Q 11-11-045JEE MainJEE Main 2026 (28 Jan, Shift 1)Hard
In the following $p - V$ diagram the equation of state along the curved path is given by $(V-2)^2 = 4ap$ where $a$ is a constant. The total work done in the closed path is
Answer: (D) $-\dfrac1{3a}$
Along the curve $p = \dfrac{(V-2)^2}{4a}$. At $A$ ($V = 1$) and $C$ ($V = 3$), $p = \dfrac1{4a}$.
Path $A\to B\to C$ (along the curve, $V$ from 1 to 3):
$$W_1 = \int_1^3\frac{(V-2)^2}{4a}\,dV = \frac1{4a}\left[\frac{(V-2)^3}{3}\right]_1^3 = \frac1{4a}\cdot\frac23 = \frac1{6a}$$
Path $C\to A$ (constant $p = \frac1{4a}$, $V$ from 3 to 1):
$$W_2 = \frac1{4a}(1 - 3) = -\frac1{2a}$$
$$W = \frac1{6a} - \frac1{2a} = -\frac1{3a}$$
(The cycle runs anticlockwise on the $p$–$V$ diagram, so the net work is negative.)
Solution by Sreeraj P, M.Sc Physics