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JEE Main 20253 AprilMorning Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 3 April · Morning Shift · Physics · Work, Power and Energy

A piston of mass M is hung from a massless spring whose restoring...

A piston of mass MM is hung from a massless spring whose restoring force law goes as F=−kx3\mathrm{F}=-\mathrm{kx}^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 ' nn ' 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 L0\mathrm{L}_0 to L1\mathrm{L}_1, the total energy delivered by the filament is (Assume spring to be in its natural length before heating)
A piston of mass M is hung from a massless spring whose restoring force law goes as F=-kx^3 , where k is the spring

Options

  1. A3nRTln⁡(L1L0)+2Mg(L1−L0)+k3(L13−L03)3 n R T \ln \left(\frac{L_1}{L_0}\right)+2 \mathrm{Mg}\left(L_1-L_0\right)+\frac{k}{3}\left(L_1^3-L_0^3\right)
  2. BnRTln⁡(L12L02)+Mg2(L1−L0)+k4(L14−L04)n R T \ln \left(\frac{L_1^2}{L_0^2}\right)+\frac{M g}{2}\left(L_1-L_0\right)+\frac{k}{4}\left(L_1^4-L_0^4\right)
  3. CnRTln⁡(L1L0)+Mg(L1−L0)+k4(L14−L04)n R T \ln \left(\frac{L_1}{L_0}\right)+M g\left(L_1-L_0\right)+\frac{k}{4}\left(L_1^4-L_0^4\right)
  4. DnRTln⁡(L1L0)+Mg(L1−L0)+3k4(L14−L04)n R T \ln \left(\frac{L_1}{L_0}\right)+M g\left(L_1-L_0\right)+\frac{3 k}{4}\left(L_1^4-L_0^4\right)

Correct answer

C. nRTln⁡(L1L0)+Mg(L1−L0)+k4(L14−L04)n R T \ln \left(\frac{L_1}{L_0}\right)+M g\left(L_1-L_0\right)+\frac{k}{4}\left(L_1^4-L_0^4\right)

Step-by-step solution

Using WET
Total energy supplied = gravitational potential energy + spring potential energy + work done by gas
<br/>Mg⁡(L1−L0)+∫L0L1k3dx+nRTℓn[L1 A L0 A]+Wext=0 K4[x4]L0L1+Mg⁡(L1−L0)+∫L0L1kx3dx+nRTℓn<br/>\begin{aligned}<br/>& \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<br/>\end{aligned}
[L1L0]+Wext=0k4( L14−L04)+Mg⁡(L1−L0)+nRTℓn[L1 L0]+Wext=0 Wext=k4( L14−L04)+Mg⁡(L1−L0)+nRTℓn[L1 L0]\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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