JEE Main 2025 · 3 April · morning · Single correct · +4 / −1
The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is
- (a) 3 × 10⁻⁸ Pascals
- (b) 0
- (c) 1.5 × 10⁻⁸ Pascals
- (d) 6 × 10⁻⁸ Pascals
Answer and solution
Answer (d) 6 × 10⁻⁸ Pascals
Solution
Physics sir · 3 steps
Spread the 450 W over a sphere of radius 2 m, then double the momentum push because the surface reflects.
- Intensity at 2 m. A point source spreads its power evenly over a sphere, so I = P/(4π r²) = 450/(4π × 2²) = 450/16π ≈ 8.95 W m⁻².
- Reflection doubles it. Light carries momentum. An absorbing surface takes in the momentum once, giving pressure I/c. A perfect reflector sends the light straight back, so the momentum reverses and the change is twice as large.
p = 2I/c = (2 × 8.95)/(3 × 10⁸) ≈ 5.97 × 10⁻⁸ Pa
- So. The radiation pressure is about 6 × 10⁻⁸ Pa, which is option (d).
The idea to keep. Radiation pressure is I/c on a perfect absorber and 2I/c on a perfect reflector. For a point source, get I from P/4π r² first.
Answer from the NTA final answer key. Solution by PrepWiser.