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Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 9 Mechanical Properties of Solids Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Solids

PSEB 11th Class Physics Guide Mechanical Properties of Solids Textbook Questions and Answers

Question 1.
A steel wire of length 4.7 m and cross-sectional area 3.0 x 10^-5 m² stretches by the same amount as a copper wire of length 3.5 m and cross-sectional area of 4.0 x 10^-5 m² under a given load. What is the ratio of the Young’s modulus of steel to that of copper?
Solution:
Given, length of the steel wire, L₁ = 4.7 m
Area of cross-section of the steel wire,A₁ = 3.0 x 10^-5 m²
Length of the copper wire, L₂ = 3.5 m
Area of cross-section of the copper wire, A₂ = 4.0 x 10^-5 m²

Change in length = ΔL₁ = ΔL₂ = ΔL
Force applied in both the cases = F
Young’s modulus of the steel wire,
Y₁ = \(\frac{F_{1}}{A_{1}} \times \frac{L_{1}}{\Delta L_{1}} \)
= \(\frac{F \times 4.7}{3.0 \times 10^{-5} \times \Delta L} \) ………………………………. (i)
Young’s modulus of the copper wire,
Y₂ = \(\frac{F_{2}}{A_{2}} \times \frac{L_{2}}{\Delta L_{2}}=\frac{F \times 3.5}{4.0 \times 10^{-5} \times \Delta L}\) ………………………………. (ii)
Dividing eq. (i) by eq. (ii), we get:
\(\frac{Y_{1}}{Y_{2}}=\frac{4.7 \times 4.0 \times 10^{-5}}{3.0 \times 10^{-5} \times 3.5}\) = 1.79:1
The ratio of Young’s modulus of steel to that of copper is 1.79: 1.

Question 2.
Figure given below shows the strain-stress curve for a given material. What are (a) Young’s modulus and (b) approximate yield strengths for this material?

Solution:
(a) It is clear from the given graph that for stress 150 x 10⁶ N/m², strain is 0.002.
∴ Young s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }}\)
= \(\frac{150 \times 10^{6}}{0.002}\) = 7.5 x 10¹⁰ N/m²
Hence, Young’s modulus for the given material is 7.5 x10¹⁰ 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 x 10⁶ N/m² or 3 x 10⁸ N/m².

Question 3.
The stress-strain graphs for materials A and B are shown in figure given below.

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?
Solution:
(a) A, for a given strain, the stress for material A is more than it is for > material B, as shown in the two graphs.
Young’s modulus = \(\frac{\text { Stress }}{\text { Strain }}\)
For a given strain, if the stress for a material is more, then Young’s modulus is also greater for that material. Therefore, Young’s modulus for material A is greater than it is for material B.

(b) A, the amount of stress required for fracturing a material, corresponding to its fracture point, gives the strength of that material. Fracture point is the extreme point in a stress-strain curve. It can be observed that material A can withstand more strain than material B. Hence, material A is stronger than material B.

Question 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.
Solution:
(a) False.
Reason: For a given stress, the strain in rubber is more than it is in steel.
Young’s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }} \)
For a constant stress Y ∝ \(\frac{1}{\text { Strain }}\)
Hence, Young’s modulus for rubber is less than it is for steel.

(b) True.
Reason: Shear modulus is the ratio of the applied stress to the change in the shape of a body. The stretching of a coil changes its shape. Hence, shear modulus of elasticity is involved in this process.

Question 5.
Two wires of diameter 0.25 cm, one made of steel and the other made of brass are loaded as shown in figure given below. 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.

Solution:
Given, diameter of the wires, d = 0.25 cm
Hence, the radius of the wires, r = \(\frac{d}{2} \) = 0.125cm = 0.125 x 10^-2m
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 = 98N .

Young’s modulus for steel
Y₁ = \(\frac{\left(\frac{F_{1}}{A_{1}}\right)}{\left(\frac{\Delta L_{1}}{L_{1}}\right)} \)
where, ΔL₁ = Change in the length of the steel wire
A₁ = Area of cross-section of the steel wire = πr²₁

Young’s modulus of steel, Y₁ = 2.0 x 10¹¹ Pa
∴ ΔL₁ = \(\frac{F_{1} \times L_{1}}{A_{1} \times Y_{1}}=\frac{F_{1} \times L_{1}}{\pi r_{1}^{2} \times Y_{1}}\)
= \(\frac{98 \times 1.5}{3.14\left(0.125 \times 10^{-2}\right)^{2} \times 2 \times 10^{11}}\)
= 1.5 x 10^-4 m

Total force on the brass wire
F₂ =6 x 9.8=58.8N
Young’s modulus for brass
Y₂ = \(\frac{\left(\frac{F_{2}}{A_{2}}\right)}{\left(\frac{\Delta L_{2}}{L_{2}}\right)}\)

where, ΔL₂ = Change in length of the steel wire
A₂ = Area of cross-section of the brass wire
∴ ΔL₂ = \(\frac{F_{2} \times L_{2}}{A_{2} \times Y_{2}}=\frac{F_{2} \times L_{2}}{\pi r_{2}^{2} \times Y_{2}}\)
= \(\frac{58.8 \times 1.0}{3.14 \times\left(0.125 \times 10^{-2}\right)^{2} \times\left(0.91 \times 10^{11}\right)}\)
= 1.3 x 10^-4 m
Hence, elongation of the steel wire =1.49 x 10^-4 m
and elongation of the brass wire = 1.3 x 10^-4 m

Question 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 the attached to the opposite face of the cube. The shear modulus of aluminium is 25 GPa. What is the vertical deflection of this face?
Solution:
Given, edge of the aluminium cube, L = 10cm = 0.1 m
The mass attached to the cube, m =100 kg
Shear modulus (ri) of aluminium = 25GPa =25 x 10⁹ Pa
Shear modulus, η = \(\frac{\text { Shear stress }}{\text { Shear strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\)
where, F = Applied force = mg = 100 x 9.8 = 980 N
A = Area of one of the faces of the cube = 0.1 x 0.1 = 0.01 m²

ΔL = Vertical deflection of the cube
ΔL = \(\frac{F L}{A \eta}=\frac{980 \times 0.1}{10^{-2} \times\left(25 \times 10^{9}\right)}\)
= 3.92 x 10^-7 m
The vertical deflection of this face of the cube is 3.92 x 10^-7 m.

Question 7.
Four identical tblloW 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.
Solution:
Given, mass of the big structure, M = 50000 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 x 10¹¹ Pa
Total force exerted, F =Mg = 50000 x 9.8N
Stress = Force exerted on a single column = \(\frac{50000 \times 9.8}{4}\) = 122500 N

Young’s modulus, Y = \(\frac{\text { Stress }}{\text { Strain }} \)
Strain = \(\frac{F}{\frac{A}{Y}} \)
where, Area, A = π(R² – r²) = π[(0.6)² – (0.3)²]
Strain = \(\frac{122500}{3.14\left[(0.6)^{2}-(0.3)^{2}\right] \times 2 \times 10^{11}}\) = 7.22 x 10^-7
Hence, the compressional strain of each column is 7.22 x 10^-7.
∴Compressional strain of all columns is given by
= 7.22 x 10 ^-7 x 4 = 2.88 x 10^-6.

Question 8.
A piece of copper having a rectangular cross-section of 15.2 minxes 19.2 mm is pulled in tension with 44,500 N force, producing only elastic deformation. Calculate the resulting strain?
Solution:
Given, cross-section area.of copper piece (A) = 15.2 mm x 19.1 mm
= (15.2 x 19.1) x 10 ^-6m²
Force applied (F) = 44500 N
Young’s modulus (Y) =1.1 x 10¹¹ Nm^-2

Young s modulus (Y) = \(=\frac{\text { Longitudinal stress }}{\text { Longitudinal strain }}\)
or Longitudinal strain = \(\frac{\text { Longitudinal stress }}{\text { Young’s modulus }}\)
Young’s modulus
= \(\frac{(F / A)}{Y}=\frac{F}{A Y}\)
= \(\frac{44500}{15.2 \times 19.1 \times 10^{-6} \times 1.1 \times 10^{11}}\)
= 0.0013934.

Question 9.
A steel cable with a radius of 1.5 cm supports a chairlift at a ski area. If the maximum stress is not to exceed 10⁸ Nm ^-2, what is the maximum load the cable can support?
Solution:
Radius of the steel cable, r = 1.5cm = 0.015m
Maximum allowable stress = 10⁸ N m^-2
Maximum stress = \(\frac{\text { Maximum force }}{\text { Area of cross – section }} \)
∴ Maximum force = Maximum stress x Area of cross – section
= 10⁸ x π (0.015)²
= 7.065 x 10⁴ N
Hence, the cable can support the maximum load of 7.065 x 10⁴ N.

Question 10.
A rigid bar of mass 15 kg is supported symmetrically by three wires each 2.0 m long. Those at each mid are of copper and the middle one is of iron. Determine the ratio of their diameters if each is to have the same tension.
Solution:
The tension force acting on each wire is the same. Thus, the extension in each case is the same. Since the wires are of the same length, the strain will also be the same.
The relation for Young’s modulus is given as:
Y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\text { Strain }}=\frac{\frac{4 F}{\pi d^{2}}}{\text { Strain }}\) ……………………………. (i)
where, F = Tension force
A = Area of cross-section
d = Diameter of the wire
It can be inferred from equation (i) that Y ∝ \(\frac{1}{d^{2}}\)
Young’s modulus for iron, Y₁ = 190 x 10⁹ Pa
Diameter of the iron wire = d₁
Young’s modulus for copper, Y₂ = 110 x 10⁹ Pa
Diameter of the copper wire = d₂
Therefore, the ratio of their diameters is given as:
\(\frac{d_{2}}{d_{1}}=\sqrt{\frac{Y_{1}}{Y_{2}}}=\sqrt{\frac{190 \times 10^{9}}{110 \times 10^{9}}}=\sqrt{\frac{19}{11}}\)
= 1.31:11.

Question 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.
Solution:
Given, mass, m = 14.5kg
Length of the steel wire, l = 1.0 m
Angular velocity, ω = 2 rev / s
Cross-sectional area of the wire, a = 0.065cm² = 0.065 x 10^-4 m²
Let δl be died 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 x 9.8 +14.5x 1 x (2)² = 200.1 N

Young’s modulus = \(\frac{\text { Stress }}{\text { Strain }} \)
Y = \(\frac{\frac{F}{A}}{\frac{\Delta l}{l}}=\frac{F}{A} \cdot \frac{l}{\Delta l}\)
∴ Δl = \(\frac{F l}{A Y}\)

Young’s modulus for steel = 2 x 10¹¹ Pa
∴ Δl = \(\frac{200.1 \times 1}{0.065 \times 10^{-4} \times 2 \times 10^{11}} \) = 1539.23 x 10^-7
= 1.539 x 10^-4
Hence, the elongation of the wire is 1.539 x 10^-4 m.

Question 12.
Compute the bulk modulus of water from the following data: Initial volume =100.0 litre, Pressure increase =100.0atm (1 atm = 1.013 x 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.
Solution:
Initial volume, V₁ = 100.0 l x 10^-3 m³
Final volume, V₂ = 100.5l = 100.5 x 10^-3 m^-3
Increase in volume, V = V₂ — V₁ = 0.5 x 10^-3 m³
Increase in pressure, Δ_p =100.0 atm = 100 x 1.013 x 10⁵ Pa
Bulk modulus = \( \frac{\Delta p}{\frac{\Delta V}{V_{1}}}=\frac{\Delta p \times V_{1}}{\Delta V}\)
= \(\frac{100 \times 1.013 \times 10^{5} \times 100 \times 10^{-3}}{0.5 \times 10^{-3}}\) = 2.206 x 10⁹ Pa
Bulk modulus of air = 1.0 x 10⁵ Pa
∴ \(\frac{\text { Bulk modulus of water }}{\text { Bulk modulus of air }}=\frac{2.026 \times 10^{9}}{1.0 \times 10^{5}}\)
= 2.026 x 10⁴
This ratio is very high because air is more compressible than water.

Question 13.
What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is 103 x 10³ kgm^-3?
Solution:
Let the given depth be h.
Pressure at the given depth, p = 80.0 atm = 80 x 1.01 x 10⁵ Pa
Density of water at the surface, ρ₁ = 1.03 x 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 volumé of water of mass m at the depth h.
Let ΔV be the change in volume.

ΔV = V₁ – V₂ = \(m\left(\frac{1}{\rho_{1}}-\frac{1}{\rho_{2}}\right)\)
∴ Volumetric strain= \(\frac{\Delta V}{V_{1}}=m\left(\frac{1}{\rho_{1}}-\frac{1}{\rho_{2}}\right) \times \frac{\rho_{1}}{m}\)
∴ \(\frac{\Delta V}{V_{1}}=1-\frac{\rho_{1}}{\rho_{2}}\) ………………………………. (i)

Bulk modulus, B = \(\frac{p V_{1}}{\Delta V}\)
\(\frac{\Delta V}{V_{1}}=\frac{p}{B}\)
Compressibility of water = \(\frac{1}{B}=45.8 \times 10^{-11} \mathrm{~Pa}^{-1}\)
∴ \(\frac{\Delta V}{V_{1}}=80 \times 1.013 \times 10^{5} \times 45.8 \times 10^{-11}\) = 3.71 x 10^-3 ………….(ii)
From equations (i) and (ii), we get

Therefore, the density of water at the given depth (h) is 1.034 x 10³ kg m^-3.

Question 14.
Compute the fractional change in volume of a glass slab, when subjected to a hydraulic pressure of 10 atm.
Solution:
Hydraulic pressure exerted on the glass slab, p = 10 atm = 10 x 1.013 x 10⁵ Pa
Bulk modulus of glass, B = 37 x 10⁹ Nm^-2
Bulk modulus, B = \(\frac{p}{\frac{\Delta V}{V}}\)
where, \(\frac{\Delta V}{V}\) = Fractional change in volume
∴ \(\frac{\Delta V}{V}=\frac{p}{B}=\frac{10 \times 1.013 \times 10^{5}}{37 \times 10^{9}}\)
= 2.73 x 10^-5
Hence, the fractional change in the volume of the glass slab is
2.73 x 10^-5 = 2.73 x 10^-3% = 0.0027%

Question 15.
Determine the volume contraction of a solid copper cube, 10 cm on an edge, when subjected to a hydraulic pressure of 7.0x 10⁶ Pa.
Solution:
Length of an edge of the solid copper cube, l =10 cm = 0.1 m
Hydraulic pressure, p = 7.0 x 10⁶ Pa
Bulk modulus of copper, B = 140 x 10⁹ Pa
Bulk modulus, B = \(\frac{P}{\frac{P}{\Delta V}}\)
where, \(\frac{\Delta V}{V}\) = Volumetric strain
ΔV = Change in volume
V =,Original volume
ΔV = \(\frac{p V}{B}\)
Original volume of the cube, V = l³

Therefore, the volume contraction of the solid copper cube is 0.05 cm³.

Question 16.
How much should the pressure on a litre of water be changed to compress by 0.10%?
Solution:
Volume of water, V =1 L
It is given that water is to be compressed by 0.10%.
∴ Fractional change, \(\frac{\Delta V}{V}=\frac{0.1}{100 \times 1}=10^{-3}\)
Bulk modulus, B = \(\frac{p}{\frac{\Delta V}{V}}\)
p = B x \( \frac{\Delta V}{V}\)

Bulk modulus of water, B = 2.2 x 10⁹ Nm^-2
p = 22 x 10⁹ x 10^-3
=2.2 x 10⁶ Nm^-2
Therefore, the pressure on water should be 2.2 x 10⁶ Nm^-2.

Additional Exercises

Question 17.
Anvils made of single crystals of diamond, with the shape as shown in figure given below, 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?

Solution:
Diameter of the cones at the narrow ends, d = 0.50 mm = 0.5 x 10^-3m
radius, r = \( \frac{d}{2}\) = 0.25 x 10^-3 m
Compressional force, F = 50000 N
Pressure at the tip of the anvil,
p = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{\pi r^{2}}\)
= \(\frac{50000}{3.14 \times\left(0.25 \times 10^{-3}\right)^{2}}\)
= 2.55 x 10¹¹ Pa
Therefore, the pressure at the tip of the anvil is 2.55 x 10¹¹ Pa.

Question 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 figure given below. The cross-sectional areas of wires A and B are 1.0 mm² and 2.0mm², respectively. At what point along the rod should a mass m he suspended in order to produce (a) equal stresses and (b) equal strains in both steel and aluminium wires.

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

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 = \(\frac{\text { Force }}{\text { Area }}=\frac{F}{a}\)
If the two wires have equal stresses, then,
\( \frac{F_{1}}{a_{1}}=\frac{F_{2}}{a_{2}}\)
where, F₁ = Force exerted on the steel wire

F₂ = Force exerted on the aluminium wire
\(\frac{F_{1}}{F_{2}}=\frac{a_{1}}{a_{2}}=\frac{1}{2}\) …………………………. (i)
The situation is shown in the following figure.

Taking torque about the point of suspension, we have
F₁y = F₂(1.05— y)
\(\frac{F_{1}}{F_{2}}=\frac{(1.05-y)}{y}\) ……………………….(ii)
Using equations (i) and (ii), we can write

In order to produce an equal stress in the two wires, the mass should be suspended at a distance of 0.7 m from the end where wire A is attached.

(b) Young s modulus = \(\frac{\text { Stress }}{\text { Strain }} \)
Strain = \(\frac{\text { Stress }}{\text { Young’s modulus }}=\frac{\frac{F}{a}}{Y}\)
If the strain in the two wires is equal, then,

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)
\(\frac{F_{1}}{F_{2}}=\frac{\left(1.05-y_{1}\right)}{y_{1}}\) ……………………………. (iv)
Using equations (iii) and (iv), we get
\(\frac{\left(1.05-y_{1}\right)}{y_{1}}=\frac{10}{7}\)
7(1.05 – y₁) = 10 y₁
⇒ 17 y₁ = 7.35
y₁ = 0.432 m
In order to produce an equal strain in the two wires, the mass should be suspended at a distance of 0.432 m from the end where wire A is attached.

Question 19.
A mild steel wire of length 1.0 m and cross-sectional area 0.50 x 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 mid-point.
Solution:

Length of the steel wire = 1.0 m
Area of cross-section, A = 0.50 x 10^-2 cm² = 0.50 x 10^-6 m²
A mass 100 g is suspended from its mid-point.
m = 100 g = 0.1kg
Hence, the wire dips, as shown in the given figure.

Original length = XZ
Depression = l
The length after mass m, is attached to the wire = XO +OZ
Increase in the length of the wire:
Δl = (XO + OZ)-XZ
Where

Expanding and neglecting higher terms, we get:
Δl = \(\frac{l^{2}}{0.5}\)
Strain = \(\frac{\text { Increase in length }}{\text { Original length }}\)
Let T be the tension in the wire.

∴ mg = 2Tcos θ
Using the figure, it can be written as
Cos θ = \(\frac{l}{\left[(0.5)^{2}+l^{2}\right]^{\frac{1}{2}}}=\frac{l}{(0.5)\left[1+\left(\frac{l}{0.5}\right)^{2}\right]^{\frac{1}{2}}}\)
Expanding the expression and eliminating the higher terms, we get
Cos θ = \(\frac{l}{(0.5)\left(1+\frac{l^{2}}{2(0.5)^{2}}\right)}\)
\(\left(1+\frac{l^{2}}{2(0.5)^{2}}\right)\) ≈ 1 for small l

Hence, the depression at the mid-point is 0.0107 m.

Question 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 x 10⁷ Pa?
Assume that each rivet is to carry one-quarter of the load.
Solution:
Diameter of the metal strip, d = 6.0 mm = 6.0 x 10^-3 m
Radius, r = \(\frac{d}{2}\) = 3.0 x 10^-3 m
Maximum shearing stress = 6.9 x 10⁷ Pa
Maximum stress = \(\frac{\text { Maximum load or force }}{\text { Area }}\)
Maximum force = Maximum stress x Area
= 6.9 x 10⁷ x π x (r)²
= 6.9 x 10⁷ x 3.14 x (3 x10^-3)²
= 1949.94 N
Each rivet carries one quarter of the load.
∴ Maximum tension on each rivet = 4 x 1949.94 = 7799.76 N.

Question 21.
The Marina trench is located in the Pacific Ocean, and at one place it is nearly eleven km beneath the surface of the water. The water pressure at the bottom of the trench is about 1.1 x 10⁸ Pa. A steel ball of initial volume 0.32m 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?
Solution:
Water pressure at the bottom, p = 1.1 x 10⁸ Pa
Initial volume of the steel ball, V = 0.32 m³
Bulk modulus of steel, B = 1.6 x 10¹¹ Nm^-2
The ball falls at the bottom of the Pacific Ocean, which is 11 km beneath the surface.
Let the change in the volume of the ball on reaching the bottom of the trench be ΔV.
Bulk modulus, B = \( \frac{p}{\frac{\Delta V}{V}}\)
ΔV = \(\frac{p V}{B}=\frac{1.1 \times 10^{8} \times 0.32}{1.6 \times 10^{11}}\)
= 2.2 x 10^-4 m³
Therefore, the change in volume of the ball on reaching the bottom of the trench is 2.2 x 10^-4 m³.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 6 Work, Energy and Power Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 6 Work, Energy and Power

PSEB 11th Class Physics Guide Work, Energy and Power Textbook Questions and Answers

Question 1.
The sign of work done by a force on a body is important to understand. State carefully if the following quantities are positive or negative:
(a) work done by a man in lifting a bucket out of a well by means of a rope tied to the bucket.
(b) work done by gravitational force in the above case,
(c) work done by friction on a body sliding down an inclined plane,
(d) work done by an applied force on a body moving on a rough horizontal plane with uniform velocity,
(e) work done by the resistive force of air on a vibrating pendulum in bringing it to rest.
Solution:
(a) Positive
In the given case, force and displacement are in the same direction. Hence, the sign of work done is positive. In this case, the work is done on the bucket.

(b) Negative
In the given case, the direction of force (vertically downward) and displacement (vertically upward) are opposite to each other. Hence, the sign of work done is negative.

(c) Negative
Since the direction of frictional force is opposite to the direction of motion, the work done by frictional force is negative in this case.

(d) Positive
Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.

(e) Negative
The resistive force of air acts in the direction opposite to the direction of motion of the pendulum. Hence, the work done is negative in this case.

Question 2.
A body of mass 2 kg initially at rest moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1.
Compute the
(a) work done by the applied force in 10 s,
(b) work done by friction in 10 s,
(c) work done by the net force on the body in 10 s,
(d) change in kinetic energy of the body in 10 s, and interpret your results.
Solution:
Mass of the body, m = 2 kg
Applied force, F = 7 N
Coefficient of kinetic friction, μ = 0.1
Initial velocity, u = 0
Time, t = 10 s
The acceleration produced in the body by the applied force is given by Newton’s second law of motion as:
a’ = \(\frac{F}{m}\) = \(\frac{7}{2}\) = 3.5 m/s²
Frictional force is given as:
f = μ mg = 0.1 × 2 × 9.8 = -1.96 N
The acceleration produced by the frictional force:
a” = –\(\frac{1.96}{2}\) = -0.98 m/s²
Total acceleration of the body:
a = a’ + a”
= 3.5 + (-0.98) = 2.52 m/s²
The distance travelled by the body is given by the equation of motion:
s = ut + \(\frac{1}{2}\) at²
= 0 + \(\frac{1}{2}\) × 2.52 × (10)² = 126 m
(a) Work done by the applied force, W_a = F × s = 7x 126 = 882 J
(b) Work done by the frictional force, W_f = f × s = -1.96 × 126 = -247 J
(c) Net force = 7 + (-1.96) = 5.04 N
Work done by the net force, W_net = 5.04 × 126 = 635 J
(d) From the first equation of motion, final velocity can be calculated as:
υ = u + at
= 0 + 2.52 × 10 = 25.2 m/s
Change in kinetic energy = \(\frac{1}{2}\) mυ ² – \(\frac{1}{2}\) mu²
= \(\frac{1}{2}\) × 2(υ² – u²) = (25.2)² – 0² = 635 J

Question 3.
Given in figure below are examples of some potential energy functions in one dimension. The total energy of the particle is indicated by a cross on the ordinate axis. In each case, specify the regions, if any, in which the particle cannot be found for the given energy. Also, indicate the minimum total energy the particle muqt have in each case. Think of simple physical contexts for which these potential energy shapes are relevant.


Solution:
Total energy of a system is given by the relation:
E = P.E. + K.E.
> K.E. = E – P.E.
Kinetic energy of a body is a positive quantity. It cannot be negative. Therefore, the particle will not exist in a region where K.E. becomes negative.
(i) In the given case, the potential energy (V₀) of the particle becomes greater than total energy (E) for x > a. Hence, kinetic energy becomes negative in this region. Therefore, the particle will not exist in this region. The minimum total energy of the particle is zero.

(ii) In the given case, the potential energy (V₀) is greater than total energy (E) in all regions. Hence, the particle will not exist in this region.

(iii) In the given case, the condition regarding the positivity of K.E. is satisfied only in the region between x > a and x<b.
The minimum potential energy in this case is -V₁. Therefore, K.E. = E-(-V₁) = E + V₁.
Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than -Vj. So, the minimum total energy the particle must have is -V₁.

(iv) In the given case, the potential energy (V₀) of the particle becomes greater than the total energy (E) for –\(\frac{b}{2}\)< x <\(\frac{b}{2}\) and –\(\frac{a}{2}\)< x <\(\frac{a}{2}\).
Therefore, the particle will not exist in these regions.
The minimum potential energy in this case is -V₁. Therefore, K.E. = E – (-V₁ ) = E + V₁. Therefore, for the positivity of the kinetic energy, the total energy of the particle must be greater than -V₁. So, the minimum total energy the particle must have is -V₁.

Question 4.
The potential energy function for a particle executing linear simple harmonic motion is given by V (x) = kx² / 2, where k is the force constant of the oscillator. For k = 0.5 Nm^-1, the graph of V (x) versus x is shown in figure below. Show that a particle of total energy 1 J moving under this potential must ‘turn back’ when it reaches x = ±2m.

Solution:
Total energy of the particle, E = 1 J
Force constant, k = 0.5 Nm^-1
Kinetic energy of the particle, K = \(\frac{1}{2}\)mυ²
According to the conservation law:
E = V + K
1 = \(\frac{1}{2}\)kx² + \(\frac{1}{2}\)mυ ²
At the moment of ‘turn back’, velocity (and hence K) becomes zero.
> 1 = \(\frac{1}{2}\)kx²
\(\frac{1}{2}\) × 0.5 ײ = 1
x² = 4
x = ±2
Hence, the particle turns back when it reaches x = ±2 m.

Question 5.
Answer the following:
(a) The casing of a rocket in flight burns up due to friction. At whose expense is the heat energy required for burning obtained? The rocket or the atmosphere?
(b) Comets move around the sun in highly elliptical orbits. The gravitational force on the comet due to the sun is not normal to the comet’s velocity in general. Yet the work done by the gravitational force over every complete orbit of the comet is zero. Why?
(c) An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth?
(d) In fig. (i) the man walks 2 m carrying a mass of 15 kg on his hands. In fig. (ii), he walks the same distance pulling the rope behind him. The rope goes over a pulley, and a mass of 15 kg hangs at its other end. In which case is the work done greater?

Solution:
(a) The burning of the casing of a rocket in flight (due to friction) results in the reduction of the mass of the rocket.
According to the conservation of energy:
Total Energy (T.E.) = Potential energy (P.E.) + Kinetic energy (K.E.)
= mgh + \(\frac{1}{2}\) mυ²
The reduction in the rocket’s mass causes a drop in the total energy. Therefore, the heat energy required for the burning is obtained from the rocket.

(b) Gravitational force is a conservative force. Since the work done by a conservative force over a closed path is zero, the work done by the gravitational force over every complete orbit of a comet is zero.

(c) When an artificial satellite, orbiting around earth, moves closer to earth, its potential energy decreases because of the reduction in the height. Since the total energy of the system remains constant, the reduction in P.E. results in an increase in K.E. Hence, the velocity of the satellite increases. However, due to atmospheric friction, the total energy of the satellite decreases by a small amount.

(d) Case (i)
Mass, m = 15 kg
Displacement, s = 2 m
Work done, W = Fs cosθ
where, θ = Angle between force and displacement
= mgs cosθ = 15 × 2 × 9.8 cos 90°
= 0 ( cos90° = 0)

Case (ii)
Mass, m = 15 kg Displacement, s = 2 m
Here, the direction of the force applied on the rope and the direction of the displacement of the rope are same.
Therefore, the angle between them, θ =0°
Since, cos0° = 1
Work done, W = Fs cosθ = mgs
= 15 × 9.8 × 2 = 294J
Hence, more work is done in the second case.

Question 6.
Underline the correct alternative:
(a) When a conservative force does positive work on a body, the potential energy of the body increases/decreases/ remains unaltered.
(b) Work done by a body against friction always results in a loss of its kinetic/potential energy.
(c) The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
(d) In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Solution:
(a) Decreases, A conservative force does a positive work on a body when it displaces the body in the direction of force. As a result, the body advances toward the centre of force. It decreases the separation between the two, thereby decreasing the potential energy of the body.

(b) Kinetic energy, The work done against the direction of friction reduces the velocity of a body. Hence, there is a loss of kinetic energy of the body.

(c) External force, Internal forces, irrespective of their direction, cannot produce any change in the total momentum of a body. Hence, the total momentum of a many-particle system is proportional to the external forces acting on the system.

(d) Total linear momentum, The total linear momentum always remains conserved whether it is an elastic collision or an inelastic collision.

Question 7.
State if each of the following statements is true or false. Give reasons for your answer.
(a) In an elastic collision of two bodies, the momentum and energy of each body is conserved.
(b) Total energy of a system is always conserved, no matter what internal and external forces on the body are present.
(c) Work done in the motion of a body over a closed loop is zero for every force in nature.
(d) In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system.
Solution:
(a) False, In an elastic collision, the total energy and momentum of both the bodies, and not of each individual body, is conserved.

(b) False, Although internal forces are balanced, they cause no work to be done on a body. It is the external forces that have the ability to do work. Hence, external forces are able to change the energy of a system.

(c) False, The work done in the motion of a body over a closed loop is zero for a conservation force only.

(d) True, In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system. This is because in such collisions, there is always a loss of energy in the form of heat, sound, etc.

Question 8.
Answer carefully, with reasons:
(a) In an elastic collision of two billiard balls, is the total kinetic energy conserved during the short time of collision of the balls (i. e., when they are in contact)?
(b) Is the total linear momentum conserved during the short time of an elastic collision of two balls?
(c) What are the answers to (a) and (b) for an inelastic collision?
(d) If the potential energy of two billiard balls depends only on the separation distance between their centres, is the collision elastic or inelastic? (Note, we are talking here of potential energy corresponding to the force during collision, not gravitational potential energy).
Solution:
(a) No, In an elastic collision, the total initial kinetic eneigy of the bails will be equal to the total final kinetic energy of the balls. This kinetic energy is not conserved at the instant the two balls are in contact with each other. In fact, at the time of collision, the kinetic energy of the balls will get converted into potential energy.

(b) Yes, In an elastic collision, the total linear momentum of the system always remains conserved.

(c) No, In an inelastic collision, there is always a loss of kinetic energy, i. e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision.
Yes, The total linear momentum of the system of billiards balls will remain conserved even in the case of an inelastic collision.

(d) Elastic, In the given case, the forces involved are conservation. This is because they depend on the separation between the centres of the billiard balls. Hence, the collision is elastic.

Question 9.
A body is initially at rest. It undergoes one-dimensional motion with constant acceleration. The power delivered to it at time t is proportional to
(i) \(t^{\frac{1}{2}}\)
(ii) t
(iii) \(t^{\frac{3}{2}}\)
(iv) t²
Solution:
(ii) t
Let a body of mass m which is initially at rest undergoes one-dimensional motion under a constant force F with a constant acceleration a.
Acceleration (a) = \(\frac{F}{m}\) …………. (i)
Using equation of motion, υ = u + at
⇒ υ = 0 + \(\frac{F}{m}\).t …………. (∵ u = 0)
⇒ υ = \(\frac{F}{m}\)t …………… (ii)
Power delivered (P) = Fυ
Substituting the value from eq. (ii), we get
⇒ P = F × \(\frac{F}{m}\) × t
⇒ P = \(\frac{F^{2}}{m}\)t
Dividing and multiplying by m in R.H.S.
P = \(\frac{F^{2}}{m^{2}}\) × mt= a²mt [Using eq.(i)]
As mass m and acceleration a are constant.
∴ P ∝ t

Question 10.
A body is moving unidirectionally under the influence of a source of constant power. Its displacement in time t is proportional to
(i) \(t^{\frac{1}{2}}\)
(ii) t
(iii) \(t^{\frac{3}{2}}\)
(iv) t²
Solution:
(iii) \(t^{\frac{3}{2}}\)
Power is given by the relation:
P = Fυ
= maυ = mυ \(\frac{d v}{d t}\) = Constant (say,k )
υ dv = \(\frac{k}{m}\) dt
Integrating both sides:
\(\frac{v^{2}}{2}=\frac{k}{m}\)t
υ = \(\sqrt{\frac{2 k t}{m}}\)
For displacement x of the body, we have:
υ = \(\frac{d x}{d t}=\sqrt{\frac{2 k}{m}} t^{\frac{1}{2}}\)
dx = k’\(t^{\frac{1}{2}}\)dt
where, k’ = \(\sqrt{\frac{2 k}{3}}\) = New constant
On integrating both sides, we get:
x = \(\frac{2}{3} k^{\prime} t^{\frac{3}{2}}\)
x ∝ \(t^{\frac{3}{2}}\)

Question 11.
A body constrained to move along the z-axis of a coordinate system is subject to a constant force F given by F = -î + 2ĵ + 3k̂N
where î, ĵ, k̂ are unit vectors along the x-, y- and z-axis of the system respectively. What is the work done by this force in moving the body a distance of 4 m along the z-axis?
Solution:
Force exerted on the body, F = -î + 2ĵ + 3k̂N
Displacement, s = 4 k̂ m
Work done, W = F.s
= (-î + 2ĵ + 3k̂).(4k̂)
= 0 + 0 + 3 × 4 = 12 J
Hence, 12 J of work is done by the force on the body.

Question 12.
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton? Obtain the ratio of their speeds, (electron mass = 9.11 × 10^-31 kg, proton mass = 1.67 × 10^-27 kg, 1 eV = 1.60 × 10^-19 J).
Solution:
Mass of the electron, m_e = 9.11 × 10^-31 kg
Mass of the proton, m_p =1.67 × 10^-27 kg
Kinetic energy of the electron, E_Ke =10 keV = 10⁴ eV
= 10⁴ × 1.60 × 10^-19
1.60 × 10^-15 J
Kinetic energy of the proton, E_Kp = 100 keV = 10⁵ eV = 1.60 × 10^-14 J
For the velocity of an electron v_e, its kinetic energy is given by the relation:

Question 13.
A rain drop of radius 2 mm falls from a height of500 m above the ground. It falls with decreasing acceleration (due to viscous resistance of the air) until at half its original height, it attains its maximum (terminal) speed, and moves with uniform speed thereafter. What is the work done hy the gravitational force on the drop in the first and second half of its journey? What is the work done by the resistive force in the entire journey if its speed on reaching the ground is 10 ms^-1?
Solution:
Radius of the rain drop, r = 2 mm = 2 × 10 ^-3 m
Volume of the rain drop, V = \(\frac{4}{3}\) πr³
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ m^-3
Density of water, ρ = 10³ kgm^-3
Mass of the rain drop, m = ρV
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³
Gravitational, F = mg
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³ 9.8N
The work done by the gravitational force on the drop in the first half of its journey.
W₁ = Fs
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3 )³ × 10³ × 9.8 × 250
= 0.082 J

This amount of work is equal to the work done by the gravitational force on the drop in the second half of its journey, i. e., W_II, = 0.082 J.
As per the law of conservation of energy, if no resistive force is present, then the total energy of the rain drop will remain the same.
∴ Total energy at the top:
E_T = mgh + 0
= \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³ × 9.8 × 500
= 0.164 J
Due to the presence of a resistive force, the drop hits the ground with a velocity of 10 m/s.
∴ Total energy at the ground:
E_g = \(\frac{1}{2}\) mυ² + 0
= \(\frac{1}{2}\) × \(\frac{4}{3}\) × 3.14 × (2 × 10^-3)³ × 10³ × 9.8 × (10)²
= 1.675 × 10^-3 J = 0.001675
∴ Resistive force = E_g – E_t = 0.001675 – 0.164 = -0.162 J

Question 14.
A molecule in a gas container hits a horizontal wall with speed 200 m s^-1 and angle 30° with the normal, and rebounds with the same speed. Is momentum conserved in the collision? Is the collision elastic or inelastic?
Solution:
Yes; Collision is elastic
The momentum of the gas molecule remains conserved Whether the collision is elastic or inelastic.
The gas molecule moves with a velocity of 200 m/s and strikes the stationary wall of the container, rebounding with the same speed.
It shows that the rebound velocity of the wall remains zero. Hence, the total kinetic energy of the molecule remains conserved during the collision. The given collision is an example of an elastic collision.

Question 15.
A pump on the ground floor of a building can pump up water to fill a tank of volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of the pump is 30%, how much electric power is consumed by the pump?
Solution:
Volume of the tank, V = 30 m³
Time of operation, t = 15 min = 15 × 60 = 900 s
Height of the tank, h = 40 m
Efficiency of the pump, η = 30%
Density of water, ρ = 10³ kg/m³
Mass of water, m = ρ V = 30 × 10³ kg
Output power can be obtained as:
P₀ = \(\frac{\text { Work done }}{\text { Time }}=\frac{m g h}{t}\)
= \(\frac{30 \times 10^{3} \times 9.8 \times 40}{900}\) = 13.067 × 10³ W
For input power P_i efficiency η, is given by the relation :
η = \(\frac{P_{o}}{P_{i}}\) = 30%
P_i = \(\frac{13.067}{30}\) × 100 10³
= 0.436 × 10² W = 43.6 kW

Question 16.
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed V. If the collision is elastic, which of the following figure is a possible result after collision?

Solution:
It can be observed that the total momentum before and after collision in each case is constant.
For an elastic collision, the total kinetic energy of a system remains conserved before and after collision.
For mass of each ball bearing m, we can write:
Total kinetic energy of the system before collision:
= \(\frac{1}{2}\)mV² + \(\frac{1}{2}\)(2m)0
= \(\frac{1}{2}\)mV²

Case (i)
Total kinetic energy of the system after collision:
= \(\frac{1}{2}\)m × 0 + \(\frac{1}{2}\)(2m) (\(\frac{V}{2}\))²
= \(\frac{1}{4}\)mV²
Hence, the kinetic energy of the system is not conserved in case (i).

Case (ii)
Total kinetic energy of the system after collision:
= \(\frac{1}{2}\) (2m) × 0 + \(\frac{1}{2}\)mV²
= \(\frac{1}{2}\) mV²
Hence, the kinetic energy of the system is conserved in case (ii).

Case (iii)
Total kinetic energy of the system after collision:
= \(\frac{1}{2}\)(3m) (\(\frac{V}{3}\))²
= \(\frac{1}{6}\) mV²
Hence, the kinetic energy of the system is not conserved in case (iii).

Question 17.
The bob A of a pendulum released from 30° to the vertical hits another hob B of the same mass at rest on a table as shown in’ figure given below. How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic.

Solution:
Bob A will not rise at all
In an elastic collision between two equal masses in which one is stationary, while the other is moving with some velocity, the stationary mass acquires the same velocity, while the moving mass immediately comes to rest after collision. In this case, a complete transfer of momentum takes place from the moving mass to the stationary mass. Hence, bob A of mass m, after colliding with bob B of equal mass, will come to rest, while bob B will move with the velocity of bob A at the instant of collision.

Question 18.
The bob of a pendulum is released from a horizontal position. If the length of the pendulum is 1.5 m, what is the speed with which the bob arrives at the lowermost-point, given that it dissipated 5% of its initial energy against air resistance?
Solution:
Length of the pendulum, l = 1.5 m
Mass of the bob = m
Energy dissipated = 5%
According to the law of conservation of energy, the total energy of the system remains constant.

At the horizontal position:
Potential energy of the bob, E_P = mgl
Kinetic energy of the bob, E_K = 0
Total energy = mgl …………. (i)

At the lowermost point (mean position):
Potential energy of the bob, E_P = 0
Kinetic energy of the bob, E_K = \(\frac{1}{2}\)mυ²
Total energy, E_x = \(\frac{1}{2}\)mυ² ………….. (ii)
As the bob moves from the horizontal position to the lowermost point, 5% of its energy gets dissipated.
The total energy at the lowermost point is equal to 95% of the total energy at the horizontal point, i. e.,
\(\frac{1}{2}\)mυ² = \(\frac{95}{100}\) × mgl
υ = \(\sqrt{\frac{2 \times 95 \times 1.5 \times 9.8}{100}}\) = 5.28 m/s

Question 19.
A trolley of mass 300 kg carrying a sandbag of 25 kg is moving uniformly with a speed of 27 km/h on a frictionless track. After a while, sand starts leaking out of ahole on the floor of the trolley at the rate of 0.05 kg s^-1 . What is the speed of the trolley after the entire sand bag is empty?
Solution:
The sand bag is placed on a trolley that is moving with a uniform speed of 27 km/h. The external forces acting on the system of the sand bag and the trolley is zero. When the sand starts leaking from the bag, there will be no change in the velocity of the trolley. This is because the leaking action does not produce any external force on the system. This is in accordance with Newton’s first law of motion. Hence, die speed of the trolley will remain 27 km/h.

Question 20.
A body of mass 0.5 kg travels in a straight line with velocity υ = ax^3/2 where a = 5 m^-1/2s^-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Solution:
Mass of the body, m = 0.5 kg
Velocity of the body is governed by the equation, υ = a \(x^{\frac{3}{2}}\) where,
a = 5m^-1/2s^-1
Initial velocity, u (at x = 0) = 0
Final velocity, υ (at x = 2 m) = 5 × (2)^3/2 m/s = 10√2 m/s
Work done, W = Change in kinetic energy
= \(\frac{1}{2}\)m(υ² – u²)
= \(\frac{1}{2}\) × 0.5[(10√2)² – (0)²]
= \(\frac{1}{2}\) × 0.5 × 10 × 10 × 2
= 50 J

Question 21.
The blades of a windmill sweep out a circle of area A. (a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t? (b) What is the kinetic energy of the air? (c) Assume that the windmill converts 25% of the wind’s energy into electrical energy, and that A = 30 m², υ = 36km/h and the density of air is 1.2 kg m^-3. What is the electrical power produced?
Solution:
(a) Area of the circle swept by the windmill = A
Velocity of the wind = υ
Density of air = ρ
Volume of the wind flowing through the windmill per sec = Aυ
Mass of the wind flowing through the windmill per sec = ρ Aυ
Mass m, of the wind flowing through the windmill in time t = ρ A υ t

(b) Kinetic energy of air = \(\frac{1}{2}\) mυ²
= \(\frac{1}{2}\) (ρAυt)v² = \(\frac{1}{2}\)ρ Aυ³t

(c) Area of the circle swept by the windmill, A = 30 m²
Velocity of the wind, υ =36 km/h = 36 × \(\frac{5}{18}\) m/s
= 10 m/s [1 km/s = \(\frac{5}{18}\) m/s]
Density of air, ρ = 1.2 kg m^-3
Electric energy produced = 25% of the wind energy
= \(\frac{25}{100}\) ×Kinetic energy of air
= \(\frac{1}{8}\)ρAυ³t

\(\) = \(\frac{1}{8}\)ρAυ³
= \(\frac{1}{8}\) × 1.2 × 30 × (10)³
= 4.5 × 10³ W = 4.5 kW

Question 22.
A person trying to lose weight (dieter) lifts a 10 kg mass, one thousand times, to a height of 0.5 m each time. Assume that the potential energy lost each time she lowers the mass is dissipated, (a) How much work does she do against the gravitational force? (b) Fat supplies 3.8 × 10⁷ J of energy per kilogram which is converted to mechanical energy with a 20% efficiency rate. How much fat will the dieter use up?
Solution:
(a) Mass of the weight, m = 10 kg
Height to which the person lifts the weight, h = 0.5
m Number of times the weight is lifted, n = 1000
∴ Work done against gravitational force:
= n(mgh)
=1000 × 10 × 9.8 × 0.5
= 49 × 10³ J = 49 kJ

(b) Energy equivalent of 1 kg of fat = 3.8 × 10⁷ J
Efficiency rate = 20%
Mechanical energy supplied by the person’s body
= \(\frac{20}{100}\) × 3.8 × 10⁷ J
= \(\frac{1}{5}\) × 3.8 × 107 J 5
Equivalent mass of fat lost by the dieter
= \(\frac{1}{\frac{1}{5} \times 3.8 \times 10^{7}}\) × 49 × ³
= \(\frac{245}{3.8}\)× 3.8 × 107
= × 10^-4
= 6.45 × 10^-3 kg

Question 23.
A family uses 8 kW of power, (a) Direct solar energy is incident on the horizontal surface at an average rate of 200 W per square meter. If 20% of this energy can be converted to useful electrical energy, how large an area is needed to supply 8 kW? (b) Compare this area to that of the roof of a typical house.
Solution:
(a) Power used by the family, P = 8 kW = 8 × 10³ W
Solar energy received per square meter = 200 W
Efficiency of conversion from solar to electrical energy = 20%
Area required to generate the desired electricity = A
As per the information given in the question, we have
8 × 10³ = 20% × (A × 200)
= \(\frac{20}{100}\) × A × 200
> A = \(\frac{8 \times 10^{3}}{40}\) = 200m²

(b) In order to compare this area to that of the roof of a typical house, let ‘a’ be the side of the roof
∴ area of roof =a × a = a²
Thus a² = 200 m²
or a = \(\sqrt{200 \mathrm{~m}^{2}}\) = 14.14 m
∴ area of roof = 14.14 × 14.14 m²
Thus 200 m² is comparable to the roof of a typical house of dimensions 14.14 m × 14.14 m = 14 m × 14 m.

Question 24.
A bullet of mass 0.012 kg and horizontal speed 70 ms^-1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
Solution:
Mass of the bullet, m = 0.012 kg
Initial speed of the bullet, u_b = 70 m/s
Mass of the wooden block, M = 0.4 kg
Initial speed of the wooden block, u_B = 0
Final speed of the system of the bullet and the block = υ
Applying the law of conservation of momentum,
mu_b +MU_B = (m + M)υ
012 × 70 + 0.4 × 0 = (0.012 + 0.4)υ
∴ υ = \(\frac{0.84}{0.412}\) = 2.04 m/s

For the system of the bullet and the wooden block,
Mass of the system, m’ = 0.412 kg
Velocity of the system = 2.04 m/s
Height up to which the system rises = h
Applying the law of conservation of energy to this system,
Potential energy at the highest point
= Kinetic energy at the lowest point
m’ gh = \(\frac{1}{2}\)m’ υ²
h = \(\frac{1}{2}\)(\(\frac{v^{2}}{g}\))
= \(\frac{1}{2}\) (\(\frac{(2.04)^{2}}{9.8}\)) = 0.2123m
The wooden block will rise to a height of 0.2123 m.
Heat produced = Kinetic energy of the bullet
– Kinetic energy of the system
= \(\frac{1}{2}\) mu_b– \(\frac{1}{2}\)m’ υ²
= \(\frac{1}{2}\) × 0.012 × (70)² – \(\frac{1}{2}\) × 0.412 × (2.04)²
= 29.4 – 0.857 = 28.54 J

Question 25.
Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (see figure). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ₁ = 30°, θ₂ = 60°and h = 10m, what are the speeds and times taken by the two stones?

Solution:
No; the stone moving down the steep plane will reach the bottom first Yes; the stones will reach the bottom with the same speed The given situation can be shown as in the following figure:

Here, the initial height (AD) for both the stones is the same (h).
Hence, both will have the same potential energy at point A.
As per the law of conservation of energy, the kinetic energy of the stones at points B and C will also be the same, i. e.,
\(\frac{1}{2}\) mυ₁² = \(\frac{1}{2}\)mυ₂²
υ₁ = υ₁ = υ, say
where,
m = Mass of each stone
υ = Speed of each stone at points B and C
Hence, both stones will reach the bottom with the same speed, υ.

For stone I: Net force acting on this stone is given by:
F_net = ma₁ = mgsinθ₁
a₁ = gsinθ₁

For stone II: a₂ = gsinθ₂
∵ θ₂ > θ₁
∴ sinθ₂ > sinθ₁
∴ a₂ > a₁
Using the first equation of motion, the time of slide can be obtained as:
υ = u + at
∴ t = \(\frac{v}{a}\) (∵ u = 0)
For stone I: t₁ = \(\frac{v}{a_{1}}\)
For stone II: t₂ = \(\frac{v}{a_{2}}\)
∵ a₂ > a₁
∴ t₂ < t₁
Hence, the stone moving down the steep plane will reach the bottom first. The speed (υ) of each stone at points B and C is given by the relation obtained from the law of conservation of energy.

Question 26.
A1 kg block situated on a rough incline is connected to a spring of spring constant 100 Nm^-1 as shown in figure given below. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.

Solution:
Mass of the block, m = 1 kg
Spring constant, k = 100 N m^-1
Displacement in the block, x = 10 cm = 0.1 m
The given situation can be shown as in the following figure.

At equilibrium:
Normal reaction, R = mg cos 37°
Frictional force, f = μR = p mg cos 37°
where, μ is the coefficient of friction
Net force acting on the block = mg sin 3 7° – f
= mgsin37° – μmgcos37°
= mg (sin37° – μcos37°)
At equilibrium, the work done by the block is equal to the potential energy of the spring, i.e.,
mg (sin37° – μcos37°) x = \(\frac{1}{2}\)kx ²
1 × 9.8 (sin 37° – μcos37°) = \(\frac{1}{2}\) × 100 × 0.1
0.6018 – μ × 0.7986 = 0.5102
0.7986 μ = 0.6018 – 0.5102 = 0.0916
∴ μ = \(\frac{0.0916}{0.7986}\) = 0.115

Question 27.
A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m^-1. It hits the floor of the elevator (length of the elevator = 3m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?
Solution:
Mass of the bolt, m = 0.3 kg
Speed of the elevator = 7 m/s
Height, h = 3 m
Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy.
Heat produced = Loss of potential energy
= mgh = 0.3 × 9.8 × 3 = 8.82 J
The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.

Question 28.
A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s^-1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?
Solution:
Mass of the trolley, M = 200 kg
Speed of the trolley, υ = 36 km/h = 36 × \(\frac{5}{18}\)m/s = 10 m/s
Mass of the boy, m = 20 kg
Initial momentum of the system of the boy and the trolley
= (M + m)υ
= (200 + 20) × 10
= 2200 kg-m/s
Let v’ be the final velocity of the trolley with respect to the ground.
Final velocity of the boy with respect to the ground = υ’ – 4
Final momentum = Mυ’ + m(υ’ – 4)
= 200 υ’ + 20υ’ – 80
= 220 υ’ – 80
As per the law of conservation of momentum,
Initial momentum = Final momentum
2200 =220 υ’ – 80
∴ υ’ = \(\frac{2280}{220}\) = 10.36 m/s
Length of the trolley, l = 10 m
Speed of the boy, υ” = 4 m/s
Time taken by the boy to run, t = \(\frac{10}{4}\) = 2.5 s
∴ Distance moved by the trolley = υ’ × t= 10.36 × 2.5 = 25.9 m

Question 29.
Which of the following potential energy curves in the given figure cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centres of the balls.

Solution:
(i), (ii), (iii), (iv), and (vi)
The potential energy of a system of two masses is inversely proportional to the separation between them. In the given case, the potential energy of the system of the two balls will decrease as they come closer to each other. It will become zero [i.e., V (r) = 0] when the two balls touch each other, i.e., at r = 2R, where R is the radius of each billiard ball. The potential energy curves given in figures (i), (ii), (iii), (iv), and (vi) do not satisfy these two conditions. Hence, they do not describe the elastic collisions between them.

Question 30.
Consider the decay of a free neutron at rest: n → p + e^–. Show that the two-body decay of this type must necessarily give an electron of fixed energy and, therefore, cannot account for the observed continuous energy distribution in the β-decay of a neutron or a nucleus (see figure).

[Note: The simple result of this exercise was one among the several arguments advanced by W. Pauli to predict the existence of a third particle in the decay products of P-decay. This particle is known as
neutrino. We now know that it is a particle of intrinsic spin \(\frac{1}{2}\) (like e^–, p or n), but is neutral, and either massless or having an extremely small mass
(compared to the mass of electron) and which interacts very weakly with matter. The correct decay process of neutron is: n → p + e^– + υ]
Solution:
The decay process of free neutron at rest is given as:
n → p + e^–
From Einstein’s mass-energy relation, we have the energy of electron as
Δ𝜏 mc²
where,
Δ𝜏 m = Mass defect
= Mass of neutron – (Mass of proton + Mass of electron)
c = Speed of light
Δ𝜏 m and c are constants. Hence, the given two-body decay is unable to explain the continuous energy distribution in the p-decay of a neutron or a nucleus. The presence of neutrino v on the LHS of the decay correctly explains the continuous energy distribution.
Here, Δ𝜏 m= mass of neutron – (mass of proton + mass of electron)
= 1.6747 × 10^-24 -(1.6724 × 10^-24 + 9.11 × 10^-28)
= (1.6747 – 1.6733) × 10^-24
= 0.0014 × 10^-24 gms ‘
∴ E = 0.0014 × 10^-24 × (3 × 10¹⁰)² ergs
= 0.0126 × 10^-4 ergs
= [Latex]\frac{0.0126 \times 10^{-4}}{1.6 \times 10^{-12}}[/Latex]
( ∵ 1 eV = 1.6 × 10^-19 J = 1.6 × 10^-19 × 10⁷ ergs)
= 0.007875 × 10 ⁸ eV
= 0.7875 × 10⁶ eV
= 0.79 MeV

A positron has the same mass as an electron but an opposite charge of+e. When an electron and a positron come close to each other, they destroy each other. Their masses, are converted into energy according to Einstein’s relation and the energy so obtained is released in the form of y-rays and is given by
E’ = mc² = 2 × 9.1 × 10^-31 kgx (3 × 10⁸ ms^-1)²
= 1.638 × 10^-13 J
= \(\frac{1.638 \times 10^{-13}}{1.6 \times 10^{-13}}\) MeV
= 1.02MeV (∵1 MeV = 1.6 × 10^-13 eV)
= Minimum energy a photon must possess for pair production.

Alternate method
Decay of a free neutron at rest,
n → p + e^–
Let Δ m be the mass defect during this process.
∴ Mass defect (Δ m) = (Mass of neutron) – (Mass of proton and electron) This mass defect is fixed and hence electron of fixed energy should be produced. Therefore, two-body decay of this type cannot explain the observed continuous energy distribution in the (3-decay of a neutron in a nucleus.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 12 Thermodynamics Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 12 Thermodynamics

PSEB 11th Class Physics Guide Thermodynamics Textbook Questions and Answers

Question 1.
A geyser heats water flowing at the rate of 3.0 liters 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 x 10⁴ J/g?
Solution:
Water is flowing at a rate of 3.0 liter/min.
The geyser heats the water, raising the temperature from 27°C to 77°C.
Initial temperature, T₁ = 27°C
Final temperature, T₂ = 77°C
Rise in temperature, ΔT = T₂ -T₁
= 77-27 = 50°C

Heat of combustion = 4 x 10⁴ J/g
Specific heat of water, C = 4.2 J g^-1 °C^-1
Mass of flowing water, m = 3.0 litre/min = 3000 g/min
Total heat used, ΔQ = mC ΔT
= 3000 x 4.2 x 50
= 6.3 x 10⁵ J/min
∴ Rate of consumption = \(\frac{6.3 \times 10^{5}}{4 \times 10^{4}}\) = 15.75 g/min

Question 2.
What amount of heat must be supplied to 2.0 x 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^-1K^-1).
Solution:
Mass of nitrogen, m = 2.0 x 10^-2 kg = 20g
Rise in temperature, ΔT = 45°C
Molecular mass of N₂,M =28
Universal gas constant, R = 8.3 J mol^-1K^-1
Number of moles, n = \(\frac{m}{M}\)
= \(\frac{2.0 \times 10^{-2} \times 10^{3}}{28}\) = 0.714
Molar specific heat at constant pressure for nitrogen,
C_p = \(\frac{7}{2}\) R = \(\frac{7}{2}\) x 8.3
= 29.05J mol^-1 K^-1

The total amount of heat to be supplied is given by the relation
ΔQ = nC_pΔT
= 0.714 x 29.05 x 45 = 933.38 J
Therefore, the amount of heat to be supplied is 933.38 J

Question 3.
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 + T₂) / 2.
(b) The coolant in a chemical or a nuclear plant (i.e., the liquid used to prevent the different parts of a plant from getting too hot) should have high specific heat.
(c) Air pressure in a car tyre increases during driving.
(d) The climate of a harbour town is more temperate than that of a town in a desert at the same latitude.
Answer:
(a) When two bodies at different temperatures T₁ and T₂ are brought in thermal contact, heat flows from the body at the higher temperature to the body at the lower temperature till equilibrium is achieved, i.e., the temperatures of both the bodies become equal. The equilibrium temperature is equal to the mean temperature (T₁ + T₂) / 2 only when the thermal capacities of both the bodies are equal.

(b) The coolant in a chemical or nuclear plant should have a high specific heat. This is because higher the specific heat of the coolant, higher is its heat-absorbing capacity and vice versa. Hence, a liquid having a high specific heat is the best coolant to be used in a nuclear or chemical plant. This would prevent different parts of the plant from getting too hot.

(c) When a car is in motion, the air temperature inside the car increases because of the motion of the air molecules. According to Charles’ law, temperature is directly proportional to pressure. Hence, if the temperature inside a tyre increases, then the air pressure in it will also increase.

(d) A harbour town has a more temperate climate (i.e., without the extremes of heat or cold) than a town located in a desert at the same latitude. This is because the relative humidity in a harbour town is more than it is in a desert town.

Question 4.
A cylinder with a movable piston contains 3 moles of hydrogen at standard temperature and pressure. The walls of the cylinder are made of a heat insulator, and the piston is insulated by having a pile of sand on it. By what factor does the pressure of the gas increase if the gas is compressed to half its original volume?
Solution:
The cylinder is completely insulated from its surroundings. As a result, no heat is exchanged between the system (cylinder) and its surroundings. Thus, the process is adiabatic.
Initial pressure inside the cylinder = P₁
Final pressure inside the cylinder = P₂
Initial volume inside the cylinder = V₁
Final volume inside the cylinder = V₂
Ratio of specific heats, γ = 1.4 For an adiabatic process,
we have
P₁V₁^γ = P₂V₂^γ
The final volume is compressed to half of its initial volume.

Hence, the pressure increases by a factor of 2.639.

Question 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)
Solution:
The work done (W) on the system while the gas changes from state A to state Bis 22.3 J.
This is an adiabatic process. Hence, change in heat is zero.
ΔQ = 0
ΔW = -22.3 J (since the work is done on the system)
From the first law of thermodynamics, we have:
ΔQ = ΔU + ΔW

where, ΔU = Change in the internal energy of the gas .
ΔU = ΔQ – ΔW = -(-22.3 J)
ΔU = +22.3 J
When the gas goes from state A to state B via a process, the net heat absorbed by the system is
ΔQ =9.35cal = 9.35 x 4.19 = 39.1765J
Heat absorbed, ΔQ = ΔU + ΔW
ΔW = ΔQ – ΔU = 39.1765 – 22.3 – = 16.8765J
Therefore, 16.88 J of work is done by the system.

Question 6.
Two cylinders A and B of equal capacity are connected to each other via a stopcock.
A contains a gas at standard temperature and pressure.
B is completely evacuated. The entire system is thermally insulated.
The stopcock is suddenly opened. Answer the following:
(a) What is the final pressure of the gas in A and B?
(b) What is the change in internal energy of the gas?
(c) What is the change in the temperature of the gas?
(d) Do the intermediate states of the system (before settling to the final equilibrium state) lie on its P -V – T surface?
Solution:
(a) 0.5 atm
The volume available to the gas is doused as soon as the stopcock between cylinders A and B is opened? Since volume is inversely proportional to pressure, the pressure will decrease to one-half of the original value. Since the initial pressure of the gas is 1 atm, the pressure in each cylinder will be 0.5atm.

(b) Zero
The internal energy of the gas can change only when work is done by or on the gas. Since in this case no work is done by or on the gas, the internal energy of the gas will not change.

(c) Zero
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
The given process is a case of free expansion. It is rapid and cannot be controlled. The intermediate states do not satisfy the gas equation and since they are in non-equilibrium states, they do not lie on the P-V-T surface of the system.

Question 7.
A steam engine delivers 5.4 x 10⁸ J of work per minute and services 3.6 x 10⁹ J of heat per minute from its boiler. What is the efficiency of the engine? How much heat is wasted per minute?
Solution:
Work done by the steam engine per minute, W = 5.4 x 10⁸ J
Heat supplied from the boiler, H = 3.6 x 10⁹ J
Efficiency of the engine = \(\frac{\text { Output energy }}{\text { Input energy }}\)
∴ η = \(\frac{W}{H}=\frac{5.4 \times 10^{8}}{3.6 \times 10^{9}}\) = 0.15
Hence, the percentage efficiency of the engine is 15%.
Amount of heat wasted= 3.6 x 10⁹ – 5.4 x 10⁸
= 30.6 x 10⁸ = 3.06 x 10⁹ J
Therefore, the amount of heat wasted per minute is 3.06 x 10⁹J.

Question 8.
An electric heater supplies heat to a system at a rate of 100 W. If system performs work at a rate of 75 joules per second. At what rate is the internal energy increasing?
Solution:
Heat is supplied to the system at a rate of 100 W.
∴ Heat supplied, ΔQ = 100 J/s
The system performs at a rate of 75 J/s.
∴ Work done, ΔW = 75 J/s

From the first law of thermodynamics, we have
ΔQ = ΔU + ΔW
where ΔU = Rate of change in internal energy
ΔU = ΔQ – ΔW = 100 – 75 = 25 J/s = 25W
Therefore, the internal energy of the given electric heater increases at a rate of 25 W.

Question 9.
A thermodynamic system is taken from an original state to an intermediate state by the linear process shown in figure given below.

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.
Solution:
Total work done by the gas from D to E to F = Area of ΔDEF
Area of ΔDEF = \(\frac{1}{2}\) DF x EF
where, DF = Change in pressure
=600 N/m²
= 300N/m² = 300N/m²
FE = Change in volume
5.0 m³ – 2.0 m³ = 3.0m³
Area of ADEF = \(\frac{1}{2} \) x 300 x 3 = 450 J
Therefore, the total work done by the gas from D to E to F is 450 J.

Question 10.
A refrigerator is to maintain eatables kept inside at 9°C. If the room temperature is 36°C, Calculate the coefficient of performance.
Solution:
Temperature inside the refrigerator, T₁ = 9°C = 273 + 9 = 282 K
Room temperature, T₂ = 36°C = 273+36
Coefficient of performance = \(\frac{T_{1}}{T_{2}-T_{1}}\)
= \(\frac{282}{309-282}=\frac{282}{27}\)
309-282 = 10.44
Therefore, the coefficient of performance of the given refrigerator is 10.44.

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 11 Thermal Properties of Matter Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 11 Thermal Properties of Matter

PSEB 11th Class Physics Guide Thermal Properties of Matter Textbook Questions and Answers

Question 1.
The triple points of neon and carbon dioxide are 24.57 K and 216.55K respectively. Express these temperatures on the Celsius and Fahrenheit scales.
Solution:
Kelvin and Celsius’s scales are related as
T_C =T_K -273.15
Celsius and Fahrenheit’s scales are related as …(i)
T_F = \(\frac{9}{5}\) T_C +32 ………………………….. (ii)
For neon:
T_K = 24.57K
∴ T_G = 24.57 – 273.15 = -284.58°C
T_F =\(\frac{9}{5}\) T_C +32
= \(\frac{9}{5}\) (-248.58) + 32
=-415.44° F

For carbon dioxide:
T_K = 216.55K
∴ T_C =216.55-273.15 = -56.60°C
T_F =\(\frac{9}{5}\) T_C +32
= \(\frac{9}{5}\) (-56.60) + 32 = -69.88°C

Question 2.
Two absolute scales A and B have triple points of water defined to be 200A and 350B.
What is the relation between T_Aand T_B?
Solution:
Triple point of water on absolute scale A, T₁ = 200 A
Triple point of water on absolute scale B, T₂ = 350 B
Triple point of water on Kelvin scale, T_K = 273.15K
The temperature 273.15K on Kelvein scale is equivalent to 200 A on absolute scale A.
T₁ =T_K
200 A = 273.15 K .
∴ A= \(\frac{273.15}{200}\)

The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.
T₂=T_K
350 B = 273.15
∴ B = \(\frac{273.15}{350}\)
T_A is triple point of water on scale A.
T_B is triple point of water on scale B.
∴ \(\frac{273.15}{200} \times T_{A}=\frac{273.15}{350} \times T_{B}\)
T_A = \(\frac{200}{350} T_{B}\)
T_A = \(\frac{4}{7} T_{B}\)
Therefore, the ratioT_A:T_B is given as 4 : 7.

Question 3.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:
R = R₀[l + α (T-T₀)]
The resistance is 101.6Ω at the triple-point of water 273.16 K, and 165.50 at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4Ω?
Solution:
It is given that:
R = R₀[l + 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 = 273.16 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=R₀[l + α (T-T₀)]
165.5 = 101.6[1 + α (600.5-273.16)]
1.629 = 1+α (327.34)
∴ α = \(\frac{0.629}{327.34} \) = 1.92 x10^-3K^-1
For resistance, R₁ = 123.4Ω
R₁ =R₀[l + α (T-T₀)]

where, T is the temperature when the resistance of lead is 123.4Ω
123.4 =101.6[1 +1.92 x 10^-3(T-273.16)]
1.214 =1+1.92 x 10^-3(T- 273.16)
\(\frac{0.214}{1.92 \times 10^{-3}}\) = T -273.16
111.46 = T-273.16
⇒ T =111.46 +273.16
∴ T = 384.62 K

Question 4.
Answer the following:
(a) The triple-point of water is a standard fixed point in modern thermometry. Why? What is wrong in taking the melting point of ice and the boiling point of water as standard fixed points (as was originally done in the Celsius scale)?

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the numbers 0°C and 100°C respectively. On the absolute scale, one of the fixed points is the triple-point of water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other fixed point on this (Kelvin) scale?

(c) The absolute temperature (Kelvin scale) T is related to the temperature t_c on the Celsius scale by
t_c = 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?
Solution:
(a) The triple point of water has a unique value of 273.16 K. At particular values of volume and pressure, the triple point of water is always 273.16 K. The melting point of ice and boiling point of water do not have particular values because these points depend on pressure and temperature.

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

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

(d) Let T_F be the temperature on Fahrenheit scale and T_K be the temperature on absolute scale. Both the temperatures can be related as
\(\frac{T_{F}-32}{180}=\frac{T_{K}-273.15}{100}\) ………………………….. (i)
Let T_F1 be the temperature on Fahrenheit scale and T_K1 be the temperature on absolute scale. Both the temperatures can be relates as
\(\frac{T_{F 1}-32}{180}=\frac{T_{K 1}-273.15}{100}\) ……………………….. (ii)
It is given that:
T_K1-T_K= 1K

Subtracting equation (i) from equation (ii), we get
\(\frac{T_{F 1}-T_{F}}{180}=\frac{T_{K 1}-T_{K}}{100}\)
\(T_{F 1}-T_{F}=\frac{1 \times 180}{100}=\frac{9}{5}\)
Triple point of water = 273.16 K
∴ Triple point of water on absolute scale = 273.16 x \( \frac{9}{5}\) = 491.69

Question 5.
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:

Temperature	Pressure thermometer A	Pressure thermometer B
Triple-point of water	1.250 x 105 Pa	0.200 x 105 Pa
Normal melting point of sulphur	1.797 x 105 Pa	0.287 x 105 Pa

(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B?
(b) What do you think is the reason behind the slight difference in answers of thermometers A and B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?
Solution:
Triple point of water, T = 273.16 K.
At this temperature, pressure in thermometer A, P_A = 1.250 x 10⁵ Pa
Let T₁ be the normal melting point of sulphur.
At this temperature, pressure in thermometer A,P₁ = 1.797 x 10⁵ Pa
According to Charles’ law, we have the relation
\(\frac{P_{A}}{T}=\frac{P_{1}}{T_{1}}\)
∴ T₁ = \(\frac{P_{1} T}{P_{A}}=\frac{1.797 \times 10^{5} \times 273.16}{1.250 \times 10^{5}}\)
= 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 237.16 K, the pressure in thermometer B,
P_B =0.200 x 10⁵ Pa
At temperature T₂, the pressure in thermometer B, P₂ = 0.287 x 10⁵ Pa
According to Charles’ law, we can write the relation
\(\frac{P_{B}}{T}=\frac{P_{2}}{T_{2}}\)
\(\frac{0.200 \times 10^{5}}{273.16}=\frac{0.287 \times 10^{5}}{T_{2}}\)
∴ T₂ = \(\frac{0.287 \times 10^{5}}{0.200 \times 10^{5}} \times 273.16\) = 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.

Question 6.
A steel tape lm 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 cmon a hot day when the temperature is 45°C. What is the actual length of the steel rod on that day? What is the length of the same steel rod on a day when the temperature is 27.0°C? Coefficient of linear expansion of steel = 1.20 x 10^-5 K^-1.
Solution:
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 x 10^-5K^-1

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)
∴ l’ = 100 +1.20 x 10^-5x 100(45 -27)
= 100,0216 cm

Hence, the actual length of the steel rod measured by the steel tape at 45° C can be calculated as
l₂ = \(\frac{100.0216}{100} \times 63\) = 63.0136cm
Therefore, the actual length of the rod at 45°C is 63.0136 cm. Its length at 27.0 °C is 63.0 cm

Question 7.
A large steel wheel is to be fitted onto 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 8.69cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume the coefficient of linear expansion of the steel to be constant over the required temperature range: α_steel = 1.20 x 10^-5 K^-1.
Solution:
The given temperature, T = 27°C can be written in Kelvin as
27 + 273 =300K
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, a steel = 1.20 x 10^-5 K ^-1
After the shaft is cooled using ‘dry ice’, its temperature becomes T₁.
The wheel will slip on the shaft if the change in diameter,
Δd = 8.69-8.70 =-0.01 cm

Temperature T₁; can be calculated from the relation
Δd = d₁α_steel(T₁ -T)
-0.01 =8.70 x 1.20 x 10^-5(T₁ -300)
(T₁ – 300) = \(\frac{-0.01}{8.70 \times 1.20 \times 10^{-5}}\)
(T₁ -300) = -95.78
∴ T₁ = 300 – 95.78 = 204.22 K
= (204.22-273)°C
= -68.78°C ≈ 69°C
Therefore, the wheel will slip on the shaft when the temperature of the shaft is -69°C.

Question 8.
A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0°C. What is the change in the diameter of the hole when the sheet is heated to 227°C? Coefficient of linear expansion of copper = 1.70 x 10^-5 K ^-1.
solution:
Initial temperature, T₁ = 27.0°C
Diameter of the hole at T₁, d₁ = 4.24 cm
Final temperature, T₂ = 227°C
Let, diameter of the hole at T₂=d₂
Coefficient of linear expansion of copper, α_Cu = 1.70 x 10^-5 K^-1
For coefficient of superficial expansion β, and change in temperature ΔT, we have the relation:
\(\frac{\text { Change in area }(\Delta \mathrm{A})}{\text { Original area }(\mathrm{A})}=\beta \Delta \mathrm{T}\)

Change in diameter = d₂ – d₁ = 4.2544 – 4.24 = 0.0144 cm
Hence, the diameter increases by 1.44 x 10^-2 cm.

Question 9.
A brass wire 1.8m 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? Coefficient of linear expansion of brass = 2.0 x 10^-5K^-1; Young’s modulus of brass = 0.91 x 10¹¹ Pa.
Solution:
Initial temperature, T₁ = 27°C
Length of the brass wire at T₁,l = 1.8 m
Final temperature, T₂ = -39 °C
Diameter of the wire, d = 2.0 mm = 2 x 10^-3 m
Let, tension developed in the wire = F
Coefficient of linear expansion of brass, a = 2.0 x 10^-5 K^-1

Young’s modulus of brass, Y = 0.91 x 10¹¹Pa
Young’s modulus is given by the relation ‘
Y = \(\frac{\text { Stress }}{\text { Strain }}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\)
ΔL = \(\frac{F \times L}{A \times Y}\) ……………………………….. (i)
where, F = Tension developed in the wire
A = Area of cross-section of the wire.
ΔL = Change in the length, given by the relation
ΔL = αL(T₂ -T₁) …………………………… (ii)

Equating equations (i) and (ii), we get

= -3.8 x 10² N
(The negative sign indicates that the tension is directed inward.) Hence, the tension developed in the wire is 3.8 x 10² N.

Question 10.
A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C? Is there a ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Coefficient of linear expansion of brass = 2.0 x 10^-5 K^-1, steel = 1.2 x 10^-5K^-1).
Solution:
Initial temperature, T₁ = 40°C
Final temperature, T₂ = 250 °C
Change in temperature, ΔT = T₂ – T₁ = 210°C
Length of the brass rod at T₁,l₁ = 50 cm
Diameter of the brass rod at T₂, d₁ = 3.0 mm
Length of the steel rod at T₂,l₂ = 50 cm
Diameter of the steel rod at T₂,d₂ = 3.0 mm
Coefficient of linear expansion of brass, α₁ = 2.0 x 10^-5 K^-1
Coefficient of linear expansion of steel, α₂ = 1.2 x 10^-5 K^-1

For the expansion in the brass rod, we have
\(\frac{\text { Change in length }\left(\Delta l_{1}\right)}{\text { Original length }\left(l_{1}\right)}=\alpha_{1} \Delta T\)
∴ Δl₁ = 50 x (2.1 x 10^-5)X210
= 0.2205cm
For the expansion in the steel rod, we have Change in length
\(\frac{\text { Change in length }\left(\Delta l_{2}\right)}{\text { Original length }\left(l_{2}\right)}=\alpha_{2} \Delta T\)
∴ Δl₂ =50 x (1.2 x 10^-5)x 210
= 0.126 cm

Total change in the lengths of brass and steel,
Δl = Δl₁ + Δl₂
=0.2205 + 0.126
= 0.346 cm
Total change in the length of the combined rod = 0.346 cm Since the rod expands freely from both ends, no thermal stress is developed at the junction.

Question 11.
The coefficient of volume expansion of glycerin is 49 x 10^-5K^-1.
What is the fractional change in its density for a 30°C rise in temperature?
Solution:
Coefficient of volume expansion of glycerin, α_v = 49 x 10^-5 K^-1
Rise in temperature, ΔT = 30°C
Fractional change in its volume = \(\frac{\Delta V}{V}\)

where, m = Mass of glycerine
PT₁ = Initial density at T₁
PT₂ = Final density at T₂
\(\frac{\rho_{T_{1}}-\rho_{T_{2}}}{\rho_{T_{2}}}=\alpha_{\mathrm{V}} \Delta T\)
Where, \(\frac{\rho_{T_{1}}-\rho_{T_{2}}}{\rho_{T_{2}}}\) = Fractional change in the density
∴ Fractional change in the density of glycerin
= 49 x 10^-5 x 30 = 1.47 x 10^-2

Question 12.
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.91 Jg^-1K^-1.
Solution:
Power of the drilling machine, P =10 kW = 10 x 10³ W
Mass of the aluminum block, m = 8.0 kg = 8 x 10³ g
Time for which the machine is used, t = 2.5min = 2.5x 60 = 150 s
Specific heat of aluminium, C = 0.91 J g^-1K^-1
Rise in the temperature of the block after drilling = ΔT

Total energy of the drilling machine = Pt
= 10 x 10³ x 150 = 1.5 x 10⁶ J

It is given that only 50% of the power is useful.

Useful energy, ΔQ = \(\frac{50}{100} \times 1.5 \times 10^{6}\) = 7.5×10⁵J
But ΔQ = mCΔT
∴ ΔT = \(\frac{\Delta Q}{m C}=\frac{7.5 \times 10^{5}}{8 \times 10^{3} \times 0.91}\)
=103°C
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.

Question 13.
A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? (Specific heat of copper = 0.39 Jg^-1K^-1; heat of fusion of water = 335Jg^-1).
Solution:
Mass of the copper block, m = 2.5kg = 2500 g
Rise in the temperature of the copper block, Δθ = 500 °C
Specific heat of copper, C = 0.39 J g^-1 °C^-1
The heat of fusion of water, L = 335 J g^-1

The maximum heat the copper block can lose, Q = mCΔθ
= 2500×0.39×500 =487500J
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 = \(\frac{Q}{L}=\frac{487500}{335}\) =1455.22g
Hence, the maximum amount of ice that can melt is 1.45 kg.

Question 14.
In an experiment on the specific heat of a metal, a 0.20kg block of the metal at 150°C is dropped in a copper calorimeter (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?
Solution:
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’ = 0.025kg = 25g
Volume of water, V = 150 cm³

Mass (M) of water at temperature T = 27°C,
150×1 =150g
Fall in the temperature of the metal,
ΔT =T₁ -T₂ =150-40 =110°C
Specific heat of water, C_w = 4.186 J/g/K
Specific heat of metal = C
Heat lost by the metal, Q = mCΔT …………………………….. (i)

Rise in the temperature of the water and calorimeter system,
ΔT’ = 40 -27 = 13°C
Heat gained by the water and calorimeter system,
ΔQ’ = m₁C_wΔT’
= (M + m’)C_wΔT’ ……………………………… (ii)

Heat lost by the metal = Heat gained by the water and colourimeter system
mCΔT =(M + m’)C_wΔT’
200 xC x 110 = (150+25) x 4.186 x 13
∴ C = \(\frac{175 \times 4.186 \times 13}{110 \times 200} \) = 0.43 Jg^-1K^-1
If some heat is lost to the surroundings, then the value of C will be smaller than the actual value.

Question 15.
Given below are observations on molar specific heats at room temperature of some common gases.

Gas	Molar specific heat (Cv) (cal mol-1K-1)
Hydrogen	4.87
Nitrogen	4.97
Oxygen	5.02
Nitric oxide	4.99
Carbon monoxide	5.01
Chlorine	6.17

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?
Solution:
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 = \(\frac{5}{2} R=\frac{5}{2} \times 1.98\) = 4.95 cal mol^-1K^-1 With the exception of chlorine, all the observations in the given table agree with(\(\frac{5}{2} R\)) This is because at room temperature, chlorine also has vibrational modes of motion besides rotational and translational modes of motion.

Question 16.
Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO₂ co-exist in equilibrium?
(b)What is the effect of decrease of pressure on the fusion and boiling point of CO₂?
(c) What are the critical temperature and pressure for CO₂? What is their significance?
(d) Is CO₂ solid, liquid or gas at
(a) -70°C under 1 atm,
(b) -60°C under 10 atm,
(c) 15°C under 56 atm?
Solution:
The P-T phase diagram for CO₂ is shown in the following figure.

(a) C is the triple point of the CO₂ phase diagram. This means that at the temperature and pressure corresponding to this point (i.e., at -56.6°C and 5.11 atm), the solid, liquid, and vaporous phases of CO₂ co-exist in equilibrium.
(b) The fusion and boiling point of CO₂ decrease with a decrease in pressure.
(c) The critical temperature and critical pressure of CO₂ are 31.1°C and 73 atm respectively.
Even if it is compressed to a pressure greater than 73 atm, CO₂ will not liquefy above the critical temperature.
(d) It can be concluded from the P-T phase diagram of CO₂ that:
(i) CO₂ is gaseous at -70 °C, under 1 atm pressure
(ii) CO₂ is solid at -60 °C, under 10 atm pressure
(iii) CO₂ is liquid at 15°C, under 56 atm pressure.

Question 17.
Answer the following questions based on the P-T phase diagram of CO₂:
(a) CO₂ at 1 atm pressure and temperature -60°C is compressed isothermally. Does it go through a liquid phase?
(b) What happens when C0₂ 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?
The P-T phase diagram for CO₂ is shown in the following figure:

Solution:
(a) No
Explanation:
At 1 atm pressure and at -60°C, CO₂ lies to the left of -56.6°C (triple point C). Hence, it lies in the region of vaporous and solid phases. Thus, CO₂ condenses into the solid-state directly, without going through the liquid state.
(b) It condenses to solid directly.
Explanation:
At 4 atm pressure, C0₂ lies below 5.11 atm (triple point C). Hence, it lies in the region of vaporous and solid phases. Thus, it condenses into the solid-state directly, without passing through the liquid state.

(c) The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.
Explanation:
When the temperature of a mass of solid CO₂ (at 10 atm pressure and at -65° C) is increased, it changes to the liquid phase and then to the vaporous phase. It forms a line parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and vaporisation curves.

(d) It departs from ideal gas behaviour as pressure increases.
Explanation:
If CO₂ is heated to 70 °C and compressed isothermally, then it will not exhibit any transition to the liquid state. This is because 70 °C is higher than the critical temperature of CO₂. It will remain in the vapour state but will depart from its ideal behaviour as pressure increases.

Question 18.
A child running a temperature of 101°F is given an antipyrin (i. e., a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98°F in 20 min, what is the average rate of extra evaporation caused, by the drug? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The specific heat of human body is approximately the same as that of water, and latent heat of evaporation of water at that temperature is about 580 cal g^-1.
Solution:
Initial temperature of the body of the child, T₁ = 101°F
Final temperature of the body of the child, T₂ = 98°F
Change in temperature, ΔT = \(\left[(101-98) \times \frac{5}{9}\right] \)°c
Time taken to reduce the temperature, t = 20 min
Mass of the child, m = 30 kg = 30 x 10³ g

Specific heat of the human body = Specific heat of water = C
= 1000 cal/kg/°C
Latent heat of evaporation of water, L = 580 cal g^-1
The heat lost by the child is given as:
ΔQ = mCΔT
= 30 x 1000 x (101-98)x \(\frac{5}{9}\) = 50000cal

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
ΔQ = m₁L
∴ m₁ = \(\frac{\Delta Q}{L}=\frac{50000}{580}\) = 86.2g
∴ Average rate of extra evaporation caused by the drug = \(\frac{m_{1}}{t}\)
= \(\frac{86.2}{20}\) = 4.3 g/mm.

Question 19.
A ‘thermal’ 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 coefficient of thermal conductivity of thermal is 0.01 J s^-1m^-1 K^-1.
[Heat of fusion of water = 335 x 10³ Jkg ^-1 ]
Solution:
Side of the given cubical icebox, s = 30 cm = 0.3 m
Thickness of the icebox, l = 5.0 cm = 0.05 m
Mass of ice kept in the icebox, m = 4 kg
Time gap, t=6h = 6x 60 x 60 s
Outside temperature, T = 45°C

Coefficient of thermal conductivity of thermacole,
K =0.01 Js^-1 m^-1 K^-1
Heat of fusion of water, L = 335 x 10³ J kg^-1
Let m be the total amount of ice that melts in 6 h.
The amount of heat lost by the food,
Q = \(\frac{K A(T-0) t}{l}\)

where, A = Surface area of the box = 6s² =6 x (0.3)² = 0.54 m²
Q = \(\frac{0.01 \times 0.54 \times(45) \times 6 \times 60 \times 60}{0.05}\) = 104976 J
But Q=m’L
∴ m’ = \(\frac{Q}{L}=\frac{104976}{335 \times 10^{3}}\) = 0.313 kg
Mass of ice left = 4-0.313 = 3.687kg .
Hence, the amount of ice remaining after 6 h is 3.687 kg.

Question 20.
A brass boiler has a base area of 0.15m² and a 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 Js^-1m^-1 K^-1; Heat of vaporisation of water = 2256 x 10³ Jkg^-1.
Solution:
Base area of the boiler, A = 0.15 m²
Thickness of the boiler,l = 1.0 cm = 0.01 m
Roiling rate of water, R = 6.0 kg/min
Mass, m = 6 kg
Time, t = 1 min = 60 s

Thermal conductivity of brass, K = 109 J s^-1 m^-1 K^-1
Heat of vaporisation, L = 2256 x10³ J kg^-1
The amount of heat flowing into water through the brass base of the boiler is given by
Q = \(\frac{K A\left(T_{1}-T_{2}\right) t}{l}\) ………………………. (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
Q = mL …………………………………. (ii)

Equating equations (i) and (ii), we get
∴ mL = \(\frac{K A\left(T_{1}-T_{2}\right) t}{l}\)
T₁ -T₂= \(\frac{m L l}{K A t}\)
= \(\frac{6 \times 2256 \times 10^{3} \times 0.01}{109 \times 0.15 \times 60}\) = 137.98°C
T₁ -T₂ = 137.98°C
∴ T₁=137.98+100 = 237.98°C
Therefore, the temperature of the part of the flame in contact with the boiler is 237.98°C.

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

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

(c) An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature of a red hot iron piece kept in the open. Black body radiation equation is given by
E = σ(T⁴ -T₀⁴) where, E – Energy radiation
T = Temperature of optical pyrometer
T₀ = Temperature of open space
σ = Constant
Hence, an increase in the temperature of open space reduces the radiation energy.
When the same piece of iron is placed in a furnace, the radiation energy, E =σT⁴

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

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

Question 22.
A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool from 60°C to 30°C. The temperature of the surroundings is 20 C.
Solution:
According to Newton’s law of cooling, the rate of cooling cc difference in temperature.
Here, average of 80 °C and 50 °C, T = \(\frac{T_{1}+T_{2}}{2}=\frac{80+50}{2}\) = 65°C
Temperature of surroundings, T₀ = 20 °C .
∴Difference, ΔT = T – T₀ = 65 – 20 = 45°C
Under these conditions, the body cools 30 °C in time 5 minutes.
∴ 
or \(\frac{30}{5}\) = k x 45 ……………………………….. (i)
The average of 60°C and 30 °C is 45°C which is 25°C (45 – 20) above the room temperature and the body cools by 30 °C (60 – 30) in a time t (say).
∴\(\frac{30}{t}\) = k x 25 ……………………………….. (ii)

where k is same for this situation as for the original.
Dividing eq. (i) from eq. (ii), we get
\(\frac{30 / 5}{30 / t}=\frac{k \times 45}{k \times 25}\)
or \(\frac{t}{5}=\frac{9}{5}\)
or t = 9 min

Punjab State Board PSEB 11th Class Physics Book Solutions Chapter 10 Mechanical Properties of Fluids Textbook Exercise Questions and Answers.

PSEB Solutions for Class 11 Physics Chapter 10 Mechanical Properties of Fluids

PSEB 11th Class Physics Guide Mechanical Properties of Fluids Textbook Questions and Answers

Question 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 pressure of a liquid is given by the relation
P =hρg
where, P = Pressure
h = Height of the liquid column
ρ = Density of the liquid ‘ .
g = Acceleration due to the gravity

It can be inferred that pressure is directly proportional to height. Hence, the blood pressure in human vessels depends on the height of the blood column in the body. The height of the blood column is more at the feet than it is at the brain. Hence, jthe blood pressure at the feet is more than it is at the brain.

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

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

Question 2.
Explain why
(a) The angle of contact of mercury with glass is obtuse, while that of water with glass is acute.
(b) Water on a clean glass surface tends to spread out while mercury on the same surface tends to form drops. (Put differently, water wets glass while mercury does not.)
(c) Surface tension of a liquid is independent of the area of the surface
(d) Water with detergent dissolved in it should have small angles of contact.
(e) A drop of liquid under no external forces is always spherical in shape
Solution:
(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 (0), as shown in the given figure.

S_la, S_sa, and S_sl 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.,
cos θ = \(\frac{S_{s a}-S_{s l}}{S_{l a}}\)
The angle of contact 0, is obtuse if S_sa < S_la (as in the case of mercury on glass). This angle is acute if Ss < Sa (as in the case of water on glass).

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

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

(d) Water with detergent dissolved in it has small angles of contact (0). This is because for a small 0, 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 (0). If 0 is small, then cos 0 will be large and the rise of the detergent water in the cloth will be fast.

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

Question 3.
Fill in the blanks using the word(s) from the list appended with each statement:
(a) Surface tension of liquids generally…with temperatures (increases/ decreases)
(b) Viscosity of gases …………………. with temperature, whereas viscosity of liquids ………………………… with temperature (increases/decreases)
(c) For solids with elastic modulus of rigidity, the shearing force is proportional to …………………………… while for fluids it is proportional to ………………………………… (shear strain/rate of shear strain)
(d) For a fluid in a steady flow, the increase in flow speed at a constriction follows (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)
Solution:
(a) decreases
The surface tension of a liquid is inversely proportional to temperature.

(b) increases; decreases
Most fluids offer resistance to their motion. This is like internal mechanical friction, known as viscosity. Viscosity of gases increases with temperature, while viscosity of liquids decreases with temperature.

(c) shear strain; rate of shear strain
With reference to the elastic modulus of rigidity for solids, the shearing force is proportional to the shear strain. With reference to the elastic modulus of rigidity for fluids, the shearing force is proportional to the rate of shear strain.

(d) conservation of mass/Bernoulli’s principle
For a steady-flowing fluid, an increase in its flow speed at a constriction follows the conservation of mass/Bemoulli’s principle.

(e) greater
For the model of a plane in a wind tunnel, turbulence occurs at a greater speed than it does for an actual plane. This follows from Bernoulli’s principle and different Reynolds numbers are associated with the motions of the two planes. ,

Question 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 air is blown under a paper, the velocity of air is greater under the paper than it is above it. As per Bernoulli’s principle, atmospheric pressure reduces under the paper. This makes the paper fall. To keep a piece of paper horizontal, one should blow over it. This increases the velocity of air above the paper. As per Bernoulli’s principle, atmospheric pressure reduces above the paper and the paper remains horizontal.

(b) According to the equation of continuity,
Area x Velocity = Constant
For a smaller opening, the velocity of flow of a fluid is greater than it is when the opening is bigger. When we try to close a tap of water with our fingers, fast jets of water gush through the openings between our fingers. This is because very small openings are left for the water to flow out of the pipe. Hence, area and velocity are inversely proportional to each other.

(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 x 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.

Question 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?
Solution:
Mass of the girl, m = 50 kg
Diameter of the heel, d = 1 cm = 0.01 m
Radius of the heel, r = \(\frac{d}{2}\) = 0.005 m
Area of the heel = πr²
= 3.14 x (0.005)²
= 7.85 x 10^-5 m²

Force exerted by the heel on the floor,
F = mg
= 50 x 9.8 = 490 N
Pressure exerted by the heel on the floor,
P = \(\frac{\text { Force }}{\text { Area }}\)
= \(\frac{490}{7.85 \times 10^{-5}}\) = 6.24 x 10⁶Nm^-2
Therefore, the pressure exerted by the heel on the horizontal floor is 6.24 x 10⁶Nm^-2 .

Question 6.
Torieelli’s barometer used mercury. Pascal duplicated it using French wine of density 984kg m3. Determine the height of the wine column for normal atmospheric pressure.
Solution:
Density of mercury, ρ₁ = 13.6 x 10³ kg / m³
Height of the mercury column, h₁ = 0.76 m
Density of French wine, ρ₂ = 984 kg / m³
Height of the French wine column = h₂
Acceleration due to gravity, g = 9.8 m / s²

The pressure in both the columns is equal, i. e.,
Pressure in the mercury column = Pressure in the French wine column
ρ₁h₁g = ρ₂h₂g
h₂ = \(\frac{\rho_{1} h_{1}}{\rho_{2}}\)
= \(\frac{13.6 \times 10^{3} \times 0.76}{984}\)
= 10.5m
Hence, the height of the French wine column for normal atmospheric pressure is 10.5 m.

Question 7.
A vertical off-shore structure is built to withstand a maximum stress of 10⁹ Pa. Is the structure suitable for putting up on top of an oil well in the ocean? Take the depth of the ocean to bet roughly 3 km, and ignore ocean currents.
Solution:
Yes The maximum allowable stress for the structure, P = 10⁹Pa
Depth of the ocean, d = 3 km = 3 x 10³ m
Density of water, ρ = 10³ kg / m³
Acceleration due to gravity, g = 9.8 m / s²

The pressure exerted because of the sea water at depth, d = ρdg
= 3 x 10³ x 10³ x 9.8 = 2.94 x 10⁷ Pa
The maximum allowable stress for the structure (10⁹ Pa) is greater than the pressure of the seawater (2.94 x 10⁷ Pa). The pressure exerted by the ocean is less than the pressure that the structure can withstand. Hence, the structure is suitable for putting up on top of an oil well in the ocean.

Question 8.
A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000 kg. The area of cross-section of the piston carrying the load is 425cm2. What maximum pressure would the smaller piston have to bear?
Solution:
The maximum mass of a car that can be lifted, m = 3000 kg
Area of cross-section of the load-carrying piston, A = 425 cm² = 425x 10^-4m²
The maximum force exerted by the load, F = mg
= 3000 x 9.8 = 29400N
The maximum pressure exerted on the load-carrying piston, P = \(\frac{F}{A}\)
= \(\frac{29400}{425 \times 10^{-4}}\)
= 6.917 x 10⁵ Pa
Pressure is transmitted equally in all directions in a liquid. Therefore, the maximum pressure that the smaller piston would have to bear is 6.917 x 10⁵Pa.

Question 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?
Solution:
The given system of water, mercury, and methylated spirit is shown as follows:

Height of the spirit column, h₁ = 12.5cm = 0.125m
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 + h₁ρ₁g
Pressure at point D = P₀ + h₂ρ₂g
Pressure at points B and D is the same.

Therefore, the specific gravity of spirit is 0.8.

Question 10.
In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms? (Specific gravity of mercury = 13.6)
Solution:
Height of the water column, h₁ =10+15 = 25cm
Height of the spirit column, h₂ = 12.5 +15 = 27.5cm
Density of water, ρ₁ = 1 g cm^-3
Density of spirit, ρ₂ = 0.8 g cm^-3
Density of mercury = 13.6 g cm^-3

Let h be the difference between the levels of mercury in the two arms. Pressure exerted by height h, of the mercury column:
= hρg = h x 13.6g ……………………………….. (i)
Difference between the pressures exerted by water and spirit
h₁ρ₁g – h₂ρ₁g
= g (25 x 1 – 27.5 x 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.

Question 11.
Can Bernoulli’s equation be used to describe the flow of water through a rapid in a river? Explain.
Answer:
No Explanation: 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 streamlined flow.

Question 12.
Does it matter if one uses gauge instead of absolute pressures in applying Bernoulli’s equation? Explain.
Answer:
No Explanation: 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.

Question 13.
Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 x 10^-3 kgs^-1, what is the pressure difference between the two ends of the tube?(Density of glycerine = 1.3 x 10³ kg m ^-3 and viscosity of glycerine = 0.83Pas). [You may also like to check if the assumption of laminar flow in the tube is correct].
Solution:
Length of the horizontal tube, l = 1.5m
Radius of the tube, r = 1 cm = 0.01 m
Diameter of the tube, d = 2r = 0.02 m
Glycerine is flowing at a rate of 4.0 x 10 kgs .
M = 4.0 x 10^-3 kgs^-1
Density of glycerine, ρ = 1.3 x 10^-3 kg m^-3
Viscosity of glycerine, η = 0.83Pas
Volume of glycerine flowing per sec,
V = \(\frac{M}{\rho}=\frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}}\)
= 3.08 x 10^-6 m³ s^-1
According to Poisevelle’s formula, we have the relation for the rate of flow,
V = \(\frac{\pi p r^{4}}{8 \eta l} \)
where, p is the pressure difference between the two ends of the tube
∴ p = \(\frac{V 8 \eta l}{\pi r^{4}}\)
= \(\frac{3.08 \times 10^{-6} \times 8 \times 0.83 \times 1.5}{3.14 \times(0.01)^{4}} \)
= 9.8 x 10² Pa
Reynold’s number is given by the relation,
R = \(\frac{4 \rho V}{\pi d \eta}=\frac{4 \times 1.3 \times 10^{3} \times 3.08 \times 10^{-6}}{3.14 \times(0.02) \times 0.83}\)
= 0.3
Reynold’s number is about 0.3. Hence, the flow is laminar.

Question 14.
In a test experiment on a model aeroplane in a wind tunnel, the flow speeds on the upper and lower surfaces of the wing are 70ms^-1 and 63 ms^-1 respectively. What is the lift on the wing if its area is 2.5m²? Take the density of air to be 1.3 kg m^-3.
Solution:
Speed of wind on the upper surface of the wing, V₁ = 70 m/s
Speed of wind on the lower surface of the wing, V₂ = 63 m/s
Area of the wing, A = 2.5 m²
Density of air, ρ = 1.3 kg m
According to Bernoulli’s theorem, we have the relation:

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

Therefore, the lift on the wing of the aeroplane is 1.51 x 10³N.

Question 15.
Figures (a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect? Why?

Solution:
Figure (a) is incorrect.
Take the case given in figure (b).

where, A₁ = Area of pipe 1
A₂ = Area of pipe 2
V₁ = Speed of the fluid in pipe 1
V₂ = Speed of the fluid in pipe 2
From the law of continuity, we have
A₁V₁ = A₂V₂
When the area of a cross-section in the middle of the venturi meter is small, the speed of the flow of liquid through this part is more. According to Bernoulli’s principle, if speed is more, then pressure is less. Pressure is directly proportional to height. Hence, the level of water in pipe 2 is less. Therefore, figure (a) is not possible.

Question 16.
The cylindrical tube of a spray pump has a cross-section of 8.0 cm² one end of which has 40 fine holes each of diameter 1.0 mm. If the liquid flow inside the tube is 1.5 m min^-1, what is the speed of ejection of the liquid through the holes?
Solution:
Area of cross-section of the spray pump. A = 8 cm² = 8 x 10^-4 m²
number of holes, n = 40
Diameter of each hole, d = 1 mm = 1 x 10^-3 m
Radius of each hole,r = d/2 = 0.5 x 10^-3 m
Area of cross-section of each hole, a = πr² = π(0.5 x 10^-3)²m²
Total area of 40 holes, A₂ = n x a
= 40 x 3.14 x (0.5 x 10^-3)² m²
= 31.41 x 10^-6 m²

Speed of flow of liquid inside the tube, V₁ = 1.5 m/min = 0.025m/s
Speed of ejection of liquid through the holes = V₂
According to the law of continuity, we have A₁V₁ = A₂V₂
V₂ = \(\frac{A_{1} V_{1}}{A_{2}}=\frac{8 \times 10^{-4} \times 0.025}{31.41 \times 10^{-6}}\)
= 0.636 m/s
Therefore, the speed of ejection of the liquid through the holes is 0.636 m/s.

Question 17.
A U-shaped wire is dipped in a soap solution and removed. The thin soap film formed between the wire and the light slider supports a weight of 1.5 x 10^-2 N (which includes the small weight of the slider). The length of the slider is 30 cm. What is the surface tension of the film?
Solution:
The weight that the soap film supports, W = 1.5 x 10^-2 N
Length of the slider, l = 30 cm = 0.3 m
A soap film has two free surfaces.
∴ Total length = 2l = 2 x 0.3 = 0.6 m
Surface tension, T = \(\frac{\text { Force or Weight }}{2 l} \)
= \(\frac{1.5 \times 10^{-2}}{0.6}\) =  2.5 x10^-2  N/m
Therefore, the surface tension of the film is 2.5 x10^-2Nm^-1.

Question 18.
Figure (a) shows a thin liquid film supporting a small weight = 4.5 x 10^-2 N.
What is the weight supported by a film of the same liquid at the same temperature in fig. (b) and (c)? Explain your answer physically.

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

Question 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 x 10^-1 Nm^-1. The atmospheric pressure is 1.01 x 10⁵ Pa. Also give the excess pressure inside the drop.
Solution:
Radius of the mercury drop, r = 3.00 mm = 3 x 10^-3 m
Surface tension of mercury, T = 4.65 x 10^-1 N m^-1
Atmospheric pressure, P₀ = 1.01 x 10⁵ Pa
Total pressure inside the mercury drop = Excess pressure inside mercury + Atmospheric pressure
= \(\frac{2 T}{r}+P_{0}\)

Excess pressure = \(\frac{2 T}{r}=\frac{2 \times 4.65 \times 10^{-1}}{3 \times 10^{-3}}\) = 310 Pa

Question 20.
What is the excess pressure inside a bubble of soap solution of ‘ radius 5.00 mm, given that the surface tension of soap solution at the temperature (20°C) is 2.50 x 10^-2 Nm^-1? If an air bubble of the same dimension were formed at depth of 40.0 cm inside a container containing the soap solution (of relative density 1.20), what would be the pressure inside the bubble? (1 atmospheric pressure is 1.01 x 10⁵ Pa).
Solution:
Soap bubble is of radius, r = 5.00 mm = 5 x 10^-3 m
Surface tension of the soap solution, T = 2.50 x 10^-2 Nm^-1
Relative density of the soap solution = 1.20
∴ Density of the soap solution, ρ = 1.2 x 10³ kg/m³
Air bubble formed at a depth, h = 40 cm = 0.4 m
Radius of the air bubble, r = 5 mm = 5 x 10^-3 m
1 atmospheric pressure = 1.01 x 10⁵Pa

Acceleration due to gravity, g = 9.8 m/s²
Hence, the excess pressure inside the soap bubble is given by the relation
P = \(\frac{4 T}{r}=\frac{4 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\) = 20 Pa
Therefore, the excess pressure inside the soap bubble is 20 Pa.
The excess pressure inside the air bubble is given by the relation
P’ = \(\frac{2 T}{r}\)
= \(\frac{2 \times 2.5 \times 10^{-2}}{5 \times 10^{-3}}\)
=10 Pa

Therefore, the excess pressure inside the air bubble is 10 Pa.
At a depth of 0.4 m, the total pressure inside the air bubble =Atmospheric pressure + hρg + P’
= 1.01 x 10⁵ + 0.4 x 1.2 x 10³ x 9.8 + 10 ,
= 1.057 x 10⁵ Pa = 1.06 x 10⁵ Pa
Therefore, the pressure inside the air bubble is 1.06 x 10⁵ Pa.

Additional Exercises

Question 21.
A tank with a square base of area 1.0 m² is divided by a vertical partition in the middle. The bottom of the partition has a small-hinged door of area 20 cm². The tank is filled with water in one compartment, and an acid (of relative density 1.7) in the other, both to a height of 4.0 m. compute the force necessary to keep the door close.
Solution:
Base area of the given tank, A = 1.0 m²
Area of the hinged door, a = 20 cm² = 20 x 10^-4 m²
Density of water, ρ₁ = 10³ kg/m³
Density of acid, ρ₂ = 1.7 x 10³ kg/m³
Height of the water column, h₁ = 4 m
Height of the acid column, h₂ = 4 m
Acceleration due to gravity, g = 9.8 m/s²
Pressure due to water is given as
P₁ =h₁ρ₁g = 4 x 10³ x 9.8 = 3.92 x 10⁴Pa
Pressure due to acid is given as, P₂ = h₂ρ₂g
= 4 x 1.7 x10³ x 9.8
= 6.664 x 10⁴ Pa

Pressure difference between the water and acid columns,
ΔP=P₂– P₁
= 6.664 x 10⁴ -3.92 x10⁴
= 2.744 x10⁴ Pa
Hence, the force exerted on the door = ΔP x a
= 2.744 x 10⁴ x 20 x 10^-4 = 54.88N
Therefore, the force necessary to keep the door closed is 54.88N.

Question 22.
A manometer reads the pressure of a gas in an enclosure as shown in figure (a). When a pump removes some of the gas, the manometer reads as in figure (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).

Solution:
(a) For figure (a)
Atmospheric pressure, P₀ = 76 cm of Hg
The 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-18cm = 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,
P_R = 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,
P_L = 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.

Question 23.
Two vessels have the same base area but different shapes. The first vessel takes twice the volume of water that the second vessel requires to fill upto a particular common height. Is the force exerted by the water on the base of the vessel the same in the two cases? If so, why do the vessels filled with water to that same height give different readings on a weighing scale?
Answer:
Yes.
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.

Question 24.
During blood transfusion the needle is inserted in a vein where the gauge pressure is 2000Pa. At what height must the blood container be placed so that blood may just enter the vein? [Take the density of whole blood = 1.06 x 10³ kg m^-3 ].
Solution:
Given, gauge pressure, P = 2000 Pa
Density of whole blood, p = 1.06 x 10³ kg m^-3
Acceleration due to gravity, g = 9.8 m/s²
Height of the blood container = h
Pressure of the blood container, P = hρg
h = \(\frac{P}{\rho g}=\frac{2000}{1.06 \times 10^{3} \times 9.8}\)
= 0.1925 m
The blood may enter the vein if the blood container is kept at a height greater than 0.1925m, i. e., about 0.2 m.

Question 25.
In deriving Bernoulli’s equation, we equated the work done on the fluid in the tube to its change in the potential and kinetic energy,
(a) What is the largest average velocity of blood flow in an artery of diameter 2 x 10^-3 m if the flow must remain laminar?
(b) Do the dissipative forces become more important as the fluid velocity increases? Discuss qualitatively.
Solution:
(a) Diameter of the artery, d = 2×10^-3 m
Viscosity of blood, η = 2.084 x 10^-3 Pas
Density of blood, ρ = 1.06 x 10³ kg/m³
Reynolds’ number for laminar flow, N_R = 2000
The largest average velocity of blood is given as
V_avg = \(\frac{N_{R} \eta}{\rho d}\)
= \(\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 2 \times 10^{-3}}\)
= 1.966 m/s
Therefore, the largest average velocity of blood is 1.966 m/s
(b) Yes, as the fluid velocity increases, the dissipative forces become more important. This is because of the rise of turbulence. Turbulent flow causes dissipative loss in a fluid.

Question 26.
(a) What is the largest average velocity of blood flow in an artery of radius 2 x 10^-3 m if the flow must remain laminar?
(b) What is the corresponding flow rate? (Take viscosity of blood to be 2.084 x 10^-3 Pas).
Solution:
(a) Radius of the artery, r = 2 x 10^-3 m
Diameter of the artery, d=2 x 2x 10^-3 m = 4 x 10^-3m
Viscosity of blood, η = 2.084 x 10^-3 Pa s
Density of blood, ρ = 1.06 x 10³ kg/m³
Reynolds’ number for laminar flow, N_R = 2000
The largest average velocity of blood is given by the relation
V_Avg = \(\frac{N_{R} \eta}{\rho d}=\frac{2000 \times 2.084 \times 10^{-3}}{1.06 \times 10^{3} \times 4 \times 10^{-3}}\)
= 0.983 m/s
Therefore, the largest average velocity of blood is 0.98,3 m/s.

(b) Flow rate is given by the relation
R = πr² V_avg
= 3.14 x (2 x 10^-3)² x 0.983
= 1.235 x 10^-5m³s^-1
Therefore, the corresponding flow rate is 1.235 x 10^-5m³s^-1.

Question 27.
A plane is in level flight at constant speed and each of its two wings has an area of 25m². If the speed of the air is 180km/h over the lower wing and 234 km/h over the upper wing surface, determine the plane’s mass. (Take air density to be 1kg m^-3).
Solution:
The area of the wings of the plane, A = 2 x 25 = 50 m²
Speed of air over the lower wing,
V₁ = 180 km/h = 180 x \(\frac{5}{18}\) m/s = 50 m/s
Speed of air over the upper wing,
V₂ = 234 km/h = 234 x \(\frac{5}{18}\) m/s = 65 m/s
Density of air, ρ = 1 kg m^-3
Pressure of air over the lower wing = P₁
Pressure of air over the upper wing = P₂
The upward force on the plane can be obtained using Bernoulli’s equation as

The upward force (F) on the plane can be calculated as

Using Newton’s force equation, we can obtain the mass (m) of the plane as
F = mg
m = \(\frac{43125}{9.8}\)
= 4400.51 kg ≈ 4400 kg
Hence, the mass of the plane is about 4400 kg.

Question 28.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 x 10^-5 m and density 1.2 x 10^-5 kg m?
Take the viscosity of air at the temperature of the experiment to be 1.8x 10⁵ Pas. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.
Solution:
Terminal speed = 5.8cm/s; Viscous force = 3.9 x 10^-10 N
Radius of the given uncharged drop, r = 2.0 x 10^-5 m
Density of the uncharged drop, ρ = 1.2 x 10³ kg m^-3
Viscosity of air, η = 1.8 x 10^-5 Pa s
Density of air (ρ₀) can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, g = 9.8 m/s²
Terminal velocity (ν) is given by the relation
ν = \(\frac{2 r^{2} \times\left(\rho-\rho_{0}\right) g}{9 \eta}\)
= \(\frac{2 \times\left(2.0 \times 10^{-5}\right)^{2}\left(1.2 \times 10^{3}-0\right) \times 9.8}{9 \times 1.8 \times 10^{-5}}\)
= 5.807 x 10^-2ms^-1
= 5.8 cm s^-1
Hence, the terminal speed of the drop is 5.8 cms^-1.
The viscous force on the drop is given by:
F = 6πηrν
∴ F = 6 x 3.14 x 1.8 x 10^-5 x 2.0 x 10^-5 x 5.8 x 10^-2
= 3.9 x 10^-10N
Hence, the viscous force on the drop is 3.9 x 10^-10N.

Question 29.
Mercury has an angle of contact equal to 140° with soda-lime glass. A narrow tube of radius 1.00 mm made of this glass is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside? Surface tension of mercury at the temperature of the experiment is 0.465 Nm^-1. Density of mercury = 13.6 x 10³ kgm^-3.
Solution:
Angle of contact between mercury and soda-lime glass, θ = 140°
Radius of the narrow tube, r = 1 mm = 1 x 10^-3 m
Surface tension of mercury at the given temperature, T = 0.465N m^-1
Density of mercury, ρ = 13.6 x 10³ kg/m³
Dip in the height of mercury = h

Acceleration due to gravity, g = 9.8 m/s²
Surface tension is related with the angle of contact and the dip in the height as
T = \(\frac{h \rho g r}{2 \cos \theta}\)
∴ h = \(\frac{2 T \cos \theta}{r \rho g}\)

= -5.34 mm
Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm

Question 30.
Two narrow bores of diameters 3.0mm and 6.0mm 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 x 10^-2Nm^-1.
Take the angle of contact to be zero and density of water to be 1.0x 10³ kg m^-3 (g = 9.8ms^-2).
Solution:
Diameter of the first bore, d₁ = 3.0 mm = 3 x 10^-3 m
Hence, the radius of the first bore, r₁ = \(\frac{d_{1}}{2}\) =1.5 x 10^-3m
Diameter of the second bore, d₂ =6.0 mm
Hence, the radius of the second bore, r₂ = \(\frac{d_{2}}{2} \) = 3 x 10^-3 m
Surface tension of water, T = 7.3 x 10^-2 N m^-1
Angle of contact between the bore surface and water, θ=0
Density of water, ρ = 1.0 x 10³ kg/m^-3
Acceleration due to gravity, g =9.8 m/s²
Let h₁ and h₂ be the heights to which water rises in the first and second tubes respectively.

These heights are given by the relations
h₁ = \(\frac{2 T \cos \theta}{r_{1} \rho g}\) …………………..(i)
h₂ = \(\frac{2 T \cos \theta}{r_{2} \rho g}\) …………………… (ii)
The difference between the levels of water in the two limbs of the tube can be calculated as
= h₁ – h₂
= \(\frac{2 T \cos \theta}{r_{1} \rho g}-\frac{2 T \cos \theta}{r_{2} \rho g}\)

= 4.966 x 10^-3m = 4.97 mm
Hence, the difference between levels of water in the two bores is 4.97 mm.

Question 31.
(a) It is known that density p of air decreases with height y as ρ = ρ_oe^-y/yo
where ρ_o = 1.25kg m^-3 is the density at sea level, and 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 1425m³ 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 =8000m and ρ_He = 018 kg m^-3]
Solution:
Volume of the balloon, V = 1425m³
Mass of the payload, m = 400 kg
Acceleration due to gravity, g = 9.8 m/s²
y_o =8000m
ρ_He =0.18kgm^-3
ρ_o =1.25kg/m³

Density of the balloon = ρ
Height to which the balloon rises = y
Density (ρ) of air decreases with height (y) as
ρ = ρ₀e^-y/yo
\(\frac{\rho}{\rho_{0}}=e^{-y / y_{0}}\) …………………………… (i)

This density variation is called the law of atmospheres.
It can be inferred from equation (i) that the rate of decrease of density with height is directly proportional to ρ, i. e.,

where, k is the constant of proportionality
Height changes from 0 to y, while density changes from ρ_o to ρ).
Integrating the sides between these limits, we get

Comparing equations (i) and (ii) we get
y₀ = \(\frac{1}{k}\)
k = \(\frac{1}{y_{0}}\) ……………………………………. (iii)

From equations (ii) and (iii), we get
ρ = ρ₀e^-y/yo
(b) Density,
ρ = \(\frac{\text { Mass }}{\text { Volume }}\)

= 0.46 kg/m³
From equations (ii) arid (iii), we can obtain y as
ρ = ρ₀e^-y/yo
log e\(\frac{\rho}{\rho_{0}}=-\frac{y}{y_{0}}\)
∴ y =-8000 x log_e \(\frac{0.46}{1.25}\)
=-8000 x-1=8000m8 km
Hence, the balloon will rise to a height of 8 km.