If masses are released from the position shown in the figure then time elapsed before mass $m_{1}$ collides with the floor will be

\sqrt{\frac{2 m_{1} g d}{m_{1} + m_{2}}}

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\sqrt{\frac{2 (m_{1} + m_{2}) d}{(m_{1} - m_{2}) g}}

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\sqrt{\frac{2 (m_{1} - m_{2}) d}{(m_{1} + m_{2}) g}}

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None of these

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Solution

The correct option is B $\sqrt{\frac{2 (m_{1} + m_{2}) d}{(m_{1} - m_{2}) g}}$
We have acceleration of the given system as
$a = \frac{(Supporting force-opposing force)}{Total mass}$
$\Rightarrow a = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) g$
Also, mass $m_{1}$ is released from the rest and acceleration is constant
$\Rightarrow d = \frac{1}{2} a t^{2}$
$\Rightarrow t = \sqrt{\frac{2 d}{a}} = \sqrt{\frac{2 (m_{1} + m_{2}) d}{(m_{1} - m_{2}) g}}$

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