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 (a) +1/3a (b) -1/a (c) 1/2a (d) -1/3a

Answer: (d). Curve: p = ((V-2)²)/(4a), so at V = 1 and V = 3, p = 1/4a (line AC). Enclosed area = ∫₁³ [1/4a - ((V-2)²)/(4a)] dV = 1/4a(2 - 2/3) = 1/3a. A to B to C (along the bottom, rightwards) then C to A (along the top, leftwards) is anticlockwise, so the work is negative. So the answer is (d) -1/3a.

In the following p - V diagram the equation of state along the curved… | JEE Main 2026 (28 Jan, Shift 1)

Q16JEE Main 2026 · 28 JAN · SHIFT 1Difficult

P-V Diagrams & Cyclic Processes

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

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Answer: (d)

Curve: $p = \frac{(V-2)^2}{4a}$ , so at $V = 1$ and $V = 3$ , $p = \frac{1}{4a}$ (line AC).
Enclosed area $= \int_1^3 \left[\frac{1}{4a} - \frac{(V-2)^2}{4a}\right] dV = \frac{1}{4a}\left(2 - \frac{2}{3}\right) = \frac{1}{3a}$ .
A to B to C (along the bottom, rightwards) then C to A (along the top, leftwards) is anticlockwise, so the work is negative.
So the answer is (d) $-\frac{1}{3a}$ .

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