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(C) Given: Intensity $I_0 = 3.3 \, W m^{-2}$,Area $A = 3 	imes 10^{-4} \, m^2$,Angular speed $\omega = 31.4 \, rad/s$.The intensity of light passing

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(C) Given: Intensity $I_0 = 3.3 \, W m^{-2}$,Area $A = 3 	imes 10^{-4} \, m^2$,Angular speed $\omega = 31.4 \, rad/s$.The intensity of light passing through a rotating polariser is given by Malus' Law: $I(t) = I_0 \cos^2(\omega t)$.The power transmitted at any instant is $P(t) = I(t) 	imes A = I_0 A \cos^2(\omega t)$.The energy $E$ transmitted in one revolution (time period $T = \frac{2\pi}{\omega}$) is the integral of power over the period:$E = \int_{0}^{T} P(t) dt = \int_{0}^{2\pi/\omega} I_0 A \cos^2(\omega t) dt$.Using the average value of $\cos^2(	heta)$ over a full cycle,which is $\frac{1}{2}$:$E = I_0 A 	imes \frac{1}{2} 	imes T = I_0 A 	imes \frac{1}{2} 	imes \frac{2\pi}{\omega} = \frac{I_0 A \pi}{\omega}$.Substituting the values:$E = \frac{3.3 	imes 3 	imes 10^{-4} 	imes 3.14}{31.4} = \frac{3.3 	imes 3 	imes 10^{-4} 	imes 3.14}{10 	imes 3.14} = \frac{9.9 	imes 10^{-4}}{10} = 0.99 	imes 10^{-4} \, J$.Rounding to the nearest integer,$E \approx 1 	imes 10^{-4} \, J$.

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