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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.At a height $h = R$ (where $R$ is the radius of the Earth),the

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.At a height $h = R$ (where $R$ is the radius of the Earth),the acceleration due to gravity $g'$ is given by $g' = g \left( \frac{R}{R+h} ight)^2 = g \left( \frac{R}{R+R} ight)^2 = \frac{g}{4}$.Given $g = \pi^2 \ m/s^2$ at the surface,the effective gravity at height $h$ is $g' = \frac{\pi^2}{4} \ m/s^2$.Substituting the values $\ell = 4.0 \ m$ and $g' = \frac{\pi^2}{4} \ m/s^2$ into the time period formula:$T = 2\pi \sqrt{\frac{4}{\pi^2 / 4}} = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi 	imes \frac{4}{\pi} = 8 \ s$.

$A$ simple pendulum is taken to a place where its distance from the Earth's surface is equal to the radius of the Earth. Calculate the time period of small oscillations if the length of the string is $4.0 \ m$. (Take $g = \pi^2 \ m/s^2$ at the surface of the Earth.) (in $s$)

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