JEE MAIN - Physics (2025 - 3rd April Morning Shift - No. 12)
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A piston of mass $M$ is hung from a massless spring whose restoring force law goes as $F=-k x^3$, where k is the spring constant of appropriate dimension. The piston separates the vertical chamber into two parts, where the bottom part is filled with ' $n$ ' moles of an ideal gas. An external work is done on the gas isothermally (at a constant temperature T) with the help of a heating filament (with negligible volume) mounted in lower part of the chamber, so that the piston goes up from a height $\mathrm{L}_0$ to $\mathrm{L}_1$, the total energy delivered by the filament is:(Assume spring to be in its natural length before heating)
Explanation
Using WET
Total energy supplied $=$ gravitational potential energy + spring potential energy + work done by gas
$$ \begin{aligned} & \mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\int_{\mathrm{L}_0}^{\mathrm{L}_1} \mathrm{kx}^3 \mathrm{dx}+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{~L}_1 \mathrm{~A}}{\mathrm{~L}_0 \mathrm{~A}}\right]+\mathrm{W}_{\text {ext }}=0} \\ & \frac{\mathrm{~K}}{4}\left[\mathrm{x}^4\right]_{\mathrm{L}_0}^{\mathrm{L}_1}+\mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\int_{\mathrm{L}_0}^{\mathrm{L}_1} k x^3 \mathrm{dx}+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_0}\right]+\mathrm{W}_{\text {ext }}=0} \\ & \frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_0}\right]+\mathrm{W}_{\text {ext }}=0} \\ & \mathrm{~W}_{\text {ext }}=\frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\mathrm{Mg}\left(\mathrm{~L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n}\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_0}\right] \end{aligned}$$
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