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Physics · Atomic Physics

JEE Main 2025 — 3 April, Morning Shift — Question 48

The radiation pressure exerted by a 450 W light source on a perfectly reflecting surface placed at 2 m away from it, is

  1. Option A:

    1.5×10−81.5 \times 10^{-8} Pascals

  2. Option B:

    0

  3. Option C:

    3×10−83 \times 10^{-8} Pascals

    Correct
  4. Option D:

    6×10−86 \times 10^{-8} Pascals

Answer: C

Step-by-step solution

P=ICP=\frac{I}{C}

I=PA=4504π×22I=\frac{P}{A}=\frac{450}{4 \pi \times 2^{2}}

Pressure =45016π×3×108=\frac{450}{16 \pi \times 3 \times 10^{8}}

≈3×10−8\approx 3 \times 10^{-8}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
3 April, Morning Shift
Subject
Physics
Chapter
Atomic Physics
Topic
Photon Theory of Light and Radiation Pressure

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