Physics Q35 – JEE-MAIN 2025

Home Physics ThermodynamicsQuestion #528

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

L_{0}

L_{1}

, the total energy delivered by the filament is :(Assume spring to be in its natural length before heating)

Clue 1 of 3: Energy Conservation

The total energy delivered by the filament is converted into various forms of energy within the system, including work done by the gas and changes in potential energies.

🥷

Ninja Strategy•Term-by-Term Verification

Calculate each component of the total energy (work done by gas, change in gravitational potential energy, change in spring potential energy) separately and compare them term-by-term with the given options to quickly eliminate incorrect choices.

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Step 1: Calculate Work Done by Gas✦ Active

For an ideal gas undergoing an isothermal process, the work done by the gas is given by:

W_{g a s} = n R T \ln (\frac{V_{1}}{V_{0}})

Since the volume is proportional to the height of the piston (

V = A L

), we have:

W_{g a s} = n R T \ln (\frac{L_{1}}{L_{0}})

Step 2: Calculate Changes in Potential Energy○ Expand

The change in gravitational potential energy of the piston of mass $M$ as it moves from height $L_{0}$ to $L_{1}$ is:

Δ P E_{p i s t o n} = M g (L_{1} - L_{0})

The restoring force of the spring is $F = - k x^{3}$ . The potential energy stored in such a spring is $U_{s} = - \int F d x = \int k x^{3} d x = \frac{1}{4} k x^{4}$ . Given the options, we interpret $x$ as the absolute height $L$ of the piston, and the change in spring potential energy is:

Δ U_{s p r i n g} = U_{s} (L_{1}) - U_{s} (L_{0}) = \frac{1}{4} k L_{1}^{4} - \frac{1}{4} k L_{0}^{4} = \frac{k}{4} (L_{1}^{4} - L_{0}^{4})

💡 Teacher's Secret Hint

Note the interpretation of the spring potential energy term $U_{s} = \frac{1}{4} k x^{4}$ where $x$ is taken as the height $L$ , which is consistent with the options provided, despite the 'natural length at $L_{0}$ ' statement.

Step 3: Calculate Total Energy Delivered by Filament○ Expand

The total energy delivered by the filament is the sum of the work done by the gas and the changes in potential energies of the piston and spring. This represents the total energy transferred and stored in the system due to the filament's action:

Q_{t o t a l} = W_{g a s} + Δ P E_{p i s t o n} + Δ U_{s p r i n g}

Substituting the calculated terms:

Q_{t o t a l} = n R T \ln (\frac{L_{1}}{L_{0}}) + M g (L_{1} - L_{0}) + \frac{k}{4} (L_{1}^{4} - L_{0}^{4})

This expression matches option 2.

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Physics Question 35 – JEE-MAIN 2025

Clue 1 of 3: Energy Conservation

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