Physics Circle home Student Login

Home›Chapters›Electromagnetic Waves PYQs›Momentum & Radiation Pressure›Question 11

JEE Main 2019 (9 Apr, Shift 2) · Momentum & Radiation Pressure

Momentum & Radiation Pressure question from JEE Main 2019 (9 Apr, Shift 2)

Q11JEE Main 2019 · 9 APR · SHIFT 2Medium

Momentum & Radiation Pressure

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) :-

Show answer and solution

Answer: (a)

  1. Absorbed part gives F=IAcF = \dfrac{IA}{c} per unit fraction, reflected part gives 2IAc\dfrac{2IA}{c}
  2. F=IAc(0.75+2×0.25)=50×1×1.253×108F = \dfrac{IA}{c}(0.75 + 2 \times 0.25) = \dfrac{50 \times 1 \times 1.25}{3 \times 10^8}
  3. F≈20.8×10−8 N≈20×10−8 NF \approx 20.8 \times 10^{-8}\ \text{N} \approx 20 \times 10^{-8}\ \text{N}, so the answer is (a).
Report a mistake

More questions on Momentum & Radiation Pressure