JEE Physics Thermal Properties Of Matter MCQs Set 19

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Multiple Choice Questions

Question 1: Which law is obeyed when the temperature difference between the body and the surroundings is small?
(a) Stefan's law
(b) Newton's law of cooling
(c) Planck's law
(d) All of these
Show Answer & Explanation

Answer: (b) Newton's law of cooling

Explanation:
1. Newton's law of cooling holds for a small excess of temperature over the surroundings.
2. The cooling rate is proportional to that excess.

Teacher's Note:
1. Small difference, Newton's law.

Question 2: A beaker is filled with water at 4°C. At one time the temperature is increased by a few degrees above 4°C and at another time it is decreased by a few degrees below 4°C. One shall observe that
(a) the level remains constant in each case
(b) in the first case water flows while in the second case its level comes down
(c) in the second case water overflows while in the first case its level comes down
(d) water overflows in both the cases
Show Answer & Explanation

Answer: (d) water overflows in both the cases

Explanation:
1. Water has its minimum volume at 4°C.
2. Heating or cooling from 4°C increases the volume, so water overflows in both cases.

Teacher's Note:
1. Volume is minimum at 4°C.

Question 3: A certain amount of heat is given to 100 g of copper to increase its temperature by 21°C. If the same amount of heat is given to 50 g of water, then the rise in its temperature is (specific heat capacity of copper = 400 J kg⁻¹ K⁻¹ and that for water = 4200 J kg⁻¹ K⁻¹)
(a) 4°C
(b) 5.25°C
(c) 8°C
(d) 6°C
Show Answer & Explanation

Answer: (a) 4°C

Explanation:
1. Q = 0.1 × 400 × 21 = 840 J.
2. ΔT = 840/(0.05 × 4200) = 4°C.

Teacher's Note:
1. Same Q for both.

Question 4: The coefficient of apparent expansion of mercury in a glass vessel is 153 × 10⁻⁶ /°C and in a steel vessel is 144 × 10⁻⁶ /°C. If α for steel is 12 × 10⁻⁶ /°C, then that of glass is
(a) 9 × 10⁻⁶ /°C
(b) 6 × 10⁻⁶ /°C
(c) 36 × 10⁻⁶ /°C
(d) 27 × 10⁻⁶ /°C
Show Answer & Explanation

Answer: (a) 9 × 10⁻⁶ /°C

Explanation:
1. γ_real = γ_app + γ_vessel.
2. For steel: γ_real = 144 × 10⁻⁶ + 36 × 10⁻⁶ = 180 × 10⁻⁶.
3. For glass: γ_glass = 180 − 153 = 27 × 10⁻⁶, so α = 9 × 10⁻⁶ /°C.

Teacher's Note:
1. Real expansion is the same in both vessels.

Question 5: ...A... and ...B... of heat energy require some material as a transport medium. Here A and B refer to
(a) conduction, radiation
(b) radiation, convection
(c) conduction, convection
(d) radiation, evaporation
Show Answer & Explanation

Answer: (c) conduction, convection

Explanation:
1. Conduction and convection need a material medium.
2. Radiation can travel through a vacuum.

Teacher's Note:
1. Radiation needs no medium.

Question 6: A pendulum clock is 5 seconds fast at a temperature of 15°C and 10 seconds slow at a temperature of 30°C. At what temperature does it give the correct time? (take the time interval = 24 hours)
(a) 18°C
(b) 20°C
(c) 22°C
(d) 25°C
Show Answer & Explanation

Answer: (b) 20°C

Explanation:
1. The time error is proportional to (θ − θ₀), where θ₀ is the correct temperature.
2. (θ₀ − 15) : (30 − θ₀) = 5 : 10, so 2θ₀ − 30 = 30 − θ₀ and θ₀ = 20°C.

Teacher's Note:
1. Linear error in temperature.

Question 7: If two rods A and B of equal length L, different areas A₁ and A₂, have one end each at temperatures T₁ and T₂, have equal rates of flow of heat, then
(a) \( A_1=A_2 \)
(b) \( \frac{A_1}{A_2}=\frac{K_1}{K_2} \)
(c) \( \frac{A_1}{A_2}=\frac{K_2}{K_1} \)
(d) \( K_1=K_2 \)
Show Answer & Explanation

Answer: (c) \( \frac{A_1}{A_2}=\frac{K_2}{K_1} \)

Explanation:
1. Rate = KAΔT/L.
2. For the same L and ΔT, equal rates need K₁A₁ = K₂A₂, so A₁/A₂ = K₂/K₁.

Teacher's Note:
1. Rate is proportional to KA.

Question 8: Match Column I with Column II.
Column IColumn II
(A) Bimetallic strip(1) Change in length
(B) Steam engine(2) Energy conversion
(C) Linear expansion(3) Change in area
(D) Area expansion(4) Thermal expansion

(a) A-4, B-1, C-2, D-3
(b) A-4, B-2, C-1, D-3
(c) A-3, B-2, C-1, D-4
(d) A-2, B-4, C-1, D-3
Show Answer & Explanation

Answer: (b) A-4, B-2, C-1, D-3

Explanation:
1. A bimetallic strip works through differential thermal expansion (A-4).
2. A steam engine converts heat into work (B-2).
3. Linear expansion is a change in length (C-1) and area expansion a change in area (D-3).

Teacher's Note:
1. Link each device with its principle.

Question 9: The length of a metallic rod is 5 m at 0°C and becomes 5.01 m on heating up to 100°C. The linear expansion coefficient of the metal will be
(a) \( 2.33\times10^{-5}/^\circ\text{C} \)
(b) \( 6.0\times10^{-5}/^\circ\text{C} \)
(c) \( 4.0\times10^{-5}/^\circ\text{C} \)
(d) \( 2.0\times10^{-5}/^\circ\text{C} \)
Show Answer & Explanation

Answer: (d) \( 2.0\times10^{-5}/^\circ\text{C} \)

Explanation:
1. α = ΔL/(LΔT) = 0.01/(5 × 100).
2. = 2.0 × 10⁻⁵ /°C.

Teacher's Note:
1. α = ΔL/LΔT.

Question 10: Good absorbers of heat are
(a) poor emitters
(b) non-emitters
(c) good emitters
(d) highly polished
Show Answer & Explanation

Answer: (c) good emitters

Explanation:
1. By Kirchhoff's law, a good absorber is also a good emitter.
2. Polished surfaces are poor absorbers.

Teacher's Note:
1. Kirchhoff's law.

Question 11: Consider the following statements and select the correct statement(s). I. Water can never be boiled without heating. II. Water can be boiled below room temperature by lowering the pressure. III. On releasing the excess pressure, water refreezes into ice.
(a) I only
(b) II only
(c) I and II
(d) II and III
Show Answer & Explanation

Answer: (d) II and III

Explanation:
1. Lowering the external pressure lowers the boiling point, so water can boil without heating (II is true, I is false).
2. Regelation: ice melted by pressure refreezes when the pressure is released (III).

Teacher's Note:
1. Boiling point depends on pressure.

Question 12: A body A is at a temperature T_A and a body B is at temperature T_B such that T_A > T_B. Bodies A and B are connected. Which of the following statements is/are true? I. Body A is hotter than body B. II. Heat flows from A to B. III. Heat flows from B to A.
(a) I and II
(b) Only I
(c) II and III
(d) I and III
Show Answer & Explanation

Answer: (a) I and II

Explanation:
1. A has the higher temperature, so it is hotter.
2. Heat flows from higher to lower temperature, from A to B.

Teacher's Note:
1. Heat flows downhill.

Question 13: 19 g of water at 30°C and 5 g of ice at −20°C are mixed together in a calorimeter. What is the final temperature of the mixture? Given the specific heat of ice = 0.5 cal g⁻¹ (°C)⁻¹ and latent heat of fusion of ice = 80 cal g⁻¹
(a) 0°C
(b) −5°C
(c) 5°C
(d) 10°C
Show Answer & Explanation

Answer: (c) 5°C

Explanation:
1. Heat to warm the ice to 0°C = 5 × 0.5 × 20 = 50 cal, and to melt it = 400 cal.
2. Water cooling to 0°C releases 19 × 30 = 570 cal, so 120 cal remains.
3. It warms 24 g of water: 120/24 = 5°C.

Teacher's Note:
1. Check whether all the ice melts.

Question 14: Which of the following statements is/are correct? I. Convection is a mode of heat transfer by actual motion of matter. II. Convection is possible only in gases. III. Convection can be a natural or forced process in nature.
(a) Only I
(b) I and III
(c) Only II
(d) I, II and III
Show Answer & Explanation

Answer: (b) I and III

Explanation:
1. Convection occurs in both liquids and gases, so II is false.
2. I and III are true.

Teacher's Note:
1. Convection happens in fluids.

Question 15: In an energy recycling process, 100 g of steam at 100°C becomes water at 100°C, which converts y g of ice at 0°C into water at 100°C. The numeric value of y is (L_v = 540 cal/g, L_f = 80 cal/g)
(a) 100
(b) 200
(c) 300
(d) 400
Show Answer & Explanation

Answer: (c) 300

Explanation:
1. Heat released = 100 × 540 = 54000 cal.
2. Heat needed per gram of ice = 80 + 100 = 180 cal.
3. y = 54000/180 = 300.

Teacher's Note:
1. Heat released equals heat absorbed.

Question 16: A solid ball of metal has a spherical cavity inside it. If the ball is heated, the volume of the cavity will
(a) decrease
(b) increase
(c) remain unchanged
(d) have its shape changed
Show Answer & Explanation

Answer: (b) increase

Explanation:
1. The cavity expands like the surrounding material would.
2. So its volume increases.

Teacher's Note:
1. Cavity expands with the material.

Question 17: On heating a liquid of coefficient of cubical expansion γ in a container having a coefficient of linear expansion γ/3, the level of the liquid in the container will
(a) rise
(b) fall
(c) remain almost stationary
(d) it is difficult to say
Show Answer & Explanation

Answer: (c) remain almost stationary

Explanation:
1. The container's coefficient of volume expansion is 3 × (γ/3) = γ.
2. Container and liquid expand equally, so the level stays the same.

Teacher's Note:
1. Equal volume expansions.

Question 18:
Assertion (A): The temperature at which Centigrade and Fahrenheit thermometers read the same is −40°.
Reason (R): There is no relation between Fahrenheit and Centigrade temperature.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (c) A is true, but R is false.

Explanation:
1. Setting C = F in the conversion formula gives −40°.
2. The two scales are linearly related, so the reason is false.

Teacher's Note:
1. The scales are linearly related.

Question 19: Ice starts forming in a lake with water at 0°C when the atmospheric temperature is −10°C. If the time taken for the first 1 cm of ice to be formed is 7 hours, then the time taken for the thickness of the ice to change from 1 cm to 2 cm is
(a) 7 hours
(b) 14 hours
(c) 21 hours
(d) 3.5 hours
Show Answer & Explanation

Answer: (c) 21 hours

Explanation:
1. The time to form thickness x is proportional to x².
2. Total time for 2 cm = 4 × 7 = 28 h, so the extra time = 28 − 7 = 21 hours.

Teacher's Note:
1. t ∝ x².

Question 20: Expansion during heating
(a) occurs only in solids
(b) increases the weight of a material
(c) generally decreases the density of a material
(d) occurs at the same rate for all liquids and solids
Show Answer & Explanation

Answer: (c) generally decreases the density of a material

Explanation:
1. Heating usually increases the volume at fixed mass, so the density decreases.
2. Weight does not change and the rates differ between materials.

Teacher's Note:
1. Density = mass/volume.

Question 21: A solid copper cube of edges 1 cm is suspended in an evacuated enclosure. Its temperature is found to fall from 100°C to 99°C in 100 s. Another solid copper cube of edges 2 cm, with similar surface nature, is suspended in a similar manner. The time required for this cube to cool from 100°C to 99°C will be approximately
(a) 25 s
(b) 50 s
(c) 200 s
(d) 400 s
Show Answer & Explanation

Answer: (c) 200 s

Explanation:
1. Rate of cooling ∝ A/(mc) ∝ 1/(edge).
2. The larger cube has double the edge, so it cools half as fast and takes 200 s.

Teacher's Note:
1. Surface-to-volume ratio halves.

Question 22: Which of the following is a poor conductor of heat?
(a) Copper
(b) Concrete
(c) Mercury
(d) Air
Show Answer & Explanation

Answer: (d) Air

Explanation:
1. Gases have the lowest thermal conductivity.
2. Air is a very poor conductor and a good insulator.

Teacher's Note:
1. Gases conduct poorly.

Question 23: In a surrounding medium of temperature 10°C, a body takes 7 min for a fall of temperature from 60°C to 40°C. In what time will the temperature of the body fall from 40°C to 28°C?
(a) 7 min
(b) 11 min
(c) 14 min
(d) 21 min
Show Answer & Explanation

Answer: (a) 7 min

Explanation:
1. First case: rate = 20/7, mean excess = 40, so k = 1/14 per minute.
2. Second case: mean excess = 34 − 10 = 24, rate = 24/14, time = 12/(24/14) = 7 min.

Teacher's Note:
1. Find k from the first cooling.

Question 24: Match the quantities in Column I with the units in Column II.
Column IColumn II
(A) Amount of substance(1) J kg⁻¹ K⁻¹
(B) Coefficient of volume expansion(2) J s⁻¹ K⁻¹
(C) Specific heat(3) K⁻¹
(D) Thermal conductivity(4) mol

(a) A-4, B-1, C-3, D-2
(b) A-3, B-4, C-1, D-2
(c) A-4, B-3, C-1, D-2
(d) A-4, B-2, C-1, D-3
Show Answer & Explanation

Answer: (c) A-4, B-3, C-1, D-2

Explanation:
1. Amount of substance is measured in mol (A-4).
2. The volume expansion coefficient has unit K⁻¹ (B-3).
3. Specific heat is J kg⁻¹ K⁻¹ (C-1).
4. Thermal conductivity involves heat per second per kelvin (D-2).

Teacher's Note:
1. Check the units.

Question 25: A 2 kg copper block is heated to 500°C and then placed on a large block of ice at 0°C. If the specific heat capacity of copper is 400 J/kg°C and the latent heat of fusion of water is 3.5 × 10⁵ J/kg, the amount of ice that can melt is
(a) \( \frac78\ \text{kg} \)
(b) \( \frac75\ \text{kg} \)
(c) \( \frac87\ \text{kg} \)
(d) \( \frac57\ \text{kg} \)
Show Answer & Explanation

Answer: (c) \( \frac87\ \text{kg} \)

Explanation:
1. Heat given by copper = 2 × 400 × 500 = 4 × 10⁵ J.
2. Mass of ice = 4 × 10⁵/3.5 × 10⁵ = 8/7 kg.

Teacher's Note:
1. m = Q/L.

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