50 W/m2 energy density of sunlight is normally incident on the surface

1.	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):1. 15 × 10-8 N2. 20 × 10 -8 N3. 10 × 10-8 N4. 35 × 10-8 N
Answer» Correct Answer - Option 2 : 20 × 10 -8 N Concept: Force exerted on the surface is given by the following formula \(F = \left( {1 + r} \right)\frac{{IA}}{C}\) where r is the Reflected radiation A is the Surface area c is the speed of the sound Calculation: Given, Energy density, I = 50 W/m2 Speed of the sound, c = 3 × 108 m/s Surface area, A = 1 m2 Reflected radiation, r = 25% = 25/100 = 0.25 Force on the surface (25% is reflecting back and the remaining 75% has been absorbed from the surface) Force exerted on the surface is given by, \(F = \left( {1 + r} \right)\frac{{IA}}{C}\) \(= \frac{{\left( {1 + 0.25} \right) \times 50 \times 1}}{{3 \times {{10}^8}}}\) \(= 1.25 \times \frac{{50}}{{3 \times {{10}^8}}}\) \(= \frac{{125}}{{100}} \times \frac{{50}}{{3 \times {{10}^8}}}\) \(= \frac{{125}}{6} \times {10^{ - 8}}\) ≃ 20 × 10-8 N Thus, the force exerted on 1m2 surface area will be close to 20 × 10-8 N

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):1. 15 × 10-8 N2. 20 × 10 -8 N3. 10 × 10-8 N4. 35 × 10-8 N

Answer» Correct Answer - Option 2 : 20 × 10 -8 N

Concept:

Force exerted on the surface is given by the following formula

\(F = \left( {1 + r} \right)\frac{{IA}}{C}\)

where

r is the Reflected radiation

A is the Surface area

c is the speed of the sound

Calculation:

Given,

Energy density, I = 50 W/m2

Speed of the sound, c = 3 × 108 m/s

Surface area, A = 1 m2

Reflected radiation, r = 25% = 25/100 = 0.25

Force on the surface (25% is reflecting back and the remaining 75% has been absorbed from the surface)

Force exerted on the surface is given by,

\(F = \left( {1 + r} \right)\frac{{IA}}{C}\)

\(= \frac{{\left( {1 + 0.25} \right) \times 50 \times 1}}{{3 \times {{10}^8}}}\)

\(= 1.25 \times \frac{{50}}{{3 \times {{10}^8}}}\)

\(= \frac{{125}}{{100}} \times \frac{{50}}{{3 \times {{10}^8}}}\)

\(= \frac{{125}}{6} \times {10^{ - 8}}\)

≃ 20 × 10-8 N

Thus, the force exerted on 1m2 surface area will be close to 20 × 10-8 N