JEE Main 2026 — Jan 28, Shift 1 — Physics solution

JEE Main2026Jan 28, Shift 1Physics

In the following $p$ - $V$ diagram the equation of state along the curved path is given by $(V - 2)^{3} = 4 a p$ where $a$ is a constant. The total work done in the closed path is

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

Solution

$w = Area of parabola$

$= \dfrac{2}{3} \left( \text{Area of rectangle AC(3)(1)A} \right)$

$= \dfrac{2}{3}P_0(3-1) = \dfrac{4P_0}{3}$

$When V = 1$

$(1-2)^2 = 4aP_0$

$P_0 = \dfrac{1}{4a}$

$w = \dfrac{4}{3}P_0 = \dfrac{4}{3}\cdot\dfrac{1}{4a} = \dfrac{1}{3a}$

$W_{\text{gas}} = -\dfrac{1}{3a}$

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