$50 \:\:W/m^{2}$ energy density of sunlight is normally incident on the surface of a solar panel . Some part of incident energy $(25 \:^{o}/_{o})$ is reflected from the surface snd the rest is absorbed . The force exerted on $1\:m^{2}$ surface area will be close to $(c=3\times10^{8}\:m/s)$ :

Option 1)
$15\times10^{-8}\:\:N$

Option 2)
$20\times10^{-8}\:\:N$

Option 3)
$10\times10^{-8}\:\:N$

Option 4)
$35\times10^{-8}\:\:N$

Answers (1)

S solutionqc

Energy density = $50 \:\:W/m^{2}$

Total energy / time for $1\:m^{2}$ area = $50\times1=50\:\:W$

$T.E.=50\:\:W\\\\F=0.25\left ( 2\times\frac{T.E.}{C} \right )+0.75\left ( \frac{T.E.}{C} \right )$

$\Rightarrow 1.25\left ( \frac{T.E.}{C} \right )\\\\F\Rightarrow 1.25\times\frac{50}{3\times10^{8}}=20\times10^{-8}\:\:N$

Option 1)

$15\times10^{-8}\:\:N$

Option 2)

$20\times10^{-8}\:\:N$

Option 3)

$10\times10^{-8}\:\:N$

Option 4)

$35\times10^{-8}\:\:N$

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energy density of sunlight is normally incident on the surface of a solar panel . Some part of incident energy is reflected from the surface snd the rest is absorbed . The force exerted on surface area will be close to : Option 1) Option 2) Option 3) Option 4)

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