JEE Physics Thermal Properties Of Matter MCQs Set 05

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Multiple Choice Questions

Question 1: The temperature of a furnace is 2324°C and the intensity in its radiation spectrum is maximum at nearly 12000 Å. If the intensity in the spectrum of a star is maximum at nearly 4800 Å, the surface temperature of the star is
(a) 8400°C
(b) 7200°C
(c) 6219.5°C
(d) 5900°C
Show Answer & Explanation

Answer: (c) 6219.5°C

Explanation:
1. Wien's law: \( \lambda_mT = \text{constant} \).
2. Furnace \( T = 2324 + 273 = 2597 \) K.
3. Star \( T = 2597\times\dfrac{12000}{4800} = 6492.5 \) K.
4. In Celsius \( 6492.5 - 273 = 6219.5 \)°C.

Teacher's Note:
1. Use kelvin in Wien's law and convert back at the end.
2. A shorter peak wavelength means a hotter source.

Question 2: The original temperature of a black body is 727°C. The temperature at which the total radiant energy from this black body becomes double is
(a) 971 K
(b) 1190 K
(c) 2001 K
(d) 1458 K
Show Answer & Explanation

Answer: (b) 1190 K

Explanation:
1. Initial \( T = 1000 \) K.
2. \( E \propto T^4 \), so \( T' = 1000\times2^{1/4} = 1000\times1.189 \approx 1190 \) K.

Teacher's Note:
1. Doubling the radiant energy raises T only by the fourth root of 2.

Question 3: The energy radiated per minute from the filament of an incandescent lamp at 2000 K, if its surface area is 5×10⁻⁵ m² and its relative emittance is 0.85 (σ = 5.7×10⁻⁸ in MKS units), is
(a) 1230 J
(b) 2215 J
(c) 2115 J
(d) 2325 J
Show Answer & Explanation

Answer: (d) 2325 J

Explanation:
1. Power \( = e\sigma AT^4 = 0.85\times5.7\times10^{-8}\times5\times10^{-5}\times(2000)^4 \).
2. \( = 38.8 \) W.
3. Energy in 60 s \( = 38.8\times60 \approx 2325 \) J.

Teacher's Note:
1. Multiply by 60 s to get energy per minute.
2. Emittance e is the same as emissivity.

Question 4: The energy emitted per second by a black body at 27°C is 10 J. If the temperature of the body is increased to 327°C, the energy emitted per second will be
(a) 20 J
(b) 40 J
(c) 80 J
(d) 160 J
Show Answer & Explanation

Answer: (d) 160 J

Explanation:
1. \( T_1 = 300 \) K and \( T_2 = 600 \) K.
2. \( E \propto T^4 \), so \( E_2 = 10\times2^4 = 160 \) J.

Teacher's Note:
1. Always convert to kelvin before taking the ratio.

Question 5: Energy is emitted from the surface of a black body at 127°C at the rate of 1.0×10⁶ J/s m². The temperature of the black body at which the rate of energy emission is 16.0×10⁶ J/s m² is
(a) 754°C
(b) 527°C
(c) 254°C
(d) 508°C
Show Answer & Explanation

Answer: (b) 527°C

Explanation:
1. \( T_1 = 400 \) K and \( \dfrac{E_2}{E_1} = 16 = \left(\dfrac{T_2}{T_1}\right)^4 \).
2. \( T_2 = 2T_1 = 800 \) K = 527°C.

Teacher's Note:
1. The fourth root of 16 is 2.
2. Convert back to Celsius.

Question 6: The solar constant for the earth is 2 cal/min cm². If the distance of Mercury from the sun is 0.4 times that of the earth, the solar constant for Mercury will be
(a) 25 cal/min cm²
(b) 0.32 cal/min cm²
(c) 12.5 cal/min cm²
(d) 2 cal/min cm²
Show Answer & Explanation

Answer: (c) 12.5 cal/min cm²

Explanation:
1. Intensity \( \propto \dfrac{1}{d^2} \).
2. \( I_M = 2\times\dfrac{1}{0.4^2} = \dfrac{2}{0.16} = 12.5 \) cal/min cm².

Teacher's Note:
1. Radiant intensity from a point source follows the inverse square law.

Question 7: Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperatures \( T_1 \) and \( T_2 \). The maximum intensity in the emission spectrum of A is at 500 nm and of B at 1500 nm. Considering them to be black bodies, the ratio of the rates of total energy radiated by A and B is
(a) 6
(b) 3
(c) 9
(d) 12
Show Answer & Explanation

Answer: (c) 9

Explanation:
1. By Wien's law \( T \propto \dfrac{1}{\lambda_m} \), so \( \dfrac{T_1}{T_2} = \dfrac{1500}{500} = 3 \).
2. Power \( \propto r^2T^4 \): \( \dfrac{P_A}{P_B} = \left(\dfrac{6}{18}\right)^2\times3^4 = \dfrac{1}{9}\times81 = 9 \).

Teacher's Note:
1. Combine Wien's law (for T) and Stefan's law (for power).

Question 8: Star S₁ emits maximum radiation at wavelength 420 nm and star S₂ at 560 nm. The ratio of the temperatures of S₁ and S₂ is
(a) \( \dfrac{4}{3} \)
(b) \( \left(\dfrac{3}{4}\right)^{1/2} \)
(c) \( \dfrac{3}{4} \)
(d) \( \left(\dfrac{4}{3}\right)^{1/4} \)
Show Answer & Explanation

Answer: (a) \( \dfrac{4}{3} \)

Explanation:
1. \( T \propto \dfrac{1}{\lambda_m} \).
2. \( \dfrac{T_1}{T_2} = \dfrac{560}{420} = \dfrac{4}{3} \).

Teacher's Note:
1. The star with the shorter peak wavelength is hotter.

Question 9: A bucket full of hot water cools from 75°C to 70°C in time \( T_1 \), from 70°C to 65°C in time \( T_2 \) and from 65°C to 60°C in time \( T_3 \). Then
(a) \( T_1 = T_2 = T_3 \)
(b) \( T_1 > T_2 > T_3 \)
(c) \( T_1 \lt T_2 \lt T_3 \)
(d) \( T_1 > T_2 \lt T_3 \)
Show Answer & Explanation

Answer: (c) \( T_1 \lt T_2 \lt T_3 \)

Explanation:
1. The rate of cooling decreases as the excess temperature over the surroundings decreases (Newton's law).
2. The same 5°C drop therefore takes longer each time, so \( T_1 \lt T_2 \lt T_3 \).

Teacher's Note:
1. Cooling slows down as the body approaches the room temperature.

Question 10: A body cools from 60°C to 50°C in 10 minutes. If the room temperature is 25°C and Newton's law of cooling holds, the temperature of the body at the end of the next 10 minutes is
(a) 45°C
(b) 42.85°C
(c) 40°C
(d) 38.5°C
Show Answer & Explanation

Answer: (b) 42.85°C

Explanation:
1. First interval: \( \dfrac{10}{10} = k(55 - 25) \), so \( k = \dfrac{1}{30} \) per min.
2. Next interval: \( \dfrac{50 - T}{10} = \dfrac{1}{30}\left(\dfrac{50+T}{2} - 25\right) = \dfrac{T}{60} \).
3. \( 50 - T = \dfrac{T}{6} \), so \( T = 42.86 \)°C.

Teacher's Note:
1. Use the mean temperature of each interval.
2. The body cools by less than 10°C in the second interval.

Question 11: The radii of two spheres made of the same metal are r and 2r. They are heated to the same temperature and placed in the same surroundings. The ratio of their rates of decrease of temperature is
(a) 1 : 1
(b) 4 : 1
(c) 1 : 4
(d) 2 : 1
Show Answer & Explanation

Answer: (d) 2 : 1

Explanation:
1. Rate of fall of temperature \( \propto \dfrac{\text{surface area}}{\text{mass}} \propto \dfrac{r^2}{r^3} = \dfrac{1}{r} \).
2. Ratio \( = \dfrac{1/r}{1/2r} = 2 : 1 \).

Teacher's Note:
1. Smaller bodies have more surface per unit mass, so they cool faster.

Question 12: If E is the total energy emitted by a body at a temperature T K and \( E_{\lambda max} \) is the maximum spectral energy emitted by it at the same temperature, then
(a) \( E \propto T^4 \); \( E_{\lambda max} \propto T^5 \)
(b) \( E \propto T^4 \); \( E_{\lambda max} \propto T^{-5} \)
(c) \( E \propto T^{-4} \); \( E_{\lambda max} \propto T^4 \)
(d) \( E \propto T^5 \); \( E_{\lambda max} \propto T^4 \)
Show Answer & Explanation

Answer: (a) \( E \propto T^4 \); \( E_{\lambda max} \propto T^5 \)

Explanation:
1. Stefan's law gives \( E \propto T^4 \).
2. The peak spectral emissive power grows as the fifth power of temperature, \( E_{\lambda max} \propto T^5 \).

Teacher's Note:
1. The area under the curve goes as T⁴, while the height of the peak goes as T⁵.

Question 13: A liquid takes 5 min to cool from 80°C to 50°C. The time it will take to cool from 60°C to 30°C is (temperature of the surroundings 20°C)
(a) 15 min
(b) 20 min
(c) 100 min
(d) 9 min
Show Answer & Explanation

Answer: (d) 9 min

Explanation:
1. First: \( \dfrac{30}{5} = k(65 - 20) \), so \( k = \dfrac{2}{15} \).
2. Second: \( \dfrac{30}{t} = \dfrac{2}{15}(45 - 20) = \dfrac{10}{3} \).
3. \( t = 9 \) min.

Teacher's Note:
1. The mean excess temperature is smaller in the second case, so cooling is slower.

Question 14: A cup of tea cools from 80°C to 60°C in one minute. The ambient temperature is 30°C. In cooling from 60°C to 50°C it will take
(a) 50 s
(b) 90 s
(c) 60 s
(d) 48 s
Show Answer & Explanation

Answer: (d) 48 s

Explanation:
1. First: \( \dfrac{20}{1} = k(70 - 30) \), so \( k = \dfrac{1}{2}\ \text{min}^{-1} \).
2. Second: \( \dfrac{10}{t} = \dfrac{1}{2}(55 - 30) = 12.5 \).
3. \( t = 0.8 \) min = 48 s.

Teacher's Note:
1. Convert minutes to seconds at the end.

Question 15: Two spheres P and Q of the same colour having radii 8 cm and 2 cm are maintained at temperatures 127°C and 527°C respectively. The ratio of the energy radiated by P to that by Q is
(a) 0.054
(b) 0.0034
(c) 1
(d) 2
Show Answer & Explanation

Answer: (c) 1

Explanation:
1. Power \( \propto r^2T^4 \).
2. \( \dfrac{P}{Q} = \left(\dfrac{8}{2}\right)^2\left(\dfrac{400}{800}\right)^4 = 16\times\dfrac{1}{16} = 1 \).

Teacher's Note:
1. Use kelvin temperatures, 400 K and 800 K.

Question 16: A solid cube and a solid sphere are made of the same substance and both have the same surface area. If both are at the same temperature of 120°C, then
(a) the rate of loss of heat of the sphere will be more
(b) the rate of loss of heat of the cube will be more
(c) the rate of loss of heat will be more for the body whose mass is more
(d) both will lose heat at the same rate
Show Answer & Explanation

Answer: (d) both will lose heat at the same rate

Explanation:
1. Radiated power \( = e\sigma AT^4 \) depends only on the surface area, emissivity and temperature.
2. These are identical for both, so the rates of loss of heat are equal.

Teacher's Note:
1. Radiation does not depend on shape or mass for a given area.

Question 17: A black body at 200 K is found to emit maximum energy at a wavelength of 14 µm. When its temperature is raised to 1000 K, the wavelength at which maximum energy is emitted is
(a) 14 µm
(b) 15 µm
(c) 2.8 µm
(d) 28 µm
Show Answer & Explanation

Answer: (c) 2.8 µm

Explanation:
1. \( \lambda_m T \) is constant.
2. \( \lambda_m' = 14\times\dfrac{200}{1000} = 2.8 \) µm.

Teacher's Note:
1. Higher temperature moves the peak to shorter wavelength.

Question 18: The temperature of a perfect black body is 727°C and its area is 0.1 m². If Stefan's constant is 5.67×10⁻⁸ W/m² K⁴, the heat radiated by it in 1 minute is (1 cal = 4.2 J)
(a) 8100 cal
(b) 81000 cal
(c) 810 cal
(d) 81 cal
Show Answer & Explanation

Answer: (b) 81000 cal

Explanation:
1. \( T = 1000 \) K.
2. Power \( = 5.67\times10^{-8}\times0.1\times10^{12} = 5670 \) W.
3. Energy in 60 s \( = 3.402\times10^5 \) J \( = \dfrac{3.402\times10^5}{4.2} = 81000 \) cal.

Teacher's Note:
1. Remember to convert to calories at the end.

Question 19: When the temperature of the surface of a black body is reduced to one third (in kelvin), the amount of radiation emitted becomes
(a) \( \dfrac{1}{27} \)
(b) \( \dfrac{1}{81} \)
(c) \( \dfrac{1}{9} \)
(d) \( \dfrac{1}{3} \)
Show Answer & Explanation

Answer: (b) \( \dfrac{1}{81} \)

Explanation:
1. \( E \propto T^4 \), so the new energy is \( \left(\dfrac{1}{3}\right)^4 = \dfrac{1}{81} \).

Teacher's Note:
1. The fourth power gives 3⁴ = 81.

Question 20: A black body radiates energy at the rate of E W/m² at a high temperature T K. When the temperature is reduced to T/2 K, the radiant energy will be
(a) \( \dfrac{E}{16} \)
(b) \( \dfrac{E}{4} \)
(c) \( 16E \)
(d) \( 4E \)
Show Answer & Explanation

Answer: (a) \( \dfrac{E}{16} \)

Explanation:
1. \( E \propto T^4 \), so at \( T/2 \), the energy is \( E\left(\dfrac{1}{2}\right)^4 = \dfrac{E}{16} \).

Teacher's Note:
1. Halving the temperature reduces radiation by a factor of 16.

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Thermal Properties Of Matter Objective Questions & Solutions for JEE Physics

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