JEE Main 2024 · 30 January · evening · Single correct · +4 / −1
If the total energy transferred to a surface in time t is 6.48 × 10⁵ J, then the magnitude of the total momentum delivered to this surface for complete absorption will be:
- (a) 2.16 × 10⁻³ kg m / s
- (b) 2.46 × 10⁻³ kg m / s
- (c) 1.58 × 10⁻³ kg m / s
- (d) 4.32 × 10⁻³ kg m / s
Answer and solution
Answer (a) 2.16 × 10⁻³ kg m / s
Solution
Physics sir · 3 steps
Light carries momentum U/c; if the surface absorbs it all, that is exactly what it receives.
- Momentum of radiation. An electromagnetic wave carrying energy U also carries momentum p = U/c. When the surface absorbs the wave completely, all of this momentum is handed over to the surface.
p = U/c = (6.48 × 10⁵ J)/(3 × 10⁸ m s⁻¹)
- Put in numbers. 6.48 ÷ 3 = 2.16, and 10⁵ ÷ 10⁸ = 10⁻³. So p = 2.16 × 10⁻³ kg m s⁻¹.
- So. The total momentum delivered is 2.16 × 10⁻³ kg m s⁻¹, which is option (a).
The idea to keep. Momentum delivered by radiation is U/c for complete absorption and 2U/c for complete reflection, because a reflected wave reverses its momentum.
Answer from the NTA final answer key. Solution by PrepWiser.