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Physics · Gravitation

JEE Main 2024 — 4 April, Shift 2 — Question 49

A 90 kg body placed at 2 R distance from surface of earth experiences gravitational pull of : ( R=\mathrm{R}= Radius of earth, g=10 ms−2\mathrm{g}=10 \mathrm{~ms}^{-2} )

  1. Option A:

    300 N

  2. Option B:

    225 N

  3. Option C:

    120 N

  4. Option D:

    100 N

    Correct

Answer: D

Step-by-step solution

Value of g=gs(1+hR)−2g=g_{s}\left(1+\frac{h}{R}\right)^{-2}

=gs(1+2)−2=gs9=\mathrm{g}_{\mathrm{s}}(1+2)^{-2}=\frac{\mathrm{g}_{\mathrm{s}}}{9}

Here gs=g_{s}= gravitational acceleration at surface Force

=mg=90×gs9=100 N=\mathrm{mg}=90 \times \frac{\mathrm{g}_{\mathrm{s}}}{9}=100 \mathrm{~N}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
4 April, Shift 2
Subject
Physics
Chapter
Gravitation
Topic
Motion of Satellites and Escape Speed

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