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JEE Main 2025 (3 Apr, Shift 1) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2025 (3 Apr, Shift 1)

Q90JEE Main 2025 · 3 APR · SHIFT 1Difficult

Thermodynamic Processes

Figure

A piston of mass MM is hung from a massless spring whose restoring force law goes as F=−kx3F = -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 L0L_0 to L1L_1, the total energy delivered by the filament is:(Assume spring to be in its natural length before heating)

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Answer: (b)

  1. Isothermal gas: ΔU=0\Delta U = 0, and gas work =nRTln⁡V1V0=nRTln⁡L1L0= nRT\ln\frac{V_1}{V_0} = nRT\ln\frac{L_1}{L_0} (area constant).
  2. Piston rises by L1−L0L_1 - L_0: gain in PE =Mg(L1−L0)= Mg(L_1 - L_0).
  3. Spring energy for F=−kx3F = -kx^3: ∫kx3 dx=k4x4\int kx^3\,dx = \frac{k}{4}x^4, written in the official answer as k4(L14−L04)\frac{k}{4}(L_1^4 - L_0^4).
  4. Adding the three terms gives option (b).
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