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 ![JEE Main 2026 Physics, Thermodynamics — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Thermodynamics/6983c1ac8bfdbb24356397b1/question_1__q_6983c1ac8bfdbb24356397b1__cdn-question-pool.getmarks.app__28S1_q_42_1_v2__8340a8ba67_final_ppt_sync.png)

From the equation of state $(V-2)^2 = 4ap$, we get $p = \frac{(V-2)^2}{4a}$. At V = 1: $p = \frac{1}{4a}$ and at V = 3: $p = \frac{1}{4a}$. The curved path A→B→C follows the given equation, while C→A is a straight line at constant pressure. Work along the curved path: $W_{ABC} = \int_1^3 \frac{(V-2)^2}{4a} dV = \frac{1}{4a}\left[\frac{(V-2)^3}{3}\right]_1^3 = \frac{1}{4a} \cdot \frac{2}{3} = \frac{1}{6a}$. Work along C→A at constant pressure: $W_{CA} = \frac{1}{4a}(1-3) = -\frac{1}{2a}$. Total work: $W_{total} = \frac{1}{6a} - \frac{1}{2a} = -\frac{1}{3a}$.

JEE Main 2026 — Physics Thermodynamics

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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

JEE Main 2026 Physics, Thermodynamics — question figure

Official previous-year question

Held on 28 Jan 2026 · Verified 6 Jul 2026.

Options

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

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Solution

From the equation of state $(V-2)^2 = 4ap$ , we get $p = \frac{(V-2)^2}{4a}$ .

At V = 1: $p = \frac{1}{4a}$ and at V = 3: $p = \frac{1}{4a}$ .

The curved path A→B→C follows the given equation, while C→A is a straight line at constant pressure.

Work along the curved path:

$\displaystyle W_{ABC} = \int_1^3 \frac{(V-2)^2}{4a} dV = \frac{1}{4a}\left[\frac{(V-2)^3}{3}\right]_1^3 = \frac{1}{4a} \cdot \frac{2}{3} = \frac{1}{6a}$ .

Work along C→A at constant pressure:

$W_{CA} = \frac{1}{4a}(1-3) = -\frac{1}{2a}$ .

Total work: $W_{total} = \frac{1}{6a} - \frac{1}{2a} = -\frac{1}{3a}$ .

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Solution

From the equation of state $(V-2)^2 = 4ap$ , we get $p = \frac{(V-2)^2}{4a}$ .

At V = 1: $p = \frac{1}{4a}$ and at V = 3: $p = \frac{1}{4a}$ .

The curved path A→B→C follows the given equation, while C→A is a straight line at constant pressure.

Work along the curved path:

$\displaystyle W_{ABC} = \int_1^3 \frac{(V-2)^2}{4a} dV = \frac{1}{4a}\left[\frac{(V-2)^3}{3}\right]_1^3 = \frac{1}{4a} \cdot \frac{2}{3} = \frac{1}{6a}$ .

Work along C→A at constant pressure:

$W_{CA} = \frac{1}{4a}(1-3) = -\frac{1}{2a}$ .

Total work: $W_{total} = \frac{1}{6a} - \frac{1}{2a} = -\frac{1}{3a}$ .

JEE Main 2026 — Physics Thermodynamics

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JEE Main 2026 — Physics Thermodynamics

Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics