JEE Main 2025 — Work Power Energy Question with Solution
JEE Main 2025 (3 Apr Shift 1)
Question
A piston of mass is hung from a massless spring whose restoring force law goes as , 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 ' ' 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 to , the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
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Show full solutionCorrect option: C
Correct answer
C
Step-by-step explanation
Using WET
Total energy supplied gravitational potential energy + spring potential energy + work done by gas
$\begin{aligned}
& \operatorname{Mg} \quad\left(L_1-L_0\right)+\int_{L_0}^{L_1} k^3 d x+n R T \ell n \\ & {\left[\frac{L_1 \mathrm{~A}}{\mathrm{~L}_0 \mathrm{~A}}\right]+\mathrm{W}_{\mathrm{ext}}=0} \\ & \frac{\mathrm{~K}}{4}\left[\mathrm{x}^4\right]_{\mathrm{L}_0}^{\mathrm{L}_1}+\operatorname{Mg}\left(\mathrm{L}_1-\mathrm{L}_0\right)+\int_{\mathrm{L}_0}^{\mathrm{L}_1} k x^3 d x+n R T \ell n
\end{aligned}\begin{aligned} & {\left[\frac{L_1}{L_0}\right]+\mathrm{W}_{\mathrm{ext}}=0} \\ & \frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\operatorname{Mg} \quad\left(\mathrm{L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{L}_1}{\mathrm{~L}_0}\right]+\mathrm{W}_{\mathrm{ext}}=0} \\ & \mathrm{~W}_{\mathrm{ext}}=\frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\operatorname{Mg}\left(\mathrm{L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n}\left[\frac{\mathrm{L}_1}{\mathrm{~L}_0}\right]\end{aligned}$
Total energy supplied gravitational potential energy + spring potential energy + work done by gas
$\begin{aligned}
& \operatorname{Mg} \quad\left(L_1-L_0\right)+\int_{L_0}^{L_1} k^3 d x+n R T \ell n \\ & {\left[\frac{L_1 \mathrm{~A}}{\mathrm{~L}_0 \mathrm{~A}}\right]+\mathrm{W}_{\mathrm{ext}}=0} \\ & \frac{\mathrm{~K}}{4}\left[\mathrm{x}^4\right]_{\mathrm{L}_0}^{\mathrm{L}_1}+\operatorname{Mg}\left(\mathrm{L}_1-\mathrm{L}_0\right)+\int_{\mathrm{L}_0}^{\mathrm{L}_1} k x^3 d x+n R T \ell n
\end{aligned}\begin{aligned} & {\left[\frac{L_1}{L_0}\right]+\mathrm{W}_{\mathrm{ext}}=0} \\ & \frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\operatorname{Mg} \quad\left(\mathrm{L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n} \\ & {\left[\frac{\mathrm{L}_1}{\mathrm{~L}_0}\right]+\mathrm{W}_{\mathrm{ext}}=0} \\ & \mathrm{~W}_{\mathrm{ext}}=\frac{\mathrm{k}}{4}\left(\mathrm{~L}_1^4-\mathrm{L}_0^4\right)+\operatorname{Mg}\left(\mathrm{L}_1-\mathrm{L}_0\right)+\mathrm{nRT} \ell \mathrm{n}\left[\frac{\mathrm{L}_1}{\mathrm{~L}_0}\right]\end{aligned}$
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This is a previous-year question from JEE Main 2025, covering the Work Power Energy chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.