Physics · Thermodynamics

JEE Main 2026 — 28 January, Morning Shift — Question 32

In the following $p-V$ diagram the equation of state along the curved path is given by $(V-2)^{2}=4$ ap where $a$ is a constant. The total work done in the closed path is

Option A:
$-\frac{1}{a}$
Option B:
$+\frac{1}{3 a}$
Option C:
$\frac{1}{2 a}$
Option D:
$-\frac{1}{3 a}$
Correct

Answer: D

Step-by-step solution

$\mathrm{w}=$ Area of parabola $=\frac{2}{3}($ Area of rectangle AC31A $)$ $=\frac{2}{3} \mathrm{P}_{0}(3-1)=\frac{4 \mathrm{P}_{0}}{3}$ When $\mathrm{V}=1$ $(1-2)^{2}=4 \mathrm{aP}_{0}$ $\mathrm{P}_{0}=\frac{1}{4 \mathrm{a}}$ $\mathrm{w}=\frac{4}{3} \mathrm{P}_{0}=\frac{4}{3} \frac{1}{4 \mathrm{a}}=\frac{1}{3 \mathrm{a}}$ $\mathrm{w}_{\text {gas }}=\frac{-1}{3 \mathrm{a}}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 28 January, Morning Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Calculation of Work, Heat and Internal Energy