JEE Physics Thermal Properties Of Matter MCQs Set 17

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

Explanation:
1. Every dimension of the object expands as if it were enlarged photographically, including the gap.
2. So the rod length increases (1), the gap increases and does not decrease (3), and the diameter increases in the same ratio as the length (4).
3. Statement (2) is false.

Teacher's Note:
1. A gap in a ring widens on heating, like a hole in a plate.
2. Think of the heated ring as a scaled-up copy of the original.

Answer: (c) \( \dfrac{5}{3}\alpha \)

Explanation:
1. Total expansion \( =L\alpha\,\Delta T+2L(2\alpha)\Delta T=5L\alpha\,\Delta T \).
2. This equals \( 3L\,\alpha_{avg}\Delta T \).
3. \( \alpha_{avg}=\dfrac{5}{3}\alpha \).

Teacher's Note:
1. The average coefficient is the length-weighted mean of the two coefficients.
2. Add the individual expansions first.

Answer: (c) 0.076 cm

Explanation:
1. Initial height \( =\dfrac{500}{125}=4 \) cm.
2. Final volume \( =500(1+3.2\times10^{-4}\times70)=511.2 \) cm³.
3. Final area \( =125(1+2\times2.32\times10^{-5}\times70)=125.406 \) cm².
4. Final height \( =\dfrac{511.2}{125.406}\approx4.0764 \) cm, so the level rises by about 0.076 cm.

Teacher's Note:
1. The can's cross-section expands with twice its linear coefficient.
2. The level rises because the liquid expands more than the container.

Answer: (d) 28.9 cc

Explanation:
1. Initial volume \( V=\dfrac{4}{3}\pi(10)^{3}\approx4189 \) cm³.
2. \( \Delta V=3\alpha V\,\Delta T=3\times23\times10^{-6}\times4189\times100 \).
3. \( \Delta V\approx28.9 \) cm³.

Teacher's Note:
1. Volume expansion coefficient is three times the linear coefficient.
2. The radius is 10 cm (half of the diameter).

Answer: (d) 25 °C; \( 1.85\times10^{-5}\ \text{/°C} \)

Explanation:
1. Let the clock be correct at \( \theta \). Time lost per day \( =\dfrac12\alpha\,\Delta\theta\times86400 \).
2. At 40 °C: \( \dfrac12\alpha(40-\theta)86400=12 \). At 20 °C: \( \dfrac12\alpha(\theta-20)86400=4 \).
3. Dividing: \( \dfrac{40-\theta}{\theta-20}=3 \), so \( \theta=25 \) °C.
4. Then \( \alpha=\dfrac{2\times12}{86400\times15}\approx1.85\times10^{-5}\ \text{/°C} \).

Teacher's Note:
1. A pendulum clock loses time when hot (longer pendulum) and gains time when cold.
2. The fractional change in period is \( \tfrac12\alpha\Delta\theta \).

Answer: (d) It loses about 36.3 s

Explanation:
1. A higher temperature lengthens the pendulum and so increases the period: the clock runs slow and loses time.
2. Fractional change in period \( =\tfrac12\alpha\,\Delta\theta=\tfrac12(1.2\times10^{-5})(10)=6\times10^{-5} \).
3. In one week (604800 s) the clock loses \( 6\times10^{-5}\times604800\approx36.3 \) s.

Teacher's Note:
1. A longer pendulum has a longer period, so the clock falls behind.
2. Time lost per unit time equals the fractional increase in the period.

Answer: (c) 50 °C

Explanation:
1. The scale is linear: \( \dfrac{\text{reading}-5}{95-5}=\dfrac{\text{true}}{100} \).
2. \( \dfrac{50-5}{90}=\dfrac{x}{100} \).
3. \( x=50 \) °C.

Teacher's Note:
1. Use the ratio of the position between the two fixed points.
2. The faulty thermometer happens to be right at its midpoint.

Answer: (d) \( 8.6\times10^{-3} \) K

Explanation:
1. At constant speed the loss of potential energy goes into heat: \( Q=mg\,h=0.2\times10\times(0.6\times0.6)=0.72 \) J.
2. \( \Delta T=\dfrac{Q}{ms}=\dfrac{0.72}{0.2\times420}\approx8.6\times10^{-3} \) K.

Teacher's Note:
1. Constant speed means no change in kinetic energy, so all the lost potential energy becomes heat.
2. Vertical drop is \( l\sin\theta \).

Answer: (b) 28.6 °C

Explanation:
1. Heat to bring the ice to 0 °C water: \( 10(0.5)(10)+10(80)=850 \) cal.
2. Heat released by the steam condensing and cooling to 0 °C is \( 2(540)+2(100)=1280 \) cal and by the water cooling to 0 °C is 200 cal; total 1480 cal, more than 850 cal, so everything ends up liquid.
3. Remaining heat \( 630 \) cal warms \( 22 \) g of water from 0 °C: \( T=\dfrac{630}{22}\approx28.6 \) °C.

Teacher's Note:
1. Check first whether the ice can melt completely and the steam can condense completely.
2. Then solve for the final temperature with one energy balance.

Answer: (a) 3840 J

Explanation:
1. Area \( A=0.1\times0.1=0.01 \) m², thickness \( l=0.01 \) m, \( \Delta T=80 \) °C.
2. \( H=\dfrac{KA\,\Delta T}{l}=\dfrac{0.8\times0.01\times80}{0.01}=64 \) W.
3. Heat per minute \( =64\times60=3840 \) J.

Teacher's Note:
1. Heat flows through the thickness, so the area is the large face.
2. Convert the power to energy by multiplying by time.

Answer: (d) \( 5\times10^{-8} \) kg/s

Explanation:
1. \( A=0.04\times10^{-4}=4\times10^{-6} \) m².
2. \( H=\dfrac{KA\,\Delta T}{l}=\dfrac{42\times4\times10^{-6}\times100}{1}=1.68\times10^{-2} \) W.
3. \( \dfrac{m}{t}=\dfrac{H}{L}=\dfrac{1.68\times10^{-2}}{3.36\times10^{5}}=5\times10^{-8} \) kg/s.

Teacher's Note:
1. In steady state every joule that reaches the ice melts it.
2. Convert cm² to m² carefully (1 cm² = \( 10^{-4} \) m²).

Answer: (c) 4.0 W

Explanation:
1. Each half of AB (AC and CB) has resistance \( 2.5 \) K/W.
2. Heat balance at C (with temperature \( T_C \)): \( \dfrac{100-T_C}{2.5}+\dfrac{0-T_C}{2.5}+\dfrac{25-T_C}{5}=0 \).
3. \( 2(100-T_C)-2T_C+25-T_C=0 \), so \( T_C=45 \) °C.
4. Heat current in CD \( =\dfrac{45-25}{5}=4.0 \) W from C to D.

Teacher's Note:
1. Apply a junction rule (like Kirchhoff's) to heat currents.
2. Thermal resistance is proportional to length.

Answer: (c) 65 °C

Explanation:
1. In steady state the heater power equals the heat conducted through the insulation: \( P=\dfrac{KA\,\Delta T}{d} \).
2. \( \Delta T=\dfrac{1000\times0.04}{0.2\times5}=40 \) °C.
3. Tank temperature \( =25+40=65 \) °C.

Teacher's Note:
1. Steady state means heat in equals heat out.
2. The insulation drives the temperature difference.

Answer: (d) 5 °C

Explanation:
1. The heat current per unit area is the same in every layer, \( q\propto k\,\Delta T \) (equal thickness).
2. First layer: \( k(20-10)=10k \).
3. Second layer (conductivity \( 2k \)): \( 2k(10-\theta)=10k \), so \( 10-\theta=5 \) and \( \theta=5 \) °C.
4. Check: third layer \( k(5-(-5))=10k \); fourth layer \( 2k(-5-(-10))=10k \).

Teacher's Note:
1. In series, equal flux means temperature drop is inversely proportional to conductivity.
2. Verify the answer by checking the remaining layers.

Answer: (b) \( 8.0\times10^{6} \) J

Explanation:
1. The heater emits 60 kJ per minute, so a black body (emissivity 1) at the same temperature and area emits \( \dfrac{60}{0.45}\approx133.3 \) kJ per minute.
2. In one hour: \( 133.3\times60\approx8000 \) kJ \( =8.0\times10^{6} \) J.

Teacher's Note:
1. Emissive power is proportional to emissivity at a fixed temperature and area.
2. Convert minutes to hours at the end.

Answer: (c) 10 minutes

Explanation:
1. Newton's law: \( \dfrac{\Delta T}{t}=k(T_{avg}-T_0) \).
2. First interval: \( \dfrac{10}{6}=k(75-25)=50k \), so \( k=\dfrac{1}{30} \).
3. Second interval: \( \dfrac{10}{t}=\dfrac{1}{30}(55-25)=1 \), so \( t=10 \) min.

Teacher's Note:
1. Use the average temperature of each interval.
2. Cooler bodies cool more slowly, so the time must exceed 6 minutes.

Answer: (d) 2.08

Explanation:
1. In steady state the heat lost equals the heater power: \( 250=h(37-27) \), so \( h=25 \) W/K.
2. Immediately after switching off, loss rate is still 250 W, so \( \dfrac{dT}{dt}=\dfrac{250}{120}\approx2.08 \) K/s.

Teacher's Note:
1. The heat loss at the instant of switching off equals the heater power.
2. Rate of fall of temperature is power divided by heat capacity.

Answer: (d) 10 s

Explanation:
1. For the same small drop of 0.2 °C, the time is inversely proportional to the excess temperature.
2. Excess temperatures: 20 °C in the first case and 10 °C in the second.
3. \( t=5\times\dfrac{20}{10}=10 \) s.

Teacher's Note:
1. Cooling rate is proportional to the excess over the room temperature.
2. Half the excess means half the rate.

Answer: (a) a straight line with a negative slope

Explanation:
1. In steady state the heat current \( H=-KA\dfrac{d\theta}{dx} \) is the same at every section.
2. For uniform \( K \) and \( A \), the gradient \( \dfrac{d\theta}{dx} \) is constant.
3. So \( \theta \) falls linearly with \( x \): a straight line with a negative slope.

Teacher's Note:
1. Constant gradient means a straight-line temperature profile.
2. Non-uniform conductivity or area would curve the graph.

Answer: (a) \( \dfrac{r_1r_2}{r_2-r_1} \)

Explanation:
1. Heat flow through a thin shell of radius \( r \) and thickness \( dr \): \( H=-K\,4\pi r^{2}\dfrac{dT}{dr} \).
2. Integrate: \( H\int_{r_1}^{r_2}\dfrac{dr}{r^{2}}=4\pi K(T_1-T_2) \), giving \( H\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)=4\pi K\Delta T \).
3. \( H=4\pi K\Delta T\,\dfrac{r_1r_2}{r_2-r_1} \).

Teacher's Note:
1. The area changes with radius, so the temperature gradient is not uniform.
2. Integrate across thin shells instead of using \( KA\Delta T/l \).

Answer: (b) \( \dfrac{K_1l_2T_1+K_2l_1T_2}{K_1l_2+K_2l_1} \)

Explanation:
1. Same heat current in both sections: \( \dfrac{K_1(T_1-T)}{l_1}=\dfrac{K_2(T-T_2)}{l_2} \).
2. \( K_1l_2T_1-K_1l_2T=K_2l_1T-K_2l_1T_2 \).
3. \( T=\dfrac{K_1l_2T_1+K_2l_1T_2}{K_1l_2+K_2l_1} \).

Teacher's Note:
1. Check the result by letting \( K_1\to\infty \); the interface then takes the temperature \( T_1 \).
2. Use the equal-current condition for series sections.

Answer: (d) 8.4 kJ

Explanation:
1. With negligible expansion no work is done, so \( \Delta U=Q=ms\,\Delta T \).
2. \( \Delta U=0.1\times4148\times20=8296 \) J \( \approx8.4 \) kJ.

Teacher's Note:
1. For liquids and solids the expansion work is negligible.
2. Convert grams to kilograms.

Answer: (c) \( T\propto\dfrac{1}{R} \)

Explanation:
1. Adiabatic: \( dU=-p\,dV \), with \( U=uV \) and \( p=\dfrac{u}{3} \).
2. \( V\,du+u\,dV=-\dfrac{u}{3}dV \), so \( \dfrac{du}{u}=-\dfrac43\dfrac{dV}{V} \) and \( u\propto V^{-4/3} \).
3. Since \( u\propto T^{4} \): \( T\propto V^{-1/3}\propto\dfrac{1}{R} \).

Teacher's Note:
1. Photon gas has the adiabatic law \( TV^{1/3}=\text{const} \).
2. Volume is proportional to \( R^{3} \).

Answer: (c) 0 °C

Explanation:
1. Heat to warm the ice to 0 °C: \( 0.45\times2100\times20=18900 \) J; to melt all of it: \( 0.45\times336000=151200 \) J; total 170100 J.
2. Maximum heat from the steam condensing and cooling to 0 °C: \( 0.05\times2268000+0.05\times4200\times100=134400 \) J, which is less than 170100 J.
3. So all the steam condenses but not all the ice melts, and the final temperature is 0 °C.

Teacher's Note:
1. Compare the maximum heat available with the heat needed for complete melting.
2. When ice and water coexist, the temperature is 0 °C.