JEE Physics 2026 Question with Solution

Given: The curved path satisfies

$$(V-2)^2 = 4ap$$

so

$$p = \frac{(V-2)^2}{4a}$$

The closed cycle runs from $A \to C$ along a horizontal line and returns along the curved path from $C \to A$.

Find: Total work done in the closed path.

For a closed cycle on a $p\text{–}V$ diagram,

$$W = \oint p \, dV$$

which equals the enclosed area. Since the cycle is clockwise, the work done is positive.

At points $A$ and $C$, the volumes are $V=1$ and $V=3$. Using the curve equation,

$$(1-2)^2 = 4ap \Rightarrow p = \frac{1}{4a}$$

and

$$(3-2)^2 = 4ap \Rightarrow p = \frac{1}{4a}$$

Hence the upper horizontal path is at constant pressure

$$p = \frac{1}{4a}$$

Work done along the top path $A \to C$ is

$$W_{AC} = \int_{1}^{3} p \, dV = \frac{1}{4a}\int_{1}^{3} dV = \frac{1}{4a}(3-1) = \frac{1}{2a}$$

Along the curved return path $C \to A$,

$$W_{CA} = \int_{3}^{1} \frac{(V-2)^2}{4a} \, dV = -\frac{1}{4a}\int_{1}^{3}(V-2)^2 \, dV$$

Now let $x = V-2$. Then the limits change from $V=1$ to $V=3$ into $x=-1$ to $x=1$:

$$\int_{1}^{3}(V-2)^2 \, dV = \int_{-1}^{1} x^2 \, dx = \left[\frac{x^3}{3}\right]_{-1}^{1} = \frac{2}{3}$$

Therefore,

$$W_{CA} = -\frac{1}{4a} \cdot \frac{2}{3} = -\frac{1}{6a}$$

So the total work done in the cycle is

$$W_{\text{total}} = W_{AC} + W_{CA} = \frac{1}{2a} - \frac{1}{6a} = \frac{1}{3a}$$

Therefore, the correct option is B.