A stone is dropped from a height equal to $n R$ ( $R$ is the radius of the Earth), from the surface of the Earth. The velocity of the stone on reaching the surface of the Earth is :

\sqrt{\frac{2 g (n + 1) R}{n}}

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\sqrt{\frac{2 g R}{n + 1}}

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\sqrt{\frac{2 g n R}{n + 1}}

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\sqrt{2 g n R}

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Solution

The correct option is D $\sqrt{\frac{2 g n R}{n + 1}}$
Conserving the energy between the dropping point and at the earth's surface.
$P E_{1} = K E_{2} + P E_{2}$
$\Rightarrow - \frac{G M m}{R + h} = \frac{1}{2} m V^{2} - \frac{G M m}{R}$
$\Rightarrow \frac{1}{2} V^{2} = \frac{G M}{R} - \frac{G M}{R + h}$
But $h = n R$
$\Rightarrow \frac{1}{2} V^{2} = G M (\frac{1}{R} - \frac{1}{R + n R})$
$\Rightarrow \frac{1}{2} V^{2} = G M (\frac{n}{R (1 + n)})$
But $G M = g R^{2}$
$\Rightarrow \frac{1}{2} V^{2} = \frac{g n R}{1 + n}$
$\Rightarrow V = \sqrt{\frac{2 g n R}{1 + n}}$

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