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 $4 m$, then the time period of small oscillations will be __________ s. [take $g=\pi^2 m s^{-2}$]
Answer (integer)
8
Solution
<p>Acceleration due to gravity g' $=\frac{g}{4}$</p>
<p>$$\begin{aligned}
& \mathrm{T}=2 \pi \sqrt{\frac{4 \ell}{\mathrm{g}}} \\
& \mathrm{T}=2 \pi \sqrt{\frac{4 \times 4}{\mathrm{~g}}} \\
& \mathrm{~T}=2 \pi \frac{4}{\pi}=8 \mathrm{~s}
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Gravitational Field and Potential
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