JEE MAIN - Physics (2024 - 30th January Evening Shift - No. 28)
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
8
Explanation
Acceleration due to gravity g' $$=\frac{g}{4}$$
$$\begin{aligned} & \mathrm{T}=2 \pi \sqrt{\frac{4 \ell}{\mathrm{g}}} \\ & \mathrm{T}=2 \pi \sqrt{\frac{4 \times 4}{\mathrm{~g}}} \\ & \mathrm{~T}=2 \pi \frac{4}{\pi}=8 \mathrm{~s} \end{aligned}$$
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