1. Given data

• Intensity of incident plane polarised light: $I_0 = 3.3\ \text{W m}^{-2}$
• Area of polariser: $A = 3 \times 10^{-4}\ \text{m}^2$
• Angular speed of rotation: $\omega = 31.4\ \text{rad s}^{-1}$

We need the energy transmitted in one complete revolution.

2. Instantaneous transmitted intensity

Since the incident light is plane polarised and the polariser rotates, the transmitted intensity is given by Malus' law:

$I = I_0 \cos^2 \theta$

where $\theta = \omega t$

So,

$I(t) = I_0 \cos^2(\omega t)$

3. Instantaneous transmitted power

Power transmitted through the polariser is

$P(t) = I(t) A = I_0 A \cos^2(\omega t)$

Substitute values:

$P(t) = 3.3 \times 3\times 10^{-4} \cos^2(\omega t)$

$P(t) = 9.9 \times 10^{-4} \cos^2(\omega t)\ \text{W}$

4. Time for one revolution

For one full revolution,

$T = \frac{2\pi}{\omega} = \frac{2\pi}{31.4} \approx 0.2\ \text{s}$

5. Energy transmitted in one revolution

Energy is

$E = \int_0^T P(t)\,dt = I_0 A \int_0^T \cos^2(\omega t)\,dt$

Over one full cycle, average value of $\cos^2$ is

$\langle \cos^2 \theta \rangle = \frac{1}{2}$

Therefore,

$E = I_0 A \cdot T \cdot \frac{1}{2}$

Substitute:

$E = 3.3 \times 3\times 10^{-4} \times 0.2 \times \frac{1}{2}$

$E = 9.9\times 10^{-4} \times 0.1$

$E = 9.9 \times 10^{-5}\ \text{J}$

This is approximately

$E \approx 1.0 \times 10^{-4}\ \text{J}$

6. Option matching

The closest option is:

B: $1.0 \times 10^{-4}\ \text{J}$

7. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

So they agree.