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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 $\left(\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)$

  1. (1) $15 \times 10^{-8} \mathrm{~N}$

  2. (2) $20 \times 10^{-8} \mathrm{~N}$

  3. (3) $10 \times 10^{-8} \mathrm{~N}$

  4. (4) $35 \times 10^{-8} \mathrm{~N}$


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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