JEE Main 2021 · 20 July · morning · Numerical answer · +4 / −0
In the reported figure, heat energy absorbed by a system in going through a cyclic process is ___________ πJ.
Answer and solution
Answer 100
Solution
Physics sir · 3 steps
In a full cycle Δ U = 0, so the heat absorbed equals the area enclosed by the loop.
- Full cycle. The gas returns to its starting state, so Δ U = 0. The first law Q = Δ U + W then gives Q = W. The net work in a cycle is the area enclosed by the loop on the P-V graph.
- Read the semi-axes. Pressure runs from 20 to 40 kPa, so the half-height is 10 kPa = 10⁴ Pa. Volume runs from 20 to 40 litre, so the half-width is 10 L = 10⁻² m³. The two axes carry different units, so treat the loop as an ellipse with area π a b.
W = π a b = π × (10⁴ Pa) × (10⁻² m³) = 100π J
- So. The heat absorbed in one cycle is 100π J, so the answer is 100.
The common slip: 400. taking the full ranges (20 kPa and 20 L) as the radii gives 400π J. The radius is half of each range.
The idea to keep. For any cyclic process, heat absorbed equals net work, which is the loop's area on the P-V graph. For a circular loop, use π × (half-range of P) × (half-range of V) in SI units.
Answer from the NTA final answer key. Solution by PrepWiser.