JEE PYQ: Electromagnetic Waves Question 30
Question 30 - 2019 (09 Apr 2019 Shift 2)
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 ($c = 3 \times 10^{8}$ m/s):
(1) $15 \times 10^{-8}$ N
(2) $20 \times 10^{-8}$ N
(3) $10 \times 10^{-8}$ N
(4) $35 \times 10^{-8}$ N
Show Answer
Answer: (2)
Solution
$F = (1 + r)\frac{IA}{C} = \frac{(1 + 0.25) \times 50 \times 1}{3 \times 10^{8}} = 20 \times 10^{-8}$ N.
BETA
