Step 1: Understand radiation pressure.
Radiation pressure is the pressure exerted by electromagnetic radiation on a surface. For a perfect absorber, it equals the intensity divided by the speed of light: \[ P_{rad} = \frac{I}{c} \] where $ I $ is the intensity of the radiation at that point.
Step 2: Model the bulb as a point source.
A bulb of power $ P = 330\text{ W} $ radiates uniformly in all directions. At distance $ r = 1\text{ m} $, the radiation is spread over a sphere of area $ 4\pi r^2 $, so the intensity is: \[ I = \frac{P}{4\pi r^2} = \frac{330}{4\pi(1)^2} \]
Step 3: Calculate the intensity numerically.
Using $ \pi \approx \frac{22}{7} $: \[ 4\pi \approx \frac{88}{7} \] \[ I = \frac{330 \times 7}{88} = \frac{2310}{88} = 26.25\text{ W m}^{-2} \]
Step 4: Compute radiation pressure.
With $ c = 3\times10^8\text{ m s}^{-1} $: \[ P_{rad} = \frac{26.25}{3\times10^8} = 8.75\times10^{-8}\text{ Pa} \]
Step 5: Dimensional check.
Intensity has units $ \text{W m}^{-2} = \text{J s}^{-1} \text{m}^{-2} $, divided by $ c\ [\text{m s}^{-1}] $ gives $ \text{J m}^{-3} = \text{N m}^{-2} = \text{Pa} $. Units check out.
Step 6: Final answer.
\[ \boxed{P_{rad} = 8.75\times10^{-8}\text{ Pa}} \]