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(D) The acceleration due to gravity at a height $h$ above the earth's surface is given by $g' = g \left( \frac{R}{R+h} \right)^2$.Given $h = R$,the ac

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(D) The acceleration due to gravity at a height $h$ above the earth's surface is given by $g' = g \left( \frac{R}{R+h} \right)^2$.Given $h = R$,the acceleration due to gravity becomes $g' = g \left( \frac{R}{R+R} \right)^2 = g \left( \frac{1}{2} \right)^2 = \frac{g}{4}$.The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.Substituting $\ell = 4 \ m$ and $g' = \frac{g}{4}$,we get $T = 2\pi \sqrt{\frac{4}{g/4}} = 2\pi \sqrt{\frac{16}{g}}$.Given $g = \pi^2 \ ms^{-2}$,we substitute this into the equation:$T = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi \left( \frac{4}{\pi} \right) = 8 \ s$.

$A$ 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 $g = \pi^2 \ ms^{-2}$]

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