JEE Physics Thermal Properties Of Matter MCQs Set 16

Answer: (c) \( 2^{4/3} \)

Explanation:
1. Same material means same density, so mass \( \propto r^{3} \); mass ratio 4 gives \( \dfrac{r_A}{r_B}=4^{1/3} \).
2. Rate of heat loss \( \propto \) area \( \propto r^{2} \): ratio \( =4^{2/3}=2^{4/3} \).

Teacher's Note:
1. Heat loss depends on surface area, which scales as mass to the power two-thirds.
2. The rate of cooling (not heat loss) scales as one over the radius.

Answer: (b) 0.69

Explanation:
1. \( \lambda_mT=\text{const} \), so \( \dfrac{T_{Sun}}{T_{North}}=\dfrac{\lambda_{North}}{\lambda_{Sun}}=\dfrac{350}{510}\approx0.69 \).

Teacher's Note:
1. The star with the shorter peak wavelength is the hotter one.
2. Invert the wavelength ratio to get the temperature ratio.

Answer: (c) \( \dfrac{L}{5g} \)

Explanation:
1. The loss of gravitational potential energy goes into melting: \( mgh=\dfrac{m}{5}L \).
2. \( h=\dfrac{L}{5g} \).

Teacher's Note:
1. The mass of the whole block cancels from both sides.
2. Check dimensions: \( L/g \) has dimension of length.

Answer: (d) 2 : 1

Explanation:
1. Heat capacity \( =\rho Vs \).
2. \( 8(1500)s_A=12(2000)s_B \).
3. \( \dfrac{s_A}{s_B}=\dfrac{24000}{12000}=2 \), that is 2 : 1.

Teacher's Note:
1. Heat capacity per unit volume is density times specific heat.
2. Equate the total heat capacities and cancel common factors.

Answer: (a) 0.130

Explanation:
1. Heat gained by water and calorimeter \( =(1.1+0.02)(80-15)=72.8 \) kcal.
2. Heat given by the steam per kg: \( 540+(100-80)=560 \) kcal.
3. \( m=\dfrac{72.8}{560}=0.13 \) kg.

Teacher's Note:
1. Steam gives out latent heat and then cools as water.
2. The water equivalent is simply added to the mass of water.

Answer: (c) no change in temperature takes place whether heat is taken in or given out

Explanation:
1. \( s=\dfrac{Q}{m\,\Delta T} \), which becomes infinite when \( \Delta T=0 \) for finite \( Q \).
2. So the temperature does not change whether heat is absorbed or released, as during a change of state.

Teacher's Note:
1. Infinite specific heat occurs during melting and boiling.
2. Zero temperature change with nonzero heat means infinite specific heat.

Answer: (b) -30 °C

Explanation:
1. For ice thickness \( x \), heat conducted up per second \( \dfrac{KA\,\Delta T}{x}=\rho AL\dfrac{dx}{dt} \).
2. Integrating: \( t=\dfrac{\rho L}{2K\,\Delta T}\left(x_2^{2}-x_1^{2}\right) \).
3. \( \Delta T=\dfrac{0.9\times80\times(16-4)}{2\times4\times10^{-3}\times3600}=30 \) °C, so \( T_a=-30 \) °C.

Teacher's Note:
1. As the ice thickens, the conduction slows because the thermal resistance rises.
2. The time grows with the square of the thickness.

Answer: (a) \( T_1<T_2<T_3 \)

Explanation:
1. By Newton's law the rate of cooling is proportional to the excess temperature over the surroundings.
2. As the water gets cooler the rate of fall decreases.
3. The same 5 °C drop therefore takes longer each time: \( T_1<T_2<T_3 \).

Teacher's Note:
1. Cooling slows down as the body approaches the surroundings.
2. Equal temperature drops need increasing times.

Answer: (d) 4

Explanation:
1. Average excess for the first interval \( =70-30=40 \) °C, with fall of 10 °C in 2 min: \( \dfrac{10}{2}=k(40) \).
2. Second interval: average excess \( =50-30=20 \) °C, so \( \dfrac{10}{t}=k(20) \).
3. Dividing: \( t=4 \) min.

Teacher's Note:
1. For small intervals, use the mean temperature in Newton's law.
2. Halving the excess halves the cooling rate and doubles the time.

Answer: (c) black and rough

Explanation:
1. Good emitters are good absorbers: dark and rough surfaces.
2. A black rough surface has emissivity close to 1, so it radiates most rapidly.

Teacher's Note:
1. Polished surfaces reflect most radiation and emit poorly.
2. Rough black surfaces trap radiation and emit best.

Answer: (a) The radiation emitted is in the infrared region

Explanation:
1. Body temperature is about 310 K, and \( \lambda_m=\dfrac{b}{T}\approx\dfrac{2.9\times10^{-3}}{310}\approx9.4\ \mu\text{m} \).
2. That wavelength lies in the infrared region.
3. The body radiates all the time.

Teacher's Note:
1. Bodies near room temperature radiate mainly in the infrared.
2. Only very hot objects emit visible light.

Answer: (b) Ice covered with black cloth will melt more

Explanation:
1. The black cloth absorbs more of the sunlight and heats up more.
2. It passes more heat to the ice beneath, so more ice melts there.

Teacher's Note:
1. Dark surfaces absorb a larger fraction of incident radiation.
2. This is why white clothes are preferred in summer.

Answer: (c) \( q/p \)

Explanation:
1. Coefficient of absorption (absorptive power) \( =\dfrac{\text{heat absorbed}}{\text{heat incident}} \).
2. \( a=\dfrac{q}{p} \).

Teacher's Note:
1. Absorptive power is a ratio and lies between 0 and 1.
2. A perfect black body has \( a=1 \).

Answer: (d) 3000 Å

Explanation:
1. Initial temperature \( =1227+273=1500 \) K and final \( =2500 \) K.
2. Wien's law: \( \lambda_1T_1=\lambda_2T_2 \).
3. \( \lambda_2=5000\times\dfrac{1500}{2500}=3000 \) Å.

Teacher's Note:
1. Convert to kelvin before applying Wien's law.
2. A hotter body has its intensity peak at a shorter wavelength.

Answer: (a) 320 kcal/m²-min

Explanation:
1. Intensity \( \propto T^{4} \) for the Sun's emission.
2. Doubling \( T \) multiplies the energy by \( 2^{4}=16 \).
3. \( 16\times20=320 \) kcal/m²-min.

Teacher's Note:
1. Stefan's fourth-power law makes the radiated power very sensitive to temperature.
2. The geometry (distance and radii) is unchanged.

Answer: (a) \( \sqrt3 \)

Explanation:
1. The area under the \( E_\lambda \)-\( \lambda \) curve is the total emissive power, proportional to \( T^{4} \).
2. \( \dfrac{T_2^{4}}{T_1^{4}}=9 \), so \( \dfrac{T_2}{T_1}=\sqrt3 \).
3. Wien's law: \( \dfrac{\lambda_1}{\lambda_2}=\dfrac{T_2}{T_1}=\sqrt3 \).

Teacher's Note:
1. Area under the spectral curve is the total radiated power per unit area.
2. The hotter curve peaks at the shorter wavelength.

Answer: (d) Just after putting the coffee in the cup

Explanation:
1. The cooling rate is proportional to the excess temperature, so a hotter cup loses heat faster.
2. If the cold cream is added at once, the mixture is cooler and loses less heat for the rest of the 10 minutes.
3. The final temperature is higher by \( \delta(1-e^{-kt}) \) compared with adding the cream at the end, where \( \delta \) is the temperature drop due to the cream.

Teacher's Note:
1. Reduce the temperature early to reduce the heat loss.
2. Compare two situations using exponential cooling.

Answer: (a) more than 40 °C

Explanation:
1. Excess temperatures decrease by the same factor in equal times: \( \dfrac{50-30}{60-30}=\dfrac{2}{3} \).
2. After another 10 minutes the excess is \( 20\times\dfrac{2}{3}\approx13.3 \) °C.
3. The temperature is about 43.3 °C, more than 40 °C.

Teacher's Note:
1. Cooling slows down as the body gets closer to the surroundings.
2. A linear drop of 10 °C each time would give 40 °C, which is wrong.

Answer: (a) Plate

Explanation:
1. For the same mass and material, the rate of cooling is proportional to the surface area.
2. A plate has the largest area for a given volume and a sphere the smallest.
3. So the plate cools fastest.

Teacher's Note:
1. Among shapes of equal volume the sphere has the least surface area.
2. Large surface-to-volume ratio means quick cooling.

Answer: (b) less than 25 cm

Explanation:
1. At 0 °C the steel tape has contracted, so each division is slightly shorter than its nominal length.
2. The wood is as long as 25 of these shortened divisions, so its real length is \( 25(1-\alpha_s\times20) \) cm.
3. This is less than 25 cm.

Teacher's Note:
1. A contracted scale reads larger values than the true length.
2. A scale is correct only at its calibration temperature.

Answer: (d) 10 cm

Explanation:
1. Expansion per cm of A \( =\dfrac{0.075}{20} \) and of B \( =\dfrac{0.045}{20} \) cm per cm.
2. Let A have length \( x \): \( x\dfrac{0.075}{20}+(20-x)\dfrac{0.045}{20}=0.060 \).
3. \( 0.075x+0.9-0.045x=1.2 \), so \( 0.03x=0.3 \) and \( x=10 \) cm.

Teacher's Note:
1. The total expansion of a composite rod is the sum of the expansions of its parts.
2. Set up one equation in the unknown length.

Answer: (d) 5.0 °C

Explanation:
1. For a 1 mm interval, \( \Delta l=\alpha l\,\Delta T \).
2. \( 6\times10^{-5}=12\times10^{-6}\times1\times\Delta T \).
3. \( \Delta T=5.0 \) °C.

Teacher's Note:
1. Use \( \Delta l=\alpha l\Delta T \) with the nominal interval as \( l \).
2. Units of mm cancel on both sides.

Answer: (d) \( 2\alpha I\Delta T \)

Explanation:
1. The moment of inertia is proportional to the square of a linear dimension: \( I\propto l^{2} \).
2. \( \dfrac{\Delta I}{I}=2\dfrac{\Delta l}{l}=2\alpha\Delta T \).
3. \( \Delta I=2\alpha I\Delta T \).

Teacher's Note:
1. Quantities proportional to length squared change by \( 2\alpha\Delta T \) (like area).
2. Mass does not change on heating.

Answer: (b) (1), (3) and (4) only

Explanation:
1. Every linear dimension of an isotropic solid expands by the same percentage, 0.2%.
2. Area rises by about \( 2\times0.2=0.4\% \) (statement 1 true), volume by \( 3\times0.2=0.6\% \) (statement 3 true), but the thickness rises by only 0.2%, so statement 2 is false.
3. \( \alpha=\dfrac{0.2/100}{80}=2.5\times10^{-5}=0.25\times10^{-4}\ \text{°C}^{-1} \), so statement 4 is true.

Teacher's Note:
1. All linear dimensions, including thickness, expand in the same proportion.
2. Areal expansion is twice and volume expansion three times the linear fraction.