Option A $\quad \mathrm{P}_{0} \mathrm{~V}_{0}$
Option B $\quad 2 \mathrm{P}_{0} \mathrm{~V}_{0}$
Option C $\frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{2}$
Option D zero
Correct Option is D
Solution:
Work done by the system in the cycle can be given as,
Area under $P-V$ curve & $V$ – axis
For clockwise work, done is positive
For anticlockwise work, done is negative
$\mathrm{W}_{\mathrm{ADEA}}=+\frac{1}{2}\left(2 \mathrm{P}_{0}-\mathrm{P}_{0}\right)\left(2 \mathrm{~V}_{0}-\mathrm{V}_{0}\right)$
$\mathrm{W}_{\mathrm{EBCE}}=-\left(\frac{1}{2}\right)\left(3 \mathrm{P}_{0}-2 \mathrm{P}_{0}\right)\left(2 \mathrm{~V}_{0}-\mathrm{V}_{0}\right)$
$=\frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{2}-\frac{\mathrm{P}_{0} \mathrm{~V}_{0}}{2}=0$