50Wm−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 1m2 surface area will be close to- (c=3×108ms−1)
A
15×108N
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B
10×108N
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C
35×108N
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D
20×108N
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Solution
The correct option is D20×108N For a partially reflecting and absorbing surface,
F=(1+r)IAC =(1+0.25)×50×13×108 ≃20×10−8N
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Hence, (B) is the correct answer.