Question

Work done in the cyclic process shown in figure is

• A. $4P_0{V}_0$
• B. $-4P_0{V}_0$
• C. $-\frac{22}{7}{P}_0{V}_0$
• D. $-13P_0{V}_0$

Answer 1

Answer: (C)

Cyclic process is anticlockwise then
Work done $=-($ Area of $P-V$ graph$)$
$W=-\pi R_1{R}_2$
$=-\pi \left(\frac{3P_0-P_0}{2}\right)\times \left(\frac{3V_0-V_0}{2}\right)$
$=\frac{-22}{7}{P}_0{V}_0$