JEE Physics Thermal Properties Of Matter MCQs Set 14

Answer: (b) \( \dfrac{31}{656} \)

Explanation:
1. Resistance per unit area: single \( \dfrac{6}{K_g} \); double \( \dfrac{6}{K_g}+\dfrac{3}{K_a} \) (mm).
2. Ratio of heat flows \( =\dfrac{6/K_g}{6/K_g+3/K_a}=\dfrac{24000}{24000+483871}=\dfrac{31}{656} \).

Teacher's Note:
1. The poor conductor, air, dominates the resistance of the composite.
2. The thicknesses can stay in millimetres because only ratios matter.

Answer: (a) 1 : 8

Explanation:
1. \( H\propto\dfrac{d^2}{l} \): \( \dfrac{1^2/2}{2^2/1}=\dfrac{1/2}{4}=\dfrac18 \).

Teacher's Note:
1. The cross-section area goes with the square of the diameter.
2. Longer rods carry less current.

Answer: (c) \( \sqrt{\dfrac{2d}{R}} \)

Explanation:
1. Power absorbed by the earth \( =\sigma T_s^4\,4\pi R^2\cdot\dfrac{\pi r^2}{4\pi d^2} \); power radiated \( =\sigma T_e^4\,4\pi r^2 \).
2. Equating: \( T_s^4\dfrac{R^2}{4d^2}=T_e^4\Rightarrow T_s=T_e\sqrt{\dfrac{2d}{R}} \).

Teacher's Note:
1. The earth intercepts a disc of area pi r^2 of the sun's radiation.
2. The result does not depend on the earth's radius.

Answer: (b) B

Explanation:
1. Power \( \propto r^2T^4\propto\dfrac{r^2}{\lambda_m^4} \).
2. Values (in units): A \( \dfrac{4}{300^4}=4.9\times10^{-10} \), B \( \dfrac{16}{400^4}=6.3\times10^{-10} \), C \( \dfrac{36}{500^4}=5.8\times10^{-10} \).
3. B emits the most.

Teacher's Note:
1. Wien's law: temperature is inversely proportional to wavelength of the peak.
2. A larger disc does not necessarily emit more if it is cooler.

Answer: (d) 30°C

Explanation:
1. Let the resistances be \( R_1 \) (near the 100°C end) and \( R_2 \): \( \dfrac{100-70}{R_1}=\dfrac{70-0}{R_2}\Rightarrow\dfrac{R_1}{R_2}=\dfrac37 \).
2. Interchanged, the rod of resistance \( R_1 \) is next to 0°C: \( \dfrac{100-T}{R_2}=\dfrac{T}{R_1}\Rightarrow\dfrac{T}{100-T}=\dfrac37\Rightarrow T=30^\circ\text{C} \).

Teacher's Note:
1. The junction temperature is closer to the end of the better conductor.
2. Interchanging mirrors the temperature profile: 100 - 70 = 30.

Answer: (c) 300 J

Explanation:
1. Heat current through each rod \( \dfrac{KA\Delta T}{l} \): steel \( \dfrac{46\times0.2\times10^{-4}\times50}{0.31}=0.148\text{ W} \), brass \( \dfrac{109\times0.2\times10^{-4}\times50}{0.31}=0.352\text{ W} \).
2. Total 0.5 W, so in 600 s \( 0.5\times600=300\text{ J} \).

Teacher's Note:
1. The junction is held at constant temperature by a source or sink, so both rods deliver heat to it.
2. Use 10 minutes = 600 s.

Answer: (a) 22 W

Explanation:
1. \( P=e\sigma A(T^4-T_0^4)=0.30\times6\times10^{-8}\times4\pi(10^{-2})^2\times(10^{12}-8.1\times10^{9}) \).
2. \( =1.8\times10^{-8}\times1.257\times10^{-3}\times9.92\times10^{11}\approx22\text{ W} \).

Teacher's Note:
1. The chamber walls return some radiation, so subtract the fourth power of 300 K.
2. Area of a sphere is 4 pi r^2.

Answer: (d) (1), (2) and (3) only

Explanation:
1. Kirchhoff's law: emissivity equals absorptive power, so a good absorber is a good emitter and a bad absorber is a bad emitter: (1) is true and (4) is false.
2. With no transmission, absorptivity + reflectivity = 1, so a bad absorber is a good reflector (2) and a bad reflector is a good absorber, hence a good emitter (3).

Teacher's Note:
1. A polished metal surface is a good reflector and a poor emitter.
2. Reflectivity and absorptivity are complementary when nothing is transmitted.

Answer: (b) (1) and (3) only

Explanation:
1. Wien's law: \( \nu_m\propto T \), so the peak frequency doubles.
2. Stefan's law: \( P\propto T^4 \), so power rises by \( 2^4=16 \).

Teacher's Note:
1. Peak wavelength is inversely proportional to temperature, peak frequency is directly proportional.
2. Total power depends on the fourth power of temperature.

Answer: (c) \( 2\times10^{-2}\text{ W/mK} \)

Explanation:
1. Layer 3: \( \Delta T_3=\dfrac{qL_3}{K_3}=\dfrac{20\times0.028}{0.04}=14^\circ\text{C} \), so its face towards layer 2 is at \( -3+14=11^\circ\text{C} \).
2. Layer 2 carries 20 W/m² across \( 25-11=14^\circ\text{C} \): \( K_2=\dfrac{qL_2}{\Delta T_2}=\dfrac{20\times0.014}{14}=0.02\text{ W/mK} \).

Teacher's Note:
1. Equal flux through every layer determines each unknown.
2. Work out the temperature drop across layer 2 first.

Answer: (b) 36.4°C

Explanation:
1. Excess temperature falls from 10°C to 8°C in 10 min, a factor of 0.8.
2. In the next 10 min: \( 8\times0.8=6.4 \), so the temperature is \( 30+6.4=36.4^\circ\text{C} \).

Teacher's Note:
1. In equal time intervals the excess temperature is multiplied by the same factor.
2. Do not assume equal temperature drops.

Answer: (d) 960 J

Explanation:
1. The temperature rises uniformly so the average excess temperature is \( \dfrac{0+14}{2}=7^\circ\text{C} \).
2. Average loss \( =40\times\dfrac{7}{35}=8\text{ W} \), so over 120 s the loss is \( 8\times120=960\text{ J} \).

Teacher's Note:
1. Average the heat loss rate over the interval, not the temperature itself.
2. Because loss is linear in the excess, the average temperature works.

Answer: (c) 9

Explanation:
1. Heat towards A \( \propto\dfrac{400}{\lambda x} \) melts ice: mass \( \propto\dfrac{400}{80\lambda x} \). Heat towards B \( \propto\dfrac{300}{(10-\lambda)x} \): steam mass \( \propto\dfrac{300}{540(10-\lambda)x} \).
2. Equal masses: \( \dfrac{400}{80\lambda}=\dfrac{300}{540(10-\lambda)}\Rightarrow\dfrac{5}{\lambda}=\dfrac{5}{9(10-\lambda)} \), so \( 9(10-\lambda)=\lambda \).
3. Hence \( \lambda=9 \).

Teacher's Note:
1. Mass change equals heat divided by latent heat.
2. P is closer to A, so more of the heat goes to the ice end.

Answer: (a) \( \left(\dfrac{97}{2}\right)^{1/4}T \)

Explanation:
1. The middle plate absorbs from both outer plates and radiates from both its faces: \( \sigma(2T)^4+\sigma(3T)^4=2\sigma T_2^4 \).
2. So \( T_2^4=\dfrac{16+81}{2}T^4 \) and \( T_2=\left(\dfrac{97}{2}\right)^{1/4}T \).

Teacher's Note:
1. Count both faces of the middle plate.
2. The result is about 2.64 T, between 2T and 3T.

Answer: (d) 330 K

Explanation:
1. Absorbed power \( =I\pi r^2 \); emitted minus absorbed from surroundings \( =\sigma4\pi r^2(T^4-300^4) \).
2. So \( T^4=300^4+\dfrac{I}{4\sigma}=8.1\times10^9+4\times10^9=1.21\times10^{10} \) and \( T\approx332\text{ K} \), about 330 K.

Teacher's Note:
1. The sphere intercepts a disc of area pi r^2 but radiates from the whole surface 4 pi r^2.
2. The fourth root of 1.21e10 is about 332.

Answer: (b) 9

Explanation:
1. Temperatures in kelvin: 760 K and 3040 K, a ratio of 4.
2. Power ratio \( 4^4=256 \), so \( \log_2\dfrac{P}{P_0}=1+\log_2256=1+8=9 \).

Teacher's Note:
1. Convert the temperatures to kelvin first.
2. Each factor of 2 in power adds 1 to the reading.

Answer: (a) 4

Explanation:
1. Equal temperature drops (100 K each) mean equal thermal resistances: \( \dfrac{\ell}{K_1\pi r^2}=\dfrac{\ell}{K_2\pi(2r)^2} \).
2. So \( K_1r^2=4K_2r^2\Rightarrow\dfrac{K_1}{K_2}=4 \).

Teacher's Note:
1. The current is the same in both, so equal temperature drops mean equal resistances.
2. The wider cylinder has four times the area.

Answer: (c) \( \dfrac{k_1L_2T_1+k_2L_1T_2}{k_1L_2+k_2L_1} \)

Explanation:
1. Equal heat currents: \( \dfrac{k_1(T_1-T)}{L_1}=\dfrac{k_2(T-T_2)}{L_2} \).
2. Solving for \( T \): \( T=\dfrac{k_1L_2T_1+k_2L_1T_2}{k_1L_2+k_2L_1} \).

Teacher's Note:
1. Resistance of a section is length over conductivity (per unit area).
2. Check with equal sections: T becomes the average if the conductivities are equal.

Answer: (b) 4.8 cal/s

Explanation:
1. Resistances \( R=\dfrac{l}{kA} \): copper \( \dfrac{46}{0.92\times4}=12.5 \), brass \( \dfrac{13}{0.26\times4}=12.5 \), steel \( \dfrac{12}{0.12\times4}=25 \) (CGS).
2. Brass and steel are in parallel: \( \dfrac{12.5\times25}{37.5}=8.33 \).
3. Total \( 12.5+8.33=20.83 \), so \( H=\dfrac{100}{20.83}=4.8\text{ cal/s} \).

Teacher's Note:
1. The copper rod is in series with the parallel pair.
2. The current through copper splits between brass and steel.