Question:
A bulb of power \(660\ \text{W}\) radiates uniformly in all directions. The pressure exerted by the radiation on a surface at a distance of \(5\ \text{m}\) is
A bulb of power \(660\ \text{W}\) radiates uniformly in all directions. The pressure exerted by the radiation on a surface at a distance of \(5\ \text{m}\) is
Show Hint
For a source radiating uniformly in all directions,
\[
I=\frac{P}{4\pi r^2}.
\]
Radiation pressure on a perfectly absorbing surface is
\[
p=\frac{I}{c},
\]
while for a perfectly reflecting surface it is
\[
p=\frac{2I}{c}.
\]
Updated On: Jun 26, 2026
- \(5\times10^{-8}\ \text{Pa}\)
- \(2\times10^{-9}\ \text{Pa}\)
- \(7\times10^{-9}\ \text{Pa}\)
- \(\dfrac{3}{\pi}\times10^{-8}\ \text{Pa}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Calculate the intensity of radiation at a distance of \(5\ \text{m}\).
Since the bulb radiates uniformly in all directions, the intensity at distance \(r\) is \[ I=\frac{P}{4\pi r^2} \] Given, \[ P=660\ \text{W} \] and \[ r=5\ \text{m} \] Therefore, \[ I=\frac{660}{4\pi(5)^2} \] \[ I=\frac{660}{100\pi} \] \[ I=\frac{6.6}{\pi} \] \[ I\approx 2.1\ \text{W m}^{-2} \]
Step 2: Use the relation between radiation pressure and intensity.
For complete absorption of radiation, \[ p=\frac{I}{c} \] where \[ c=3\times10^8\ \text{m s}^{-1} \] Substituting the value of intensity, \[ p=\frac{2.1}{3\times10^8} \] \[ p=0.7\times10^{-8} \] \[ p=7\times10^{-9}\ \text{Pa} \]
Step 3: Final conclusion.
Hence, the radiation pressure exerted on the surface is \[ \boxed{7\times10^{-9}\ \text{Pa}} \]
Since the bulb radiates uniformly in all directions, the intensity at distance \(r\) is \[ I=\frac{P}{4\pi r^2} \] Given, \[ P=660\ \text{W} \] and \[ r=5\ \text{m} \] Therefore, \[ I=\frac{660}{4\pi(5)^2} \] \[ I=\frac{660}{100\pi} \] \[ I=\frac{6.6}{\pi} \] \[ I\approx 2.1\ \text{W m}^{-2} \]
Step 2: Use the relation between radiation pressure and intensity.
For complete absorption of radiation, \[ p=\frac{I}{c} \] where \[ c=3\times10^8\ \text{m s}^{-1} \] Substituting the value of intensity, \[ p=\frac{2.1}{3\times10^8} \] \[ p=0.7\times10^{-8} \] \[ p=7\times10^{-9}\ \text{Pa} \]
Step 3: Final conclusion.
Hence, the radiation pressure exerted on the surface is \[ \boxed{7\times10^{-9}\ \text{Pa}} \]
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