadro’s number, N = 6.023 × 23
10 .

The root mean square speed (vrms) of a gas of mass m, and temperature T, is given by the
relation:

1/2
vrms = (3kT/m)
where,
k is Boltzmann constant
For the given gases, k and T are constants.

Hence vrms depends only on the mass of the atoms, i.e.,
1/2

vrms ∝ (1/m)
Therefore, the root mean square speed of the molecules in the three cases is not the same.

Among neon, chlorine, and uranium hexafluoride, the mass of neon is the smallest. Hence,
neon has the largest root mean square speed among the given gases.

Q.At what temperature is the root mean square speed of an atom in an argon gas
cylinder equal to the rms speed of a helium gas atom at – 20 °C? (atomic mass of Ar
= 39.9 u, of He = 4.0 u).

Ans.Temperature of the helium atom, THe = –20°C= 253 K Atomic mass of argon, MAr = 39.9u

Atomic mass of helium, MHe = 4.0 u.Let, (vrms)Ar be the rms speed of argon. Let (vrms)He be the rms speed of

helium. The rms speed of argon is given by:

3
= 2523.675 = 2.52 × 10 K

3
Therefore, the temperature of the argon atom is 2.52 × 10 K.

Q.Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder
containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be
roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two

successive collisions (Molecular mass of N2 = 28.0 u).

Answer –7

Mean free path = 1.11 × 10 m –1
9
Collision frequency = 4.58 × 10 s

Successive collision time ≈ 500 × (Collision time)

Pressure inside the cylinder containing nitrogen, P = 2.0 atm =

5
2.026 × 10 Pa

Temperature inside the cylinder, T = 17°C =290 K

10 10 10
Radius of a nitrogen molecule, r = 1.0 Å = 1 × 10 m Diameter, d = 2 × 1 × 10 = 2 × 10 m

–3
Molecular mass of nitrogen, M = 28.0 g = 28 × 10 kg The root mean square speed of

nitrogen is given by the relation:

Collision frequency = vrms / l
-7 9 -1

= 508.26 / (1.11 × 10 ) = 4.58 × 10 s Collision time is given as:
T = d / vrms

-10 -13
= 2 × 10 / 508.26 = 3.93 × 10 s

Time taken between successive collisions: T ' = l / vrms
-7 -10

= 1.11 × 10 / 508.26 = 2.18 × 10 s
-10 -13

∴ T ' / T = 2.18 × 10 / (3.93 × 10 ) = 500
Hence, the time taken between successive collisions is 500 times the time taken for a collision.

Q. A metre long narrow bore held horizontally (and closed at one end) contains a 76 cm
long mercury thread, which traps a 15 cm column of air. What happens if the tube is held
vertically with the open end at the bottom?

Ans.
Length of the narrow bore, L = 1 m = 100 cm Length of the mercury thread, l = 76 cm

Length of the air column between mercury and the closed end, la = 15 cm
Since the bore is held vertically in air with the open end at the bottom, the mercury length
that occupies the air space is: 100 – (76 + 15) = 9 cm.Hence, the total length of the air

column = 15 + 9 = 24 cm Let h cm of mercury flow out as a result of atmospheric pressure.

∴Length of the air column in the bore = 24 + h cm And, length of the me3rcury column = 76 – h cm
Initial pressure, P1 = 76 cm of mercury Initial volume, V1 = 15 cm
3
Final pressure, P2 = 76 – (76 – h) = h cm of mercury Final volume, V2 = (24 + h) cm

Temperature remains constant throughout the process.

∴ P1V1 = P2V2

76 × 15 = h (24 + h) 2 + 24h – 1140 = 0
h

= 23.8 cm or –47.8 cm

Height cannot be negative. Hence, 23.8 cm of mercury will flow out from the bore and 52.2 cm of

mercury will remain in it. The length of the air column will be 24 + 23.8 = 47.8 cm.

Q. From a certain apparatus, the diffusion rate of hydrogen
3 –1

has an average value of 28.7 cm s . The diffusion of another gas under the same

conditions is measured to have an average

3 –1
rate of 7.2 cm s . Identify the gas.

[Hint: Use Graham’s law of diffusion: R1/R2 = 1/2 where R1, R2 are diffusion rates of
(M2/M1) ,

gases 1 and 2, and M1 and M2 their

respective molecular masses. The law is a simple consequence of kinetic theory.]
3 –1

Ans. Rate of diffusion of hydrogen, R1 = 28.7 cm s

Rate of diffusion of another gas, R2 = 7.2 3 –1 According to Graham’s Law of diffusion, we have:
cm s