Q.
Work done by the system in closed path $ABCA$, is
A) Zero
B) $\left( V_{1} - V_{2} \right)(p_{1} - p_{2})$
C) $\frac{\left( p_{2} - p_{1} \right)(V_{2} - V_{1})}{2}$
D) $\frac{\left( p_{1} + p_{1} \right)(V_{2} - V_{1})}{2}$
Explanation
(c) Work done = area of $\mathrm{\Delta}ABC$
$= \frac{AB \times BC}{2} = \frac{\left( p_{2} - p_{1} \right)(V_{2} - V_{1})}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.