A bulb of power $660 \mathrm{~W}$ radiates uniformly in all directions. The pressure exerted by the…
A bulb of power $660 \mathrm{~W}$ radiates uniformly in all directions. The pressure exerted by the radiation on the surface at a distance of $5 \mathrm{~m}$ is
$5 \times 10^{-8} \mathrm{~Pa}$
$2 \times 10^{-9} \mathrm{~Pa}$
$7 \times 10^{-9} \mathrm{~Pa}$
$\frac{3}{\pi} \times 10^{-8} \mathrm{~Pa}$
Solution
Radiation pressure on a surface at a distance $r$ from a bulb of power $P$ is gives as,
Radiation pressure $=\frac{P}{4 \pi r^2 \cdot c}$(for perfectly absorbing surface)
Here, $P=660 \mathrm{~W}, r=5 \mathrm{~m}, c=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
So, radiation pressure $=\frac{660}{4 \pi \times 5^2 \times 3 \times 10^8}$
$=7 \times 10^{-9} \mathrm{~Pa}$