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Gravitation

JEE Main level, single correct, medium

72 of 223

Question 72 of 223

Assume there are two identical simple pendulum Clocks- $1$ is placed on the earth and Clock- $2$ is placed on a space station located at a height h above the earth surface. Clock- $1$ and Clock- $2$ operate at time periods $4s$ and $6s$ respectively. Then the value of $h$ is -
(consider radius of earth ${R}_{E}=6400km$ and $g$ on earth $10m{s}^{-2}$ )

Show answer and solution

Answer: (C) $3200km$

The time period of a simple pendulum is given by,

T=2\pi \sqrt{\dfrac{l}{g}}

.
Therefore,

T\propto \dfrac{1}{\sqrt{g}}

.
The acceleration due to gravity at height

h

from the surface of the earth is given by,

{g}^{'}=\dfrac{g}{{\left(R+h\right)}^{2}}

Therefore,

\dfrac{{T}_{1}}{{T}_{2}}=\sqrt{\dfrac{{g}_{2}}{{g}_{1}}} \Rightarrow \dfrac{4}{6}=\dfrac{R}{R+h} \Rightarrow h=\dfrac{R}{2}=3200km

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JEE Main 2025 Apr 2 Shift 1 question paper

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