One mole of ideal monoatomic gas is taken along a cyclic process as shown in the figure. Process $1 \to 2$ shown is 1/4^th part of a dotted circle as shown. The efficiency of the cycle to nearest integer is.. (Use π = 3.14 )

$\begin{array}{l}{\mathrm{W}}_{\text{net }}=\left(2{\mathrm{P}}_0{\mathrm{V}}_0\right)-\left({\mathrm{P}}_0{\mathrm{V}}_0\right)-\frac{{\mathrm{\pi P}}_0{\mathrm{V}}_0}{4}\\ {\mathrm{W}}_{\text{net }}={\mathrm{P}}_0{\mathrm{V}}_0-\frac{{\mathrm{\pi P}}_0{\mathrm{V}}_0}{4}=(4-\mathrm{\pi })\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{4};\text{ Put }\mathrm{\pi }=3.14\\ {\mathrm{W}}_{\text{net }}=\frac{0.86}{4}{\mathrm{P}}_0{\mathrm{V}}_0=(0.22)\left({\mathrm{P}}_0{\mathrm{V}}_0\right)\end{array}$
Now,
$\left.\begin{array}{l}{\mathrm{T}}_1=\frac{{\mathrm{P}}_0{\mathrm{V}}_0}{\mathrm{R}}\\ {\mathrm{T}}_2=\frac{4{\mathrm{P}}_0{\mathrm{V}}_0}{\mathrm{R}}\\ {\mathrm{T}}_3=\frac{2{\mathrm{P}}_0{\mathrm{V}}_0}{\mathrm{R}}\end{array}\right\}\mathrm{Thus},\begin{array}{r}{\mathrm{\Delta U}}_{1\to 2}=1\times \frac{3\mathrm{R}}{2}\left[{\mathrm{T}}_2-{\mathrm{T}}_1\right]\\ {\mathrm{\Delta U}}_{2\to 3}=1\times \frac{3\mathrm{R}}{2}\left[{\mathrm{T}}_3-{\mathrm{T}}_2\right]\\ {\mathrm{\Delta U}}_{3\to 1}=1\times \frac{3\mathrm{R}}{2}\left[{\mathrm{T}}_1-{\mathrm{T}}_3\right]\end{array}$
$\begin{array}{l}{\mathrm{\Delta Q}}_{1\to 2}=(4.5)\left({\mathrm{P}}_0{\mathrm{V}}_0\right)+(1.22)\left({\mathrm{P}}_0{\mathrm{V}}_0\right)=(5.72)\left({\mathrm{P}}_0{\mathrm{V}}_0\right)\\ {\mathrm{\Delta Q}}_{2\to 3}=-3{\mathrm{P}}_0{\mathrm{V}}_0+0=-3\left({\mathrm{P}}_0{\mathrm{V}}_0\right)\\ {\mathrm{\Delta Q}}_{3\to 1}=-1.5\left({\mathrm{P}}_0{\mathrm{V}}_0\right)-\left({\mathrm{P}}_0{\mathrm{V}}_0\right)=-2.5\left({\mathrm{P}}_0{\mathrm{V}}_0\right)\end{array}$
Thus efficiency $\mathrm{\eta }=\frac{{\mathrm{W}}_{\text{net }}}{+\text{ veheat }}$
$\mathrm{\eta }=\frac{0.22\left({\mathrm{P}}_0{\mathrm{v}}_0\right)}{(5.72)\left({\mathrm{P}}_0{\mathrm{v}}_0\right)}=0.04$
Thus efficiency is 4%