JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Heat & Thermodynamics

In the following

p - V

diagram the equation of state along the curved path is given by

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

where

a

is a constant. The total work done in the closed path is

Answer: (D)

Solution

The closed path consists of the parabolic arc

A \to B \to C

, on which

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

, and the straight return

C \to A

at the constant pressure

p_{A}

. Along the arc the volume runs from

V = 1

V = 3

; at

V = 1

(1 - 2)^{2} = 4 a p_{A}

gives

p_{A} = \frac{1}{4 a}

.

Work along the arc (volume increasing,

V : 1 \to 3

W_{A C} = \int_{1}^{3} \frac{( V - 2 ) ^{2}}{4 a} d V = \frac{1}{4 a} \int_{- 1}^{1} u^{2} d u = \frac{1}{4 a} \cdot \frac{2}{3} = \frac{1}{6 a} .

Work along the straight return at

p_{A}

(volume decreasing,

V : 3 \to 1

W_{C A} = p_{A} (1 - 3) = \frac{1}{4 a} \times (- 2) = - \frac{1}{2 a} .

The net work done by the gas over the closed path is

W = W_{A C} + W_{C A} = \frac{1}{6 a} - \frac{1}{2 a} = - \frac{1}{3 a} .

JEE Main · 2026 · 28 Jan 2026 · Shift 1 · Single correct option