If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g=π 2 m / s 2 , then the length of the second's pendulum at a height h=2 R from the surface of earth will be,:

Physics · Gravitation

JEE Main 2024 — 1 February, Shift 1 — Question 31

If R is the radius of the earth and the acceleration due to gravity on the surface of earth is $g=\pi^{2} \mathrm{~m} / \mathrm{s}^{2}$ , then the length of the second's pendulum at a height $h=2 R$ from the surface of earth will be,:

Option A:
$\frac{2}{9} m$
Option B:
$\frac{1}{9} \mathrm{~m}$
Correct
Option C:
$\frac{4}{9} \mathrm{~m}$
Option D:
$\frac{8}{9} \mathrm{~m}$

Answer: B

Step-by-step solution

$\mathrm{g}^{\prime}=\frac{\mathrm{GMe}}{(3 \mathrm{R})^{2}}=\frac{1}{9} \mathrm{~g}$

$\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}^{\prime}}}$ Since the time period of second pendulum is 2 sec .

$\mathrm{T}=2 \mathrm{sec}$ $2=2 \pi \sqrt{\frac{\ell}{\mathrm{~g}} 9}$ $\ell=\frac{1}{9} \mathrm{~m}$

Answer key and solution verified before publishing.

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