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(C) Let the length of the string be $L$. Let $x_A$ and $x_B$ be the horizontal distances of blocks $A$ and $B$ from the vertical line passing through

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$V \sqrt{3}$

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(C) Let the length of the string be $L$. Let $x_A$ and $x_B$ be the horizontal distances of blocks $A$ and $B$ from the vertical line passing through the pulley,and $h$ be the height of the pulley from the ground.The length of the string is $L = \sqrt{x_A^2 + h^2} + \sqrt{x_B^2 + h^2}$.Differentiating with respect to time $t$,we get $0 = \frac{x_A}{\sqrt{x_A^2 + h^2}} \frac{dx_A}{dt} + \frac{x_B}{\sqrt{x_B^2 + h^2}} \frac{dx_B}{dt}$.Since $\cos 	heta = \frac{x}{\sqrt{x^2 + h^2}}$,we have $v_A \cos 	heta_A + v_B \cos 	heta_B = 0$ (considering components along the string).Here,the velocity of the string along the segment connected to $B$ is $v_B \cos(30^\circ) = V \cos(30^\circ)$.The velocity of the string along the segment connected to $A$ is $v_A \cos(60^\circ)$.Since the string is inextensible,the velocity component along the string must be equal for both sides: $v_A \cos(60^\circ) = V \cos(30^\circ)$.$v_A (1/2) = V (\sqrt{3}/2)$.$v_A = V \sqrt{3}$.

For the given diagram,when block $B$ is pulled with velocity $V$,the velocity of block $A$ will be:

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