Q 11-11-156JEE MainJEE Main 2019 (12 Jan, Shift 1)Easy
For the given cyclic process $CAB$ as shown for a gas, the work done is:
Answer: (A) $10$ J
The cycle is the triangle C(1, 6) → A(5, 6) → B(5, 1) → C, traversed clockwise on the $p$–$V$ diagram, so the net work done by the gas is positive and equals the enclosed area:
$$W = \frac12\times(5 - 1)\times(6 - 1) = 10\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics