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Physics · Atomic Physics

JEE Main 2024 — 8 April, Shift 1 — Question 32

Average force exerted on a non-reflecting surface at normal incidence is 2.4×10−4 N2.4 \times 10^{-4} \mathrm{~N}. If 360 W/cm2360 \mathrm{~W} / \mathrm{cm}^{2} is the light energy flux during span of 1 hour 30 minutes. Then the area of the surface is:

  1. Option A:

    0.2 m20.2 \mathrm{~m}^{2}

  2. Option B:

    0.02 m20.02 \mathrm{~m}^{2}

    Correct
  3. Option C:

    20 m220 \mathrm{~m}^{2}

  4. Option D:

    0.1 m20.1 \mathrm{~m}^{2}

Answer: B

Step-by-step solution

Pressure =IC=FA=\frac{I}{C}=\frac{F}{A}

⇒36010−4×3×108=2.4×10−4 A\Rightarrow \frac{360}{10^{-4} \times 3 \times 10^{8}}=\frac{2.4 \times 10^{-4}}{\mathrm{~A}}

⇒A=2×10−2 m2=0.02 m2\Rightarrow \mathrm{A}=2 \times 10^{-2} \mathrm{~m}^{2}=0.02 \mathrm{~m}^{2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 1
Subject
Physics
Chapter
Atomic Physics
Topic
Photon Theory of Light and Radiation Pressure

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