JEE MAIN - Physics (2024 - 5th April Morning Shift - No. 8)
The heat absorbed by a system in going through the given cyclic process is :
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61.6 J
431.2 J
19.6 J
616 J
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$$\begin{aligned}
\Delta Q & =\Delta U+w \\
& =\pi(140)^2 \times 10^3 \times 10^{-6} \\
& =61.6 \mathrm{~J}
\end{aligned}$$