Question:
$50 \mathrm{~W} / \mathrm{m}^{2}$ energy density of sunlight is normally incident on the surface of a solar panel. Some part of incident energy $(25 \%)$ is reflected from the surface and the rest is absorbed. The force exerted on $1 \mathrm{~m}^{2}$ surface area will be close to $\left(\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$
Correct Option: , 2
Solution:
(2) $F=(1+r) \frac{I A}{C}$
$=\frac{(1+0.25) \times 50 \times 1}{3 \times 10^{8}}$
$\simeq 20 \times 10^{-8} \mathrm{~N}$
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