Question:
The radiation pressure \(1\ \text{m}\) away from a \(330\ \text{W}\) electric bulb is
The radiation pressure \(1\ \text{m}\) away from a \(330\ \text{W}\) electric bulb is
Show Hint
Radiation pressure for complete absorption is
\[
p=\frac{I}{c}
\]
where
\[
I=\frac{P}{4\pi r^2}
\]
Always calculate intensity first.
Updated On: Jun 25, 2026
- \(1.25\times 10^{-7}\ \text{Pa}\)
- \(8.75\times 10^{-8}\ \text{Pa}\)
- \(5.45\times 10^{-8}\ \text{Pa}\)
- \(8.50\times 10^{-7}\ \text{Pa}\)
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The Correct Option is B
Solution and Explanation
Step 1: Calculate intensity at \(1\ \text{m}\).
Intensity of radiation is \[ I=\frac{P}{4\pi r^2} \] Given, \[ P=330\ \text{W} \] and \[ r=1\ \text{m} \] Thus, \[ I=\frac{330}{4\pi(1)^2} \] Using \[ \pi\approx \frac{22}{7}, \] we get \[ I=\frac{330\times 7}{88} \] \[ I=26.25\ \text{Wm}^{-2} \]
Step 2: Use radiation pressure formula.
Radiation pressure is \[ p=\frac{I}{c} \] where \[ c=3\times 10^8\ \text{ms}^{-1} \] Therefore, \[ p=\frac{26.25}{3\times 10^8} \] \[ p=8.75\times 10^{-8}\ \text{Pa} \]
Step 3: Final conclusion.
Hence, the radiation pressure is \[ \boxed{8.75\times 10^{-8}\ \text{Pa}} \]
Intensity of radiation is \[ I=\frac{P}{4\pi r^2} \] Given, \[ P=330\ \text{W} \] and \[ r=1\ \text{m} \] Thus, \[ I=\frac{330}{4\pi(1)^2} \] Using \[ \pi\approx \frac{22}{7}, \] we get \[ I=\frac{330\times 7}{88} \] \[ I=26.25\ \text{Wm}^{-2} \]
Step 2: Use radiation pressure formula.
Radiation pressure is \[ p=\frac{I}{c} \] where \[ c=3\times 10^8\ \text{ms}^{-1} \] Therefore, \[ p=\frac{26.25}{3\times 10^8} \] \[ p=8.75\times 10^{-8}\ \text{Pa} \]
Step 3: Final conclusion.
Hence, the radiation pressure is \[ \boxed{8.75\times 10^{-8}\ \text{Pa}} \]
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