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) :

Question

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) : (A) 15 × 10-8 N (B) 35 × 10-8 N (C) 10 × 10-8 N (D) 20 × 10-8 N

Tardigrade Admin · Accepted Answer

Force on the surface (25% reflecting and rest absorbing) F = (25/100) ((2I/C) )+ (75/100) ((I/C)) = (125/100)((I/C)) =(125/100) ×((50/3×108) ) = 20.83 ×10-8 N

Q. $50 W / m^{2}$ 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 $1 m^{2}$ surface area will be close to $(c = 3 \times 1 0^{8} m / s)$ :

$15 \times 1 0^{- 8} N$

$35 \times 1 0^{- 8} N$

$10 \times 1 0^{- 8} N$

$20 \times 1 0^{- 8} N$

Force on the surface (25% reflecting and rest absorbing)
$F = \frac{25}{100} (\frac{2 I}{C}) + \frac{75}{100} (\frac{I}{C}) = \frac{125}{100} (\frac{I}{C})$
$= \frac{125}{100} \times (\frac{50}{3 \times 1 0 ^{8}}) = 20.83 \times 1 0^{- 8} N$

Q. 50W/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=3×108m/s) :

15×10−8N

35×10−8N

10×10−8N

20×10−8N