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Asked in JEE Main 28th Jan 1st Shift 2026 · Work as the Area of a Closed Loop
Along the curve P=((V-2)²)/(4a).
The curve runs from V=1 to V=3, where P=1/(4a) at both ends, and the
loop is closed by the horizontal line back at that pressure.
Area under the curve: ∫₁³((V-2)²)/(4a) dV=[((V-2)³)/(12a)]₁³=1/(6a).
Area under the straight leg: 1/(4a)×2=1/(2a).
Enclosed area =1/(2a)-1/(6a)=1/(3a), and the loop is traversed so
that the work is negative: W=-1/(3a).
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