JEE Main 2026 — Jan 23, Shift 2 — Physics solution

JEE Main2026Jan 23, Shift 2Physics

The internal energy of a monoatomic gas is $3 n R T$ . One mole of helium is kept in a cylinder having internal cross section area of $17$ cm $^{2}$ and fitted with a light movable frictionless piston. The gas is heated slowly by supplying $126$ J heat. If the temperature rises by $4^{\circ} C$ , then the piston will move ______ cm.

(atmospheric pressure $= 1 0^{5}$ Pa)

A
14.5
B
1.55
C
15.5
D
1.45

Solution

The internal-energy formula $U = 3 n R T$ corresponds to $C_{v} = 3 R$ (treating this gas with the given relation, not the usual $\frac{3}{2} R$ for monoatomic). So:

$Δ U = 3 n R Δ T = 3 \cdot 1 \cdot \frac{25}{3} \cdot 4 = 100$ J.

First law: $W = Δ Q - Δ U = 126 - 100 = 26$ J.

The piston is light and atmospheric pressure is constant, so $W = P Δ V = P \cdot A \cdot Δ x$ :

$26 = 1 0^{5} \cdot 17 \times 1 0^{- 4} \cdot Δ x \Rightarrow Δ x = \frac{26}{170} \approx 0.153$ m $= 15.3$ cm.

$∴$ closest option is $15.5$ cm.

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