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Friday, 9 October

JEE Main 2025 · PhysicsMultiple choiceDiagram basedHardMulti-step

JEE Main 3 April 2025, Shift 1, Physics Q35

Question 35 of 75 in this shift, Physics question 10 of 25, Section A.

A piston of mass M 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 '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 L0L_0 to L1L_1, the total energy delivered by the filament is :(Assume spring to be in its natural length before heating)

The figure, in words

Vertical chamber with a piston hung from a spring at the top; piston at height L_0 initially (lower dashed position) and raised to L_1 (upper dashed position, spring end); gas below the piston.
  1. (1)nRTln⁡(L12L02)+Mg2(L1−L0)+k4(L14−L04)nRT\ln\left(\frac{L_1^2}{L_0^2}\right) + \frac{Mg}{2}(L_1 - L_0) + \frac{k}{4}\left(L_1^4 - L_0^4\right)
  2. (2)nRTln⁡(L1L0)+Mg(L1−L0)+k4(L14−L04)nRT\ln\left(\frac{L_1}{L_0}\right) + Mg(L_1 - L_0) + \frac{k}{4}\left(L_1^4 - L_0^4\right)Official answer
  3. (3)3nRTln⁡(L1L0)+2Mg(L1−L0)+k3(L13−L03)3nRT\ln\left(\frac{L_1}{L_0}\right) + 2Mg(L_1 - L_0) + \frac{k}{3}\left(L_1^3 - L_0^3\right)
  4. (4)nRTln⁡(L1L0)+Mg(L1−L0)+3k4(L14−L04)nRT\ln\left(\frac{L_1}{L_0}\right) + Mg(L_1 - L_0) + \frac{3k}{4}\left(L_1^4 - L_0^4\right)

Official answer

Option 2

NTA final key.

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Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. This question needs a figure we do not reproduce; it is described in words above. Lumi analyses 42 of the 174 JEE Main shifts held from 2017 to 2026.