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(C) The power of the light bulb is $P = 100 \ W$. The power converted into light is $P' = 0.66 	imes 100 \ W = 66 \ W$.The radius of the sphere is $r

Vedclass · Accepted Answer

$1.75 	imes 10^{-6} \ N \ m^{-2}$

Vedclass · Answer

(C) The power of the light bulb is $P = 100 \ W$. The power converted into light is $P' = 0.66 	imes 100 \ W = 66 \ W$.The radius of the sphere is $r = 10 \ cm = 0.1 \ m$.The intensity $I$ of the light at the surface of the sphere is given by $I = \frac{P'}{A} = \frac{P'}{4 \pi r^2}$.Substituting the values: $I = \frac{66}{4 	imes 3.14 	imes (0.1)^2} = \frac{66}{4 	imes 3.14 	imes 0.01} = \frac{66}{0.1256} \approx 525.48 \ W/m^2$.For a perfectly absorbing surface,the radiation pressure $P_r$ is given by $P_r = \frac{I}{c}$,where $c = 3 	imes 10^8 \ m/s$ is the speed of light.$P_r = \frac{525.48}{3 	imes 10^8} \approx 1.75 	imes 10^{-6} \ N \ m^{-2}$.

$A$ light bulb of power $100 \ W$ is placed at the centre of a hollow sphere of radius $10 \ cm$. If $66\%$ of the energy is converted into light,then the pressure exerted by the light on the surface of the sphere will be (Assume the surface of the sphere to be perfectly absorbing).

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