JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 15)
50 W/m2 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 × 108 m/s) :-
20 × 10–8 N
35 × 10–8 N
10 × 10–8 N
15 × 10–8 N
Explanation
Radiation pressure for 100% reflection = $${{2I} \over C}$$
Radiation pressure for 0% reflection = $${I \over C}$$
Hence, in given case, radiation pressure
= $$\left( {0.25} \right)\left( {{{2I} \over C}} \right) + \left( {0.75} \right)\left( {{I \over C}} \right)$$
$$\left( {1.25} \right)\left( {{I \over C}} \right)$$
$$ \therefore $$ Force = P × (Area) = 20.83 × 10–8 N
Radiation pressure for 0% reflection = $${I \over C}$$
Hence, in given case, radiation pressure
= $$\left( {0.25} \right)\left( {{{2I} \over C}} \right) + \left( {0.75} \right)\left( {{I \over C}} \right)$$
$$\left( {1.25} \right)\left( {{I \over C}} \right)$$
$$ \therefore $$ Force = P × (Area) = 20.83 × 10–8 N
Comments (0)
