JEE Physics Thermodynamics MCQs Set 10

Answer: (d) \( \frac3T \)

Explanation:
1. For an ideal gas \( V\propto\frac TP \) and \( P\propto T^{-2} \), so \( V\propto T^3 \).
2. Volume expansion coefficient \( \gamma=\frac1V\frac{dV}{dT}=\frac3T \).

Teacher's Note:
1. Eliminate P using the given relation to find how V depends on T.
2. If V is proportional to T^n, the expansion coefficient is n/T.

Answer: (a) 12

Explanation:
1. A hole expands like the metal that was removed: \( \Delta D=D\alpha\Delta T \).
2. \( \Delta D=5\times1.6\times10^{-5}\times150=1.2\times10^{-2}\ \text{cm}=12\times10^{-3}\ \text{cm} \), so d = 12.

Teacher's Note:
1. The diameter of a hole increases on heating, as if it were filled with the same metal.
2. Temperature change is 177 - 27 = 150 °C.

Answer: (b) 1000 K

Explanation:
1. \( \eta=1-\frac{T_2}{T_1}\Rightarrow0.5=1-\frac{T_2}{600}\Rightarrow T_2=300\ \text{K} \).
2. \( 0.7=1-\frac{300}{T_1'}\Rightarrow T_1'=\frac{300}{0.3}=1000\ \text{K} \).

Teacher's Note:
1. The sink temperature is fixed, so find it from the first case.
2. Temperatures in Carnot formulae must be in kelvin.

Answer: (b) A-II, B-I, C-IV, D-III

Explanation:
1. Isothermal: T constant, so \( \Delta U=0 \) (II).
2. Adiabatic: Q = 0, so \( \Delta U=-W \) (I).
3. Isochoric: \( \Delta V=0 \), so W = 0 (IV).
4. Isobaric: \( Q=\Delta U+P\Delta V \) (III).

Teacher's Note:
1. Apply the first law Q = ΔU + W to each process.
2. The internal energy of an ideal gas depends only on temperature.

Answer: (d) C and D only

Explanation:
1. Isothermal: T is constant, so \( \Delta U=0 \) (C).
2. First law: \( Q=W \) and Q > 0, so the gas does positive work (D).

Teacher's Note:
1. Internal energy of an ideal gas depends on temperature alone.
2. All the heat absorbed is converted into work.

Answer: (d) 4.5 J

Explanation:
1. Q = 0, so \( \Delta U=-W_{by}=+W_{on} \).
2. Work done on the gas equals the area under the straight line: \( \frac{(10+50)\times10^3}{2}\times(200-50)\times10^{-6}=4.5\ \text{J} \).
3. \( \Delta U=+4.5\ \text{J} \) (the gas is compressed).

Teacher's Note:
1. The area under a P-V line is a trapezium.
2. Compression adds energy to the gas when no heat flows out.

Answer: (a) \( \frac43 \)

Explanation:
1. Adiabatic: \( P_1V_1^\gamma=P_2V_2^\gamma\Rightarrow\frac{P_2}{P_1}=\left(\frac{V_1}{V_2}\right)^\gamma=\left(\frac8{27}\right)^\gamma \).
2. \( \frac{16}{81}=\left(\frac23\right)^4=\left(\frac8{27}\right)^{4/3} \), so \( \gamma=\frac43 \).

Teacher's Note:
1. 8/27 = (2/3)³ and 16/81 = (2/3)⁴.
2. Compare the exponents of 2/3.

Answer: (d) 62 K

Explanation:
1. Hot reservoir \( T_1=99+273=372\ \text{K} \).
2. First case: \( \frac13=1-\frac{T_2}{372}\Rightarrow T_2=248\ \text{K} \).
3. Second case: \( \frac16=1-\frac{T_2'}{372}\Rightarrow T_2'=310\ \text{K} \).
4. \( x=310-248=62\ \text{K} \).

Teacher's Note:
1. Convert 99 °C to kelvin before using the Carnot formula.
2. A rise in the sink temperature lowers the efficiency.

Answer: (b) 1.4 minutes

Explanation:
1. First case: \( \frac{12}{2}=k(92-22)=70k\Rightarrow k=\frac{6}{70} \).
2. Second case: \( \frac{6}{t}=k(72-22)=50k=\frac{300}{70} \).
3. \( t=\frac{6\times70}{300}=1.4\ \text{minutes} \).

Teacher's Note:
1. Use the average temperature in each interval.
2. The cooler the soup, the slower it cools.

Answer: (a) 313 K

Explanation:
1. Linear scale: \( \frac{41-5}{95-5}=\frac{T_C}{100}\Rightarrow T_C=\frac{36}{90}\times100=40\ ^\circ\text{C} \).
2. \( T=40+273=313\ \text{K} \).

Teacher's Note:
1. Use the fixed points of the faulty thermometer to build the conversion.
2. Add 273 to convert to kelvin.