One mole of ideal monoatomic gas is taken along a cyclic process as shown in the figure. Process $\text{1}\to \text{2}$ shown is 1/4th part of a circle as shown by dotted line process $\text{2}\to \text{3}$  is isochoric while $\text{3}\to \text{1}$ is isobaric. Find the percentage efficiency of process (closest integer value).

 $W_{net}=\left(2P_0{v}_0\right)-\left(P_0{v}_0\right)-\frac{\pi P_0{v}_0}{4}$
$W_{net}=P_0{v}_0-\frac{\pi P_0{v}_0}{4}=(4-\pi )\frac{P_0{v}_0}{4}$ ; Put  $\pi =\text{ 3}.\text{14}$
 $W_{net}=\frac{0.86}{4}{P}_0{v}_0=\left(0.22\right)\left(P_0{v}_0\right)$
Now
  $\begin{array}{l}{T}_1=\frac{P_0{v}_0}{R}\\ \\ T_2=\frac{4P_0{v}_0}{R}\\ \\ T_3=\frac{2P_0{v}_0}{R}\end{array}\}$ Thus,    $\begin{array}{l}\Delta U_{1\to 2}=1\times \frac{3R}{2}[T_2-T_1]\\ \\ \Delta U_{2\to 3}=1\times \frac{3R}{2}[T_3-T_2]\\ \\ \Delta U_{3\to 1}=1\times \frac{3R}{2}[T_1-T_3]\end{array}$
 $\Delta Q_{1\to 2}=\left(4.5\right)\left(P_0{v}_0\right)+\left(1.22\right)\left(P_0{v}_0\right)=\left(5.72\right)\left(P_0{v}_0\right)$
 $\Delta Q_{2\to 3}=-3P_0{v}_0+0=-3\left(P_0{v}_0\right)$
 $\Delta Q_{3\to 1}=-1.5\left(P_0{v}_0\right)-\left(P_0{v}_0\right)=-2.5\left(P_0{v}_0\right)$
Thus efficiency  $\eta =\frac{W_{net}}{+veheat}$
 $\eta =\frac{0.22\left(P_0{v}_0\right)}{\left(5.72\right)\left(P_0{v}_0\right)}=0.04$
Thus efficiency is 4%