A beam of plane polarized light falls normally on a polarizer of cross sectional area  $3\times {10}^{-4}{m}^2$. Flux of energy of incident ray is ${10}^{-3}$ W. The polarizer rotates with an angular frequency of 31.4 rad/sec. The energy of light passing through the polarizer per revolution will be

Using Malus law,  $I=I_0{\cos }^2\theta$
As here polariser is rotating i.e. all the values of q  are possible. 
$I_{av}=\frac{1}{2\pi }{\int }_0^{2\pi }Id\theta =\frac{1}{2\pi }{\int }_0^{2\pi }{I}_0{\cos }^2\theta d\theta$
On integration we get  $I_{av}=\frac{I_0}{2}$
where  $I_0=\frac{\text{Energy}}{\text{Area }\times \text{ Time}}=\frac{P}{A}=\frac{{10}^{-3}}{3\times {10}^{-4}}=\frac{10}{3}\frac{Watt}{m^2}$
\   $I_{av}=\frac{1}{2}\times \frac{10}{3}=\frac{5}{3}Watt$
and Time period  $T=\frac{2\pi }{\omega }=\frac{2\times 3.14}{31.4}=\frac{1}{5}sec$
\ Energy of light passing through the polariser per revolution  $=I_{av}\times \text{Area}\times T=\frac{5}{3}\times 3\times {10}^{-4}\times \frac{1}{5}={10}^{-4}J.$