JEE Physics Thermal Properties Of Matter MCQs Set 22

Answer: (c) \( \dfrac{K_1+3K_2}{4} \)

Explanation:
1. Heat flows along the length through the two parts in parallel, with areas \( \pi R^2 \) and \( \pi(4R^2-R^2)=3\pi R^2 \).
2. \( K_{eff}(4\pi R^2)=K_1\pi R^2+K_2(3\pi R^2)\Rightarrow K_{eff}=\dfrac{K_1+3K_2}{4} \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (d) \( 45^\circ\text{C} \)

Explanation:
1. Thermal resistances (proportional to length): AP \( =\tfrac L2 \), QB \( =\tfrac L2 \), and between \( P \) and \( Q \) the straight part \( (L) \) and the bent rod (\( \tfrac{3L}2 \)) in parallel: \( \dfrac{L\times1.5L}{2.5L}=0.6L \).
2. Total \( =1.6L \). The temperature difference across PQ \( =120\times\dfrac{0.6}{1.6}=45^\circ C \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (b) 4.8 cal/s

Explanation:
1. Values of \( K/L \): copper \( \dfrac{0.92}{46}=0.02 \), brass \( \dfrac{0.26}{13}=0.02 \), steel \( \dfrac{0.12}{12}=0.01 \). Let the junction be at \( T \): \( 0.02(100-T)=0.02T+0.01T\Rightarrow T=40^\circ C \).
2. Rate \( =KA\dfrac{\Delta T}{L}=4(0.02)(60)=4.8 \) cal/s.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (c) a decreasing curve that is steepest at the start and flattens out, approaching \( \theta_0 \) asymptotically from above

Explanation:
1. Newton's law: \( \dfrac{dT}{dt}=-k(T-\theta_0) \), so \( T=\theta_0+(\theta-\theta_0)e^{-kt} \).
2. The rate of fall is largest when the excess temperature is largest and the curve tends to \( \theta_0 \) but never goes below it.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

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

Explanation:
1. Steady state: energy absorbed by the middle plate equals energy emitted: \( \sigma(3T)^4-\sigma T_m^4=\sigma T_m^4-\sigma(2T)^4 \).
2. \( 2T_m^4=(81+16)T^4\Rightarrow T_m=\left(\dfrac{97}{2}\right)^{1/4}T \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (b) Warming of glass of bulb due to filament

Explanation:
1. Convection needs a fluid moving in bulk because of density differences: sea and land breeze, boiling water and hot air near a furnace all involve this.
2. The glass of a bulb is mainly warmed by radiation (and conduction through the gas) from the filament, not by bulk fluid flow.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (c) \( Q_B\text{ is maximum} \)

Explanation:
1. Wien's law: \( T\propto\dfrac1{\lambda_m} \), and power \( Q\propto r^2T^4\propto\dfrac{r^2}{\lambda_m^4} \).
2. In units of 100 nm: \( Q_A\propto\dfrac{4}{81}=0.049 \), \( Q_B\propto\dfrac{16}{256}=0.0625 \), \( Q_C\propto\dfrac{36}{625}=0.058 \). So \( Q_B \) is maximum.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (a) \( E_x\gt E_y\ \text{and}\ a_x\gt a_y \)

Explanation:
1. Faster cooling by radiation means a greater emissive power, so \( E_x\gt E_y \).
2. By Kirchhoff's law a good emitter is a good absorber (\( E/a \) is the same for all bodies at one temperature), so \( a_x\gt a_y \) as well.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (b) initially it is the darkest body and at later times it cannot be distinguished

Explanation:
1. At room temperature a black body absorbs all light and emits little, so it looks darkest against the furnace walls.
2. Once in thermal equilibrium with the furnace it emits exactly what the walls emit, so it cannot be distinguished.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (c) \( 60^\circ\text{C} \)

Explanation:
1. Heat entering the junction from the two hot rods equals heat leaving through the cold rod: \( 2(90-T)=T-0 \).
2. \( 180=3T\Rightarrow T=60^\circ C \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (a) \( U_2\gt U_1 \)

Explanation:
1. \( \lambda_m=\dfrac bT=\dfrac{2.88\times10^6}{2880}=1000 \) nm, so the peak of the spectrum lies in the 999-1000 nm band.
2. A black body emits some energy at every wavelength, so \( U_1,U_3\neq0 \), and the band at the peak has the most energy: \( U_2\gt U_1 \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (b) 1800

Explanation:
1. \( P\propto r^2T^4 \).
2. New power \( =450\times\left(\dfrac12\right)^2\times2^4=450\times4=1800 W \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (d) \( \left(\dfrac13\right)^{1/3} \)

Explanation:
1. Rate of cooling \( \propto\dfrac{\text{surface area}}{\text{mass}}\propto\dfrac{r^2}{r^3}=\dfrac1r \).
2. \( m_1=3m_2\Rightarrow r_1=3^{1/3}r_2 \), so the ratio is \( 3^{-1/3}=(1/3)^{1/3} \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (a) \( \dfrac{3}{\sqrt2+1} \)

Explanation:
1. Let \( AB=BC=l \), so \( AC=\sqrt2\,l \). Heat reaches \( C \) from \( B \) through BC and leaves \( C \) towards \( A \) through CA.
2. \( \dfrac{\sqrt2T-T_c}{l}=\dfrac{T_c-T}{\sqrt2\,l}\Rightarrow2T-\sqrt2T_c=T_c-T\Rightarrow\dfrac{T_c}{T}=\dfrac3{\sqrt2+1} \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (b) (a) and (b) only

Explanation:
1. Equal power and equal area: \( 0.01T_A^4=0.81T_B^4\Rightarrow\left(\dfrac{T_A}{T_B}\right)^4=81\Rightarrow T_B=\dfrac{5802}{3}=1934 K \): \( (a) \) true.
2. \( \lambda_A=\dfrac{2.898\times10^{-3}}{5802}=0.5\ \mu m \) and \( \lambda_B=3\lambda_A=1.5\ \mu m, a \) shift of 1.0 \( \mu m \): \( (b) \) true. \( (c) \) and \( (d) \) are false.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (d) 2

Explanation:
1. \( P\propto r^2T^4 \): \( 10^4=(400)^2\left(\dfrac{T_A}{T_B}\right)^4\Rightarrow\left(\dfrac{T_A}{T_B}\right)^4=\dfrac1{16}\Rightarrow\dfrac{T_A}{T_B}=\dfrac12 \).
2. Wien: \( \dfrac{\lambda_A}{\lambda_B}=\dfrac{T_B}{T_A}=2 \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (a) 9

Explanation:
1. At 487 \( ^\circ C =760 K \): \( \log_2(P/P_0)=1\Rightarrow P=2P_0 \).
2. 2767 \( ^\circ C =3040 K =4\times760 K \), so \( P\propto T^4 \) grows 256 times: \( P=512P_0 \) and \( \log_2512=9 \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (b) 9

Explanation:
1. Wien: \( \dfrac{T_1}{T_2}=\dfrac{1500}{500}=3 \).
2. \( \dfrac{P_A}{P_B}=\left(\dfrac{r_A}{r_B}\right)^2\left(\dfrac{T_1}{T_2}\right)^4=\dfrac19\times81=9 \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (d) \( \dfrac{4\pi R^2KT}{P} \)

Explanation:
1. For a thin shell of thickness \( x \), \( P=KA\dfrac{T}{x} \) with \( A=4\pi R^2 \).
2. \( x=\dfrac{4\pi R^2KT}{P} \), so the thickness must be at least this much.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (a) \( \dfrac{7\rho cr}{72\sigma}\times10^{-6} \)

Explanation:
1. \( \dfrac43\pi r^3\rho c\,dT=-\sigma(4\pi r^2)T^4dt\Rightarrow dt=-\dfrac{\rho cr}{3\sigma}\dfrac{dT}{T^4} \).
2. \( t=\dfrac{\rho cr}{3\sigma}\cdot\dfrac13\left[\dfrac1{100^3}-\dfrac1{200^3}\right]=\dfrac{\rho cr}{9\sigma}\cdot\dfrac78\times10^{-6}=\dfrac{7\rho cr}{72\sigma}\times10^{-6} \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (c) 1 : 1

Explanation:
1. \( P\propto r^2T^4 \).
2. \( \dfrac{P_1}{P_2}=\left(\dfrac14\right)^2(2)^4=\dfrac{16}{16}=1 \): the two spheres radiate equal energy per second.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (d) 41.5 W

Explanation:
1. Series resistances: glass \( \dfrac{0.01}{0.8(1)}=0.0125 K/W \) each and air \( \dfrac{0.05}{0.08(1)}=0.625 K/W \).
2. Total \( R=0.65 K/W \), so \( H=\dfrac{27}{0.65}\approx41.5 W \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (a) \( 26.5^\circ\text{C} \)

Explanation:
1. Heat current \( H=\dfrac{27}{0.65}=41.5 W \). Temperature drop across the first glass sheet \( =41.5\times0.0125=0.52^\circ C \).
2. So this interface is at \( 27-0.52\approx26.5^\circ C \) (and the air-glass interface on the outdoor side is at about 0.5 \( ^\circ C \)).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (c) 9 kW

Explanation:
1. Resistance per unit area: \( \dfrac{0.025}{0.125}+\dfrac{0.01}{1.5}+\dfrac{0.25}{1.0}=0.2+0.0067+0.25=0.4567 m ^2 K/W \).
2. \( P=\dfrac{A\Delta T}{R}=\dfrac{137\times30}{0.4567}\approx9000 W \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (d) The hollow sphere

Explanation:
1. Rate of loss of heat \( =e\sigma A(T^4-T_0^4) \) is the same for both because the outer area is the same.
2. Rate of fall of temperature \( =\dfrac{\text{heat loss rate}}{mc} \); the hollow sphere has less mass, so it cools faster.

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.

Answer: (a) \( 15.2^\circ\text{C} \)

Explanation:
1. Heater power \( P=\dfrac{V^2}{R}=\dfrac{200^2}{20}=2000 W \), all of which flows through the window. \( K=0.2\times4.2=0.84 W m ^{-1} K ^{-1} \).
2. \( \Delta T=\dfrac{PL}{KA}=\dfrac{2000\times0.002}{0.84\times1}=4.76^\circ C \), so the outside temperature \( =20-4.76\approx15.2^\circ C \).

Teacher's Note:
1. Identify the mode of heat transfer, then use the matching law: conduction \( Q/t = K A (T1-T2)/L \), radiation \( P = e \sigma A T^4 \), Newton cooling for small differences.
2. For rods in series the same heat current flows through each; for rods in parallel the temperature difference is common.