The masses $M_{1}$ and $M_{2}$ ( $M_{2}$ > $M_{1}$ ) are released from rest.
Using work energy theorem find out velocity of the blocks when they move a distance x.

V = \sqrt{\frac{(M_{2} - M_{1}) g x}{M_{1} + M_{2}}}

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V = \sqrt{\frac{2 (M_{2} - M_{1}) g x}{M_{1} + M_{2}}}

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V = \sqrt{\frac{3 (M_{2} - M_{1}) g x}{M_{1} + M_{2}}}

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V = \sqrt{\frac{4 (M_{2} - M_{1}) g x}{M_{1} + M_{2}}}

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Solution

The correct option is B $V = \sqrt{\frac{2 (M_{2} - M_{1}) g x}{M_{1} + M_{2}}}$
$(W_{a L L F})_{s y s t e m} = (Δ K)$
$(W_{g})_{s y s} + (W_{T})_{s y s} = (Δ K)_{s y s}$ as $(W_{T})_{s y s} = 0$
$M_{2} g x - M_{1} g x = \frac{1}{2} (M_{1} + M_{2}) V^{2} - 0$ .......... (1)
$V = \sqrt{\frac{2 (M_{2} - M_{1}) g x}{M_{1} + M_{2}}}$

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