Physics · Thermodynamics

JEE Main 2025 — 7 April, Morning Shift — Question 66

An ideal gas has undergone through the cyclic process as shown in the figure. Work done by the gas in the entire cycle is $\qquad$ $\times 10^{-1} \mathrm{~J}$ . (Take $\pi=3.14$ )

Answer: 314

Numerical answer — enter this value.

Step-by-step solution

$\Delta W=\pi\left(\frac{\Delta v}{2}\right) \cdot\left(\frac{\Delta p}{2}\right)$

\begin{aligned} \Delta w & =\pi \times 100 \times 10^{-6} \times 100 \times 10^{3} & =3.14 \times 10=31.4 \mathrm{~J}=314 \times 10^{-1} \mathrm{~J} \end{aligned}

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 7 April, Morning Shift
Subject: Physics
Chapter: Thermodynamics
Topic: Efficiency of Processes and Miscellaneous Problems