Q. A 100 W light bulb is placed at the centre of a spherical chamber of radius 0.10m. Assume that 66% of the energy supplied to the bulb is converted into light and that the surface of chamber is perfectly absorbing. The pressure exerted by the light on the surface of the chamber is
A) $0.87 \times 10^{- 6}$Pa
B) $1.77 \times 10^{- 6}$Pa
C) $3.50 \times 10^{- 6}$Pa
D) None of these
Explanation
(b) Light falling per second on the surface of sphere
$E = \frac{66}{100} \times 100 = 66$ W
Momentum of the light falling per second on the surface of sphere $= \frac{E}{c}$
Momentum of the reflected light $= 0;\ $as the light is completely absorbed.
Force exerted by light, $F = \frac{E}{c} - 0 = \frac{E}{c}$
Pressure on surface, $P = \frac{F}{4\pi r^{2}} = \frac{E/c}{4\pi r^{2}}$
$= \frac{66/(3 \times 10^{8})}{4 \times (22/7) \times (0.10)^{2}} = 1.75 \times 10^{- 6}Pa$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.