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)

 

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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