1
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
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The difference of temperature in a material can convert heat energy into electrical energy. To harvest the heat energy, the material should have

A

low thermal conductivity and high electrical conductivity

B

low thermal conductivity and low electrical conductivity

C

high thermal conductivity and high electrical conductivity

D

high thermal conductivity and low electrical conductivity

2
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
+4
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Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A) : With the increase in the pressure of an ideal gas, the volume falls off more rapidly in an isothermal process in comparison to the adiabatic process.

Reason (R) : In isothermal process, PV = constant, while in adiabatic process $PV^{\gamma}$ = constant. Here $\gamma$ is the ratio of specific heats, P is the pressure and V is the volume of the ideal gas.

In the light of the above statements, choose the correct answer from the options given below:

A

(A) is true but (R) is false

B

Both (A) and (R) are true but (R) is NOT the correct explanation of (A)

C

(A) is false but (R) is true

D

Both (A) and (R) are true and (R) is the correct explanation of (A)

3
JEE Main 2025 (Online) 29th January Evening Shift
MCQ (Single Correct Answer)
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A cup of coffee cools from 90°C to 80°C in t minutes when the room temperature is 20°C. The time taken by the similar cup of coffee to cool from 80°C to 60°C at the same room temperature is:

A

$\frac{13}{5}t$

B

$\frac{10}{13}t$

C

$\frac{5}{13}t$

D

$\frac{13}{10}t$

4
JEE Main 2025 (Online) 29th January Evening Shift
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JEE Main 2025 (Online) 29th January Evening Shift Physics - Heat and Thermodynamics Question 23 English

A poly-atomic molecule $\left(C_V=3 R, C_P=4 R\right.$, where $R$ is gas constant) goes from phase space point $\mathrm{A}\left(\mathrm{P}_{\mathrm{A}}=10^5 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{A}}=4 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\mathrm{B}\left(\mathrm{P}_{\mathrm{B}}=5 \times 10^4 \mathrm{~Pa}, \mathrm{~V}_{\mathrm{B}}=6 \times 10^{-6} \mathrm{~m}^3\right)$ to point $\mathrm{C}\left(\mathrm{P}_{\mathrm{C}}=10^4\right.$ $\mathrm{Pa}, \mathrm{V}_C=8 \times 10^{-6} \mathrm{~m}^3$ ). A to $B$ is an adiabatic path and $B$ to $C$ is an isothermal path.

The net heat absorbed per unit mole by the system is :

A
$500 \mathrm{R}(\ln 3+\ln 4)$
B
$450 \mathrm{R}(\ln 4-\ln 3)$
C
$500 \mathrm{R} \ln 2$
D
$400 \mathrm{R} \ln 4$
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