If masses are released from the position shown in fig. then speed of mass $m_{1}$ just before it collides on floor will be:

[2 m_{1} g d / (m_{1} + m_{2})]^{1 / 2}

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[2 (m_{1} - m_{2}) g d / (m_{1} + m_{2})]^{1 / 2}

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[2 (m_{1} m_{2}) g d / m_{1}]^{1 / 2}

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Solution

The correct option is B $[2 (m_{1} - m_{2}) g d / (m_{1} + m_{2})]^{1 / 2}$
Equation of motion are-
$T - m_{2} g = m_{2} a$
$m_{1} g - T = m_{1} a$
adding the above equations we get,
$a = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2}}$ .
now final velocity $v = \sqrt{a d}$
which gives,velocity= $\sqrt{\frac{2 (m_{1} - m_{2}) g d}{m_{1} + m_{2}}}$

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