JEE Physics Thermal Properties of Matter MCQs Set 27

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Chapter-wise Objective Questions: Thermal Properties Of Matter

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Multiple Choice Questions

Question 1: In a mercury thermometer, the ice point (lower fixed point) is marked as 10\(^\circ\)C and the steam point (upper fixed point) is marked as 130\(^\circ\)C. At 40\(^\circ\)C temperature, what will this thermometer read?
(a) 78\(^\circ\)C
(b) 66\(^\circ\)C
(c) 62\(^\circ\)C
(d) 58\(^\circ\)C
Show Answer & Explanation

Answer: (d) 58\(^\circ\)C

Explanation:
1. The scale has 120 divisions for 100 true degrees.
2. Reading \( =10+\dfrac{40}{100}\times120=58^\circ \)C.

Teacher's Note:
1. Map the true temperature linearly between the two fixed points.

Question 2: Two temperature scales A and B are related by \( \dfrac{A-42}{110}=\dfrac{B-72}{220} \). At which temperature do the two scales have the same reading?
(a) \( -42^\circ \)C
(b) \( -72^\circ \)C
(c) 12\(^\circ\)C
(d) 40\(^\circ\)C
Show Answer & Explanation

Answer: (c) 12\(^\circ\)C

Explanation:
1. Put \( A=B=x \): \( \dfrac{x-42}{110}=\dfrac{x-72}{220} \).
2. \( 2(x-42)=x-72\Rightarrow x=12 \).

Teacher's Note:
1. Set both readings equal and solve.

Question 3: The resistance of the wire in a platinum resistance thermometer at the ice point is 5 \( \Omega \) and at the steam point is 5.25 \( \Omega \). When the thermometer is inserted in an unknown hot bath, its resistance is found to be 5.5 \( \Omega \). The temperature of the hot bath is
(a) 100\(^\circ\)C
(b) 200\(^\circ\)C
(c) 300\(^\circ\)C
(d) 350\(^\circ\)C
Show Answer & Explanation

Answer: (b) 200\(^\circ\)C

Explanation:
1. \( t=\dfrac{R_t-R_0}{R_{100}-R_0}\times100=\dfrac{0.5}{0.25}\times100=200^\circ \)C.

Teacher's Note:
1. Resistance is assumed linear in temperature.

Question 4: On a hilly region, water boils at 95\(^\circ\)C. The temperature expressed in Fahrenheit is
(a) 100\(^\circ\)F
(b) 20.3\(^\circ\)F
(c) 150\(^\circ\)F
(d) 203\(^\circ\)F
Show Answer & Explanation

Answer: (d) 203\(^\circ\)F

Explanation:
1. \( F=\dfrac95\times95+32=203^\circ \)F.

Teacher's Note:
1. \( F=\tfrac95C+32 \).

Question 5: The readings of a constant volume gas thermometer at 0\(^\circ\)C and 100\(^\circ\)C are 40 cm and 60 cm of mercury respectively. If its reading at an unknown temperature is 100 cm of mercury column, then the temperature is
(a) 100\(^\circ\)C
(b) 50\(^\circ\)C
(c) 25\(^\circ\)C
(d) 300\(^\circ\)C
Show Answer & Explanation

Answer: (d) 300\(^\circ\)C

Explanation:
1. \( t=\dfrac{100-40}{60-40}\times100=300^\circ \)C.

Teacher's Note:
1. Pressure is linear in temperature at constant volume.

Question 6: On the Centigrade scale, the temperature of a body increases by 30\(^\circ\). The increase in temperature on the Fahrenheit scale is
(a) 50\(^\circ\)
(b) 40\(^\circ\)
(c) 30\(^\circ\)
(d) 54\(^\circ\)
Show Answer & Explanation

Answer: (d) 54\(^\circ\)

Explanation:
1. \( \Delta F=\dfrac95\Delta C=54^\circ \).

Teacher's Note:
1. One Celsius degree equals 1.8 Fahrenheit degrees.

Question 7: An aluminium sphere of 20 cm diameter is heated from 0\(^\circ\)C to 100\(^\circ\)C. Its volume changes by (\( \alpha_{Al}=23\times10^{-6}/^\circ \)C)
(a) 28.9 cc
(b) 2.89 cc
(c) 9.28 cc
(d) 49.8 cc
Show Answer & Explanation

Answer: (a) 28.9 cc

Explanation:
1. \( V=\tfrac43\pi(10)^3=4189 \) cc.
2. \( \Delta V=3\alpha V\Delta T=3\times23\times10^{-6}\times4189\times100=28.9 \) cc.

Teacher's Note:
1. \( \gamma=3\alpha \) for solids.

Question 8: A metal rod of Young's modulus \( Y \) and coefficient of thermal expansion \( \alpha \) is held at its two ends such that its length remains invariant. If its temperature is raised by \( t^\circ \)C, the linear stress developed in it is
(a) \( \dfrac{\alpha t}{Y} \)
(b) \( \dfrac{Y}{\alpha t} \)
(c) \( Y\alpha t \)
(d) \( \dfrac{1}{Y\alpha t} \)
Show Answer & Explanation

Answer: (c) \( Y\alpha t \)

Explanation:
1. Thermal strain \( \alpha t \) is prevented, so stress \( =Y\times\text{strain}=Y\alpha t \).

Teacher's Note:
1. Constrained expansion produces compressive stress.

Question 9: The coefficient of volume expansion of a liquid is \( 49\times10^{-5} \) K\(^{-1}\). The fractional change in its density when the temperature is raised by 30\(^\circ\)C is
(a) \( 7.5\times10^{-3} \)
(b) \( 3.0\times10^{-3} \)
(c) \( 1.5\times10^{-2} \)
(d) \( 1.1\times10^{-3} \)
Show Answer & Explanation

Answer: (c) \( 1.5\times10^{-2} \)

Explanation:
1. \( \dfrac{\Delta\rho}{\rho}=\gamma\Delta T=49\times10^{-5}\times30=1.47\times10^{-2}\approx1.5\times10^{-2} \).

Teacher's Note:
1. Density falls by the same fraction as volume rises.

Question 10: A 33000 N tensile force is applied on a steel rod of area of cross-section \( 10^{-3} \) m\(^2\). The change of temperature required to produce the same elongation, if the steel rod is heated, is (\( Y=3\times10^{11} \) N m\(^{-2}\), \( \alpha=1.1\times10^{-5}\,^\circ \)C\(^{-1}\))
(a) 20\(^\circ\)C
(b) 15\(^\circ\)C
(c) 10\(^\circ\)C
(d) 0\(^\circ\)C
Show Answer & Explanation

Answer: (c) 10\(^\circ\)C

Explanation:
1. \( \dfrac{FL}{AY}=L\alpha\Delta T\Rightarrow\Delta T=\dfrac{F}{AY\alpha} \).
2. \( \Delta T=\dfrac{33000}{10^{-3}\times3\times10^{11}\times1.1\times10^{-5}}=10^\circ \)C.

Teacher's Note:
1. Equate mechanical and thermal strain.

Question 11: Two uniform brass rods A and B of lengths \( l \) and \( 2l \), and radii \( 2r \) and \( r \) respectively are heated to the same temperature. The ratio of the increase in the volumes of A to that of B is
(a) 1 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4
Show Answer & Explanation

Answer: (c) 2 : 1

Explanation:
1. \( \Delta V=\gamma V\Delta T \), so the ratio equals \( V_A:V_B=\pi(2r)^2l:\pi r^2(2l)=4:2=2:1 \).

Teacher's Note:
1. Same material and temperature rise, so \( \Delta V\propto V \).

Question 12: A clock which keeps correct time at 20\(^\circ\)C is subjected to 40\(^\circ\)C. If the coefficient of linear expansion of the pendulum is \( 12\times10^{-6}\,^\circ \)C\(^{-1}\), how much time will it lose per day?
(a) 10.3 s
(b) 20.6 s
(c) 5 s
(d) 20 min
Show Answer & Explanation

Answer: (a) 10.3 s

Explanation:
1. \( T\propto\sqrt l \): \( \dfrac{\Delta T}{T}=\tfrac12\alpha\Delta\theta=\tfrac12\times12\times10^{-6}\times20=1.2\times10^{-4} \).
2. Loss per day \( =1.2\times10^{-4}\times86400=10.4\approx10.3 \) s.

Teacher's Note:
1. A longer pendulum swings slower, so the clock loses time.

Question 13: A brass rod of length 500 mm and diameter 3 mm is joined to a steel rod of the same length and diameter at 50\(^\circ\)C. If the coefficients of linear expansion of brass and steel are \( 2.5\times10^{-5}\,^\circ \)C\(^{-1}\) and \( 1.25\times10^{-5}\,^\circ \)C\(^{-1}\), then the change in length of the combined rod at 200\(^\circ\)C is
(a) 2.4 mm
(b) 2.8 mm
(c) 3.2 mm
(d) 3.6 mm
Show Answer & Explanation

Answer: (b) 2.8 mm

Explanation:
1. \( \Delta T=150^\circ \)C.
2. \( \Delta L=500\times150\,(2.5+1.25)\times10^{-5}=2.81 \) mm.

Teacher's Note:
1. Add the expansions of the two rods.

Question 14: The temperature of a thin uniform circular disc of 1 m diameter is increased by 10\(^\circ\)C. The percentage increase in the moment of inertia of the disc about an axis passing through its centre and perpendicular to the circular face is (\( \alpha=11\times10^{-6}\,^\circ \)C\(^{-1}\))
(a) 0.0055
(b) 0.011
(c) 0.022
(d) 0.044
Show Answer & Explanation

Answer: (c) 0.022

Explanation:
1. \( I=\tfrac12MR^2\propto R^2 \), so \( \dfrac{\Delta I}{I}=2\alpha\Delta T=2\times11\times10^{-6}\times10=2.2\times10^{-4} \), i.e. 0.022%.

Teacher's Note:
1. Mass is constant; only \( R^2 \) changes.

Question 15: An iron bar of length 10 m is heated from 0\(^\circ\)C to 100\(^\circ\)C. If the coefficient of linear thermal expansion of iron is \( 10\times10^{-6}\,^\circ \)C\(^{-1}\), the increase in the length of the bar is
(a) 0.5 cm
(b) 1.0 cm
(c) 1.5 cm
(d) 2.0 cm
Show Answer & Explanation

Answer: (b) 1.0 cm

Explanation:
1. \( \Delta L=10\times10\times10^{-6}\times100=0.01 \) m \( =1.0 \) cm.

Teacher's Note:
1. \( \Delta L=L\alpha\Delta T \).

Question 16: A composite rod made of copper (\( \alpha=1.8\times10^{-5} \) K\(^{-1}\)) and steel (\( \alpha=1.2\times10^{-5} \) K\(^{-1}\)) is heated. Then,
(a) it bends with steel on the concave side
(b) it bends with copper on the concave side
(c) it does not expand
(d) data is insufficient
Show Answer & Explanation

Answer: (a) it bends with steel on the concave side

Explanation:
1. Copper expands more, so it ends up on the outer (convex) side and steel on the inner (concave) side.

Teacher's Note:
1. The metal with the larger \( \alpha \) is on the outside of the curve when heated.

Question 17: A scientist proposes a new temperature scale in which the ice point is 25 X (X is the new unit of temperature) and the steam point is 305 X. The specific heat capacity of water in this new scale is (in J kg\(^{-1}\) X\(^{-1}\))
(a) \( 4.0\times10^3 \)
(b) \( 3.0\times10^3 \)
(c) \( 1.2\times10^3 \)
(d) \( 1.5\times10^3 \)
Show Answer & Explanation

Answer: (d) \( 1.5\times10^3 \)

Explanation:
1. Between the fixed points 100\(^\circ\)C \( =280 \) X.
2. \( c=4200\times\dfrac{100}{280}=1500 \) J kg\(^{-1}\) X\(^{-1}\).

Teacher's Note:
1. A larger unit gives a smaller number per unit.

Question 18: Steam at 100\(^\circ\)C is passed into 1 kg of water contained in a calorimeter of water equivalent 0.2 kg at 9\(^\circ\)C till the temperature of the calorimeter and water in it is increased to 90\(^\circ\)C. The mass of steam condensed in kg is nearly (specific heat of water \( =1 \) cal/g\(^\circ\)C, latent heat of vaporisation \( =540 \) cal/g)
(a) 0.81
(b) 0.18
(c) 0.27
(d) 0.54
Show Answer & Explanation

Answer: (b) 0.18

Explanation:
1. Heat gained \( =1200\times81=97200 \) cal.
2. Heat lost \( =m(540+10) \Rightarrow m=\dfrac{97200}{550}=176.7 \) g \( \approx0.18 \) kg.

Teacher's Note:
1. The condensed steam also cools from 100\(^\circ\)C to 90\(^\circ\)C.

Question 19: 250 g of water and an equal volume of alcohol of mass 200 g are replaced successively in the same calorimeter and cool from 60\(^\circ\) to 55\(^\circ\) in 130 s and 67 s respectively. If the water equivalent of the calorimeter is 10 g, then the specific heat of alcohol in cal/g\(^\circ\)C is
(a) 1.25
(b) 0.69
(c) 0.62
(d) 0.67
Show Answer & Explanation

Answer: (c) 0.62

Explanation:
1. Equal temperature fall under the same conditions means heat lost \( \propto \) time: \( \dfrac{260}{130}=\dfrac{200s+10}{67} \).
2. \( 200s+10=134\Rightarrow s=0.62 \).

Teacher's Note:
1. Rate of heat loss is the same for both liquids.

Question 20: 22320 cal of heat is supplied to 100 g of ice at 0\(^\circ\)C. If the latent heat of fusion of ice is 80 cal g\(^{-1}\) and the latent heat of vaporisation of water is 540 cal g\(^{-1}\), the final amount of water and its temperature are
(a) 8 g, 100\(^\circ\)C
(b) 100 g, 90\(^\circ\)C
(c) 92 g, 100\(^\circ\)C
(d) 82 g, 100\(^\circ\)C
Show Answer & Explanation

Answer: (c) 92 g, 100\(^\circ\)C

Explanation:
1. Melting: 8000 cal; heating to 100\(^\circ\)C: 10000 cal; left \( =4320 \) cal.
2. Steam formed \( =4320/540=8 \) g, so water remaining \( =92 \) g at 100\(^\circ\)C.

Teacher's Note:
1. Handle melting, heating and boiling in sequence.

Question 21: In a reversible ideal engine, 650 J of heat comes from a source at 450 K. The heat rejected to the sink at 225 K is
(a) 325 J
(b) 625 J
(c) 700 J
(d) 800 J
Show Answer & Explanation

Answer: (a) 325 J

Explanation:
1. \( \dfrac{Q_2}{Q_1}=\dfrac{T_2}{T_1}\Rightarrow Q_2=650\times\dfrac{225}{450}=325 \) J.

Teacher's Note:
1. For a Carnot engine \( Q\propto T \).

Question 22: A lead bullet strikes against a steel plate with a velocity of 200 m/s. If the impact is perfectly inelastic and the heat produced is equally shared between the bullet and the target, then the rise in temperature of the bullet is (specific heat capacity of lead \( =125 \) J kg\(^{-1}\)K\(^{-1}\))
(a) 80\(^\circ\)C
(b) 60\(^\circ\)C
(c) 40\(^\circ\)C
(d) 120\(^\circ\)C
Show Answer & Explanation

Answer: (a) 80\(^\circ\)C

Explanation:
1. Heat to the bullet \( =\tfrac12\times\tfrac12mv^2=mc\Delta T \).
2. \( \Delta T=\dfrac{v^2}{4c}=\dfrac{40000}{500}=80^\circ \)C.

Teacher's Note:
1. Only half of the kinetic energy heats the bullet.

Question 23: A piece of ice (heat capacity \( =2100 \) J kg\(^{-1}\,^\circ\)C\(^{-1}\), latent heat \( =3.36\times10^5 \) J kg\(^{-1}\)) of mass \( m \) gram is at \( -5^\circ \)C at atmospheric pressure. It is given 420 J of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that 1 g of ice has melted. Assuming there is no other heat exchange, the value of \( m \) is
(a) 8
(b) 6
(c) 4
(d) 8.5
Show Answer & Explanation

Answer: (a) 8

Explanation:
1. Heat to bring \( m \) g to 0\(^\circ\)C \( =m\times2.1\times5=10.5m \) J; melting 1 g needs 336 J.
2. \( 10.5m+336=420\Rightarrow m=8 \).

Teacher's Note:
1. Split the heat into warming and melting.

Question 24: An experiment takes 10 min to raise the temperature of water from 0\(^\circ\)C to 100\(^\circ\)C and another 55 min to convert it totally into steam by a stabilised heater. The latent heat of vaporisation comes out to be
(a) 530 cal g\(^{-1}\)
(b) 540 cal g\(^{-1}\)
(c) 550 cal g\(^{-1}\)
(d) 560 cal g\(^{-1}\)
Show Answer & Explanation

Answer: (c) 550 cal g\(^{-1}\)

Explanation:
1. The heater supplies 100 cal/g in 10 min, i.e. 10 cal g\(^{-1}\) per minute; in 55 min it supplies \( 550 \) cal g\(^{-1}\).

Teacher's Note:
1. Constant power means heat is proportional to time.

Question 25: The temperatures of equal masses of three different liquids A, B and C are 12\(^\circ\)C, 19\(^\circ\)C and 28\(^\circ\)C respectively. The temperature when A and B are mixed is 16\(^\circ\)C and when B and C are mixed is 23\(^\circ\)C. The temperature when A and C are mixed is
(a) 18.2\(^\circ\)C
(b) 22\(^\circ\)C
(c) 20.2\(^\circ\)C
(d) 24.2\(^\circ\)C
Show Answer & Explanation

Answer: (c) 20.2\(^\circ\)C

Explanation:
1. A+B: \( 4c_A=3c_B \); B+C: \( 4c_B=5c_C \), so \( c_A=0.75c_B \) and \( c_C=0.8c_B \).
2. A+C: \( 0.75(T-12)=0.8(28-T)\Rightarrow T=20.26^\circ \)C.

Teacher's Note:
1. Equal masses, so specific heats are inversely proportional to temperature changes.

Question 26: The work done in converting 1 g of ice at \( -10^\circ \)C into steam at 100\(^\circ\)C is (taking 1 cal = 4.2 J)
(a) 3045 J
(b) 6056 J
(c) 721 J
(d) 616 J
Show Answer & Explanation

Answer: (a) 3045 J

Explanation:
1. Heat \( =1\times0.5\times10+80+100+540=725 \) cal.
2. \( 725\times4.2=3045 \) J.

Teacher's Note:
1. Add warming ice, melting, warming water and vaporising.

Question 27: A copper block of mass 4 kg is heated in a furnace to a temperature of 425\(^\circ\)C and then placed on a large ice block. The mass of ice that will melt in this process will be (specific heat of copper \( =500 \) J kg\(^{-1}\,^\circ\)C\(^{-1}\) and heat of fusion of ice \( =336 \) kJ kg\(^{-1}\))
(a) 0.5 kg
(b) 1 kg
(c) 1.5 kg
(d) 2.5 kg
Show Answer & Explanation

Answer: (d) 2.5 kg

Explanation:
1. Heat lost by copper \( =4\times500\times425=8.5\times10^5 \) J.
2. \( m=\dfrac{8.5\times10^5}{3.36\times10^5}=2.5 \) kg.

Teacher's Note:
1. The ice block is large, so the final temperature is 0\(^\circ\)C.

Question 28: A vessel contains 110 g of water. The heat capacity of the vessel is equal to 10 g of water. The initial temperature of water in the vessel is 10\(^\circ\)C. If 220 g of hot water at 70\(^\circ\)C is poured in the vessel, the final temperature neglecting radiation loss will be
(a) 70\(^\circ\)C
(b) 80\(^\circ\)C
(c) 60\(^\circ\)C
(d) 50\(^\circ\)C
Show Answer & Explanation

Answer: (d) 50\(^\circ\)C

Explanation:
1. Effective cold mass \( =110+10=120 \) g.
2. \( T=\dfrac{120\times10+220\times70}{340}=48.8\approx50^\circ \)C.

Teacher's Note:
1. Include the vessel's water equivalent.

Question 29: Three rods of the same dimensions form a triangle PQR as shown, with PR and RQ having thermal conductivities \( K_1 \) and \( K_2 \), and the rod PQ having thermal conductivity \( K_3 \). The points P and Q are maintained at different temperatures so that the heat flows at the same rate along PRQ and along PQ. Then
(a) \( K_3=\tfrac12(K_1+K_2) \)
(b) \( K_3=K_1+K_2 \)
(c) \( K_3=\dfrac{K_1K_2}{K_1+K_2} \)
(d) \( K_3=2(K_1+K_2) \)
Show Answer & Explanation

Answer: (c) \( K_3=\dfrac{K_1K_2}{K_1+K_2} \)

Explanation:
1. Path PRQ is two rods in series (length \( 2l \)); PQ is one rod of length \( l \).
2. Equal rates: \( \dfrac{\Delta T}{l/K_1A+l/K_2A}=\dfrac{K_3A\Delta T}{l}\Rightarrow\dfrac1{K_3}=\dfrac1{K_1}+\dfrac1{K_2} \).

Teacher's Note:
1. Series resistances add; compare with the single rod.

Question 30: A, B and C are three identical conductors but made from different materials. They are kept in contact in series. Their thermal conductivities are \( K \), \( 2K \) and \( K/2 \). The free end of A is at 100\(^\circ\)C and the free end of C is at 0\(^\circ\)C. During steady state, the temperature of the junction of A and B is nearly
(a) 37\(^\circ\)C
(b) 71\(^\circ\)C
(c) 29\(^\circ\)C
(d) 63\(^\circ\)C
Show Answer & Explanation

Answer: (b) 71\(^\circ\)C

Explanation:
1. Resistances \( \propto1/K \): \( 1:\tfrac12:2 \), total 3.5.
2. Drop across A \( =100\times\dfrac1{3.5}=28.6 \), so junction at \( 71.4^\circ \)C.

Teacher's Note:
1. Temperature drop is proportional to resistance.

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Thermal Properties Of Matter Objective Questions & Solutions for JEE Physics

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