Sunlight of intensity $50 {Wm}^{- 2}$ 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 $1 m^{2}$ surface area will be close to $(c = 3 \times 10^{8} {ms}^{- 1})$

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$21 \times 10^{- 8} N$

$35 \times 10^{- 8} N$

$15 \times 10^{- 8} N$

$10 \times 10^{- 8} N$

answer is A.

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Detailed Solution

SInce 25% energy is getting reflected, 25% extra momentum will be delivered to the panel

$\Rightarrow F_{n} = \frac{I A}{c} (1 + 0.25) = \frac{(1.25) \times (1) \times (50)}{(3 \times 10^{8})} \approx 21 \times 10^{- 8} N$

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