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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.The acceleration due to gravity at a height $h$ is $g' = \frac{

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$3T_1 = 2T_2$

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(C) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{\ell}{g'}}$.The acceleration due to gravity at a height $h$ is $g' = \frac{GM}{(R+h)^2}$.Thus,$T = 2\pi \sqrt{\frac{\ell (R+h)^2}{GM}} = 2\pi \sqrt{\frac{\ell}{GM}} (R+h)$.For height $h_1 = R$,$T_1 = 2\pi \sqrt{\frac{\ell}{GM}} (R+R) = 2\pi \sqrt{\frac{\ell}{GM}} (2R)$.For height $h_2 = 2R$,$T_2 = 2\pi \sqrt{\frac{\ell}{GM}} (R+2R) = 2\pi \sqrt{\frac{\ell}{GM}} (3R)$.Taking the ratio: $\frac{T_1}{T_2} = \frac{2R}{3R} = \frac{2}{3}$.Therefore,$3T_1 = 2T_2$.

$A$ simple pendulum performing small oscillations at a height $R$ above the Earth's surface has a time period of $T_1 = 4 \ s$. What would be its time period $T_2$ if it is brought to a point at a height $2R$ from the Earth's surface? Choose the correct relation ($R =$ radius of Earth).

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