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Physics · Gravitation

JEE Main 2024 — 29 January, Shift 1 — Question 45

At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R.)

  1. Option A:

    5R−R\sqrt{5} R-R

  2. Option B:

    3R−R2\frac{\sqrt{3} R-R}{2}

  3. Option C:

    R2\frac{R}{2}

  4. Option D:

    5R−R2\frac{\sqrt{5} R-R}{2}

    Correct

Answer: D

Step-by-step solution

gp=gR2(R+h)2g_{p}=\frac{g R^{2}}{(R+h)^{2}}

gq=g(1−hR)\mathrm{g}_{\mathrm{q}}=\mathrm{g}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)

figure

gp=gq\mathrm{g}_{\mathrm{p}}=\mathrm{g}_{\mathrm{q}} g(1+hR)2=g(1−hR)\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)^{2}}=\mathrm{g}\left(1-\frac{\mathrm{h}}{\mathrm{R}}\right)

(1−h2R2)(1+hR)=1\left(1-\frac{\mathrm{h}^{2}}{\mathrm{R}^{2}}\right)\left(1+\frac{\mathrm{h}}{\mathrm{R}}\right)=1

Take hR=x\frac{\mathrm{h}}{\mathrm{R}}=\mathrm{x}

So

x3−x+x2=0\mathrm{x}^{3}-\mathrm{x}+\mathrm{x}^{2}=0

x=5−12\mathrm{x}=\frac{\sqrt{5}-1}{2}

h=R2(5−1)\mathrm{h}=\frac{\mathrm{R}}{2}(\sqrt{5}-1)

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
29 January, Shift 1
Subject
Physics
Chapter
Gravitation
Topic
Gravitational Field and Gravity