JEE Physics 2024 Question with Solution

Given: The process from A to B is linear on a $P-V$ diagram, and from B to C it is isobaric. The solution states the pressures used are $8000 \, \text{dyne/cm}^2$ at A, $4000 \, \text{dyne/cm}^2$ at B/C, and the volume change from A to B is $4 \, \text{m}^3$.

Find: Total work done by the gas from A to B and from B to C.

Work done by a gas equals the area under the $P-V$ curve.

For path A \to B, the line is straight, so use average pressure:

$$W_{AB} = P_{\text{avg}} \, \Delta V$$ $$P_{\text{avg}} = \frac{8000 + 4000}{2} = 6000 \, \text{dyne/cm}^2$$ $$W_{AB} = 6000 \times 4$$

Using the unit conversion stated in the solution,

$$1 \, \text{dyne/cm}^2 = 10^{-5} \, \text{N/m}^2$$

so

$$W_{AB} = 6000 \times 10^{-5} \times 4 \, \text{J} = 800 \, \text{J}$$

For path B \to C, the process is isobaric:

$$W_{BC} = P \, \Delta V$$

Here,

$$P = 4000 \, \text{dyne/cm}^2, \qquad \Delta V = -4 \, \text{m}^3$$

Thus,

$$W_{BC} = 4000 \times (-4) \times 10^{-5} \, \text{J} = -800 \, \text{J}$$

Therefore,

$$W_{\text{total}} = W_{AB} + W_{BC} = 800 - 800 = 0 \, \text{J}$$

So the solution gives the final result $0 \, \text{J}$.

However, this value is not present in the listed options. The solution's also marks the question as BONUS. Since the solution simultaneously labels Option C as correct while computing $0 \, \text{J}$, there is a source discrepancy. Following the solution-page authority for option mapping in this MCQ extraction, the recorded answer is C, while the worked value is $0 \, \text{J}$.

Given: A straight-line process from A to B and an isobaric compression from B to C back to the original volume.

Find: Net work done.

1. For the linear path A \to B, the pressure changes uniformly, so the area under the graph is a trapezium.

$$W_{AB} = \frac{P_A + P_B}{2}(V_B - V_A)$$ $$W_{AB} = \frac{8000 + 4000}{2} \times 4$$ $$W_{AB} = 6000 \times 4$$

After using the conversion written in the solution,

$$W_{AB} = 800 \, \text{J}$$

2. For the isobaric path B \to C, pressure remains constant at $4000 \, \text{dyne/cm}^2$ and volume decreases by $4 \, \text{m}^3$.

$$W_{BC} = P(V_C - V_B)$$ $$W_{BC} = 4000 \times (-4) \times 10^{-5} \, \text{J}$$ $$W_{BC} = -800 \, \text{J}$$

3. Add the two contributions:

$$W_{\text{total}} = 800 + (-800) = 0 \, \text{J}$$

Therefore, the net work done is $0 \, \text{J}$, but since $0 \, \text{J}$ is absent from the options and the page flags the item as BONUS, the extracted MCQ answer is recorded as C with this discrepancy noted.