A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4m$ , then the time period of small oscillations will be _________ $s$ .
$[$ take $g = {\pi}^{2}m{s}^{- 2}$ ]

Question

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is $4m$ , then the time period of small oscillations will be _________ $s$ .

$[$ take $g = {\pi}^{2}m{s}^{- 2}$ ]

Quantrex Academy · Accepted Answer

Correct answer: 0

Acceleration due to gravity at surface of earth, $g = \dfrac{GM}{{R}^{2}}$ .
Now, at distance $R$ from the surface of the earth, ${g}^{'} = \dfrac{GM}{{\left(R + R\right)}^{2}} = \dfrac{g}{4}$
Therefore,
$T = 2\pi\sqrt{\dfrac{l}{\dfrac{g}{4}}}$
$⇒ T = 2\pi\sqrt{\dfrac{4 \times 4}{g}}$
$⇒ T = 2\pi\dfrac{4}{\pi} = 8s$

A simple pendulum is placed at a place where its distance from the...

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