JEE Physics Thermal Properties Of Matter MCQs Set 07

Answer: (b) 10.4 s

Explanation:
1. For a pendulum \( T\propto\sqrt{l} \), so \( \dfrac{\Delta T}{T}=\dfrac{1}{2}\alpha\Delta\theta=\dfrac{1}{2}\times12\times10^{-6}\times20=1.2\times10^{-4} \).
2. The warm pendulum is longer, so its period increases and the clock runs slow.
3. Loss per day = \( 1.2\times10^{-4}\times86400\approx10.4 \) s. Answer: 10.4 s per day.

Teacher's Note:
1. Time lost is the fractional change in period multiplied by the seconds in the time interval.
2. A longer pendulum always means a slower clock.

Answer: (c) 8 min 20 s

Explanation:
1. Net power going into the water = 1000 − 160 = 840 W.
2. Heat needed \( Q=ms\Delta\theta=2\times4200\times50=4.2\times10^{5} \) J.
3. Time \( t=\dfrac{4.2\times10^{5}}{840}=500 \) s = 8 min 20 s. Answer: 8 min 20 s.

Teacher's Note:
1. Always subtract the heat losses from the supply before dividing the heat required by the power.
2. Convert the final seconds into minutes and seconds at the end.

Answer: (c) is the same at every cross-section and does not depend on x

Explanation:
1. The curved surface is insulated, so no heat escapes sideways.
2. In the steady state no heat is stored in any part of the conductor, so the same heat per second must cross every section.
3. The temperature gradient adjusts itself (smaller where the area is larger) so that \( H=-KA\,\dfrac{dT}{dx} \) stays constant. Answer: H is constant along x.

Teacher's Note:
1. In a steady state, heat entering a section equals heat leaving it, so H is the same everywhere in a lagged conductor.
2. It is the temperature gradient, not H, that changes with the area.

Answer: (a) bends upward

Explanation:
1. Aluminium expands more than iron for the same rise in temperature.
2. The lower (aluminium) layer therefore becomes longer than the upper (iron) layer, so it lies on the outside (convex side) of the bend.
3. The strip curves with its centre of curvature above it, that is, it bends upward.

Teacher's Note:
1. A bimetallic strip always bends toward the metal with the smaller expansion coefficient.
2. The metal that expands more ends up on the outer side of the curve.

Answer: (b) 1.4 kW m⁻²

Explanation:
1. \( S=\dfrac{2\times4.18\ \text{J}}{(10^{-2}\ \text{m})^{2}\times60\ \text{s}} \).
2. \( S=\dfrac{8.36}{6\times10^{-3}}\approx1.4\times10^{3}\ \text{W m}^{-2} \). Answer: 1.4 kW m⁻².

Teacher's Note:
1. Convert cm² to m² by multiplying with \( 10^{-4} \) (so dividing by it when it is in the denominator) and minutes to seconds.
2. 1 kW m⁻² = 1000 J s⁻¹ m⁻².

Answer: (d) 527 °C

Explanation:
1. Stefan's law: emission rate \( \propto T^{4} \), where T is in kelvin.
2. Initial \( T_1=127+273=400 \) K, and \( \dfrac{T_2^{4}}{T_1^{4}}=16 \) gives \( T_2=2T_1=800 \) K.
3. \( 800-273=527 \) °C. Answer: 527 °C.

Teacher's Note:
1. Always convert to kelvin before applying Stefan's law.
2. A 16-fold increase in the emitted power only needs the temperature to double.

Answer: (c) Planck's law

Explanation:
1. Wien's law fits the short-wavelength part only, and the Rayleigh-Jeans law fits the long-wavelength part only.
2. Stefan's law gives only the total power emitted, and Kirchhoff's law relates emission to absorption.
3. Planck's radiation law reproduces the whole spectral distribution at all temperatures. Answer: Planck's law.

Teacher's Note:
1. Planck's quantum hypothesis was introduced exactly to explain the black body spectrum.
2. Remember what each law deals with: total power (Stefan), peak wavelength (Wien), whole curve (Planck).

Answer: (d) 1 : 12

Explanation:
1. Thermal capacity = \( m s=\dfrac{4}{3}\pi r^{3}\rho\, s \).
2. Ratio \( =\left(\dfrac{1}{2}\right)^{3}\times\dfrac{2}{1}\times\dfrac{1}{3}=\dfrac{1}{8}\times2\times\dfrac{1}{3}=\dfrac{1}{12} \). Answer: 1 : 12.

Teacher's Note:
1. Mass depends on the cube of the radius, which is the step most often forgotten.
2. Multiply the three ratios together to get the ratio of the thermal capacities.

Answer: (d) 0 °C

Explanation:
1. Heat released when the water cools to 0 °C: \( 300\times1\times25=7500 \) cal.
2. Heat needed to melt all the ice: \( 100\times80=8000 \) cal.
3. Since 7500 cal is less than 8000 cal, the ice cannot melt completely, so the mixture stays at 0 °C.

Teacher's Note:
1. Always compare the heat available with the heat needed for complete melting before assuming a final temperature.
2. When ice and water coexist, the temperature is exactly 0 °C.

Answer: (c) 25 °C

Explanation:
1. In the steady state the heat current is the same in both blocks: \( K_A(100-T)=K_B(T-0) \).
2. With \( K_B=3K_A \): \( 100-T=3T \), so \( T=25 \) °C.

Teacher's Note:
1. Blocks of equal length and area in series carry the same heat current in the steady state.
2. The junction is closer to the end of the better conductor.

Answer: (d) 60 °C

Explanation:
1. \( \dfrac{C}{5}=\dfrac{F-32}{9} \).
2. \( C=\dfrac{5\times(140-32)}{9}=\dfrac{5\times108}{9}=60 \) °C.

Teacher's Note:
1. Subtract 32 from the Fahrenheit reading first and then multiply by 5/9.
2. Check your answer by confirming that 212 °F gives 100 °C.

Answer: (d) \( \dfrac{2T}{\rho L} \)

Explanation:
1. When the radius decreases by dr, the loss of surface energy is \( T\,d(4\pi r^{2})=8\pi r T\,dr \).
2. The mass evaporated is \( dm=4\pi r^{2}\rho\,dr \), needing heat \( L\,dm=4\pi r^{2}\rho L\,dr \).
3. Equating the two: \( 8\pi rT=4\pi r^{2}\rho L \), so \( r=\dfrac{2T}{\rho L} \).

Teacher's Note:
1. The energy released by the shrinking surface must at least supply the latent heat of the vapour formed.
2. Differentiate the surface area, not the volume, for surface energy.

Answer: (b) 2 minutes 34 seconds

Explanation:
1. Rate of cooling \( =k(\theta_{avg}-\theta_0) \). First interval: average 38 °C, so \( \dfrac{4}{2}=k(38-20) \) gives \( k=\dfrac{1}{9}\ \text{min}^{-1} \).
2. Second interval: average 34 °C, so \( \dfrac{4}{t}=\dfrac{1}{9}(34-20)=\dfrac{14}{9} \).
3. \( t=\dfrac{36}{14}\approx2.57 \) min = 2 min 34 s (approximately).

Teacher's Note:
1. Use the mean temperature of the interval in Newton's law when the fall in temperature is small.
2. The body cools more slowly as it approaches the surroundings, so the second interval takes longer than the first.

Answer: (a) \( \sqrt{\dfrac{\pi}{6}} : 1 \)

Explanation:
1. Rate of cooling \( =\dfrac{\sigma(T^{4}-T_0^{4})S}{mc} \); for equal S, T and material it is proportional to \( \dfrac{1}{m}\propto\dfrac{1}{V} \).
2. Equal areas: \( 4\pi r^{2}=6a^{2} \), so \( a=r\sqrt{\dfrac{2\pi}{3}} \).
3. \( \dfrac{R_{sphere}}{R_{cube}}=\dfrac{V_{cube}}{V_{sphere}}=\dfrac{a^{3}}{\frac{4}{3}\pi r^{3}}=\dfrac{(2\pi/3)^{3/2}}{4\pi/3}=\sqrt{\dfrac{\pi}{6}} \).

Teacher's Note:
1. With equal surface area the radiated power is equal, so the body with the smaller mass cools faster.
2. For a given surface area the sphere has the largest volume, so the sphere cools more slowly.

Answer: (d) \( \nu_m\propto T \)

Explanation:
1. Wien's law: \( \lambda_mT=b \) (a constant).
2. Since \( \nu_m=\dfrac{c}{\lambda_m} \), we get \( \nu_m=\dfrac{c}{b}\,T \).
3. So \( \nu_m\propto T \): the graph of \( \nu_m \) against T is a straight line through the origin.

Teacher's Note:
1. Do not confuse Wien's law (peak position) with Stefan's law (\( \propto T^{4} \), total power).
2. A hotter body has its peak at a shorter wavelength and a higher frequency.

Answer: (d) 200 K

Explanation:
1. \( \lambda_mT=b \), so \( T=\dfrac{b}{\lambda_m} \).
2. \( T=\dfrac{2892\times10^{-6}}{14.46\times10^{-6}}=200 \) K.

Teacher's Note:
1. 1 micron is \( 10^{-6} \) m, so the powers of ten cancel here.
2. The answer is in kelvin because Wien's constant is in m K.