Two bodies begin a free fall from the same height at a time interval of $N$ s. If vertical separation between the two bodies is $1$ m after $n$ seconds from the start of the first body, then $n$ is equal to:

\sqrt{g N}

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\frac{1}{g N}

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\frac{1}{g N} + \frac{N}{2}

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\frac{1}{g N} + N

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Solution

The correct option is C $\frac{1}{g N} + \frac{N}{2}$
$s = u t + \frac{1}{2} a t^{2}$
$s_{1} - s_{2} = 1 m = \frac{1}{2} g t^{2} - \frac{1}{2} g {(t - N)}^{2}$
$= \frac{1 \times g}{2} [N (2 t - N)]$
$\frac{2}{g} = N (2 n - N)$
$\Rightarrow n = \frac{1}{2} (\frac{2}{g N} + N)$
$\Rightarrow n = \frac{1}{g N} + \frac{N}{2}$

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