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Four rods each of length l, have been hinged to form a rhombus. Vertex A is fined to a rigid support. Vertex C being pulled to the right along x-axis with a uniform speed V as shown. The speed at which vertex B moves at the moment the rhombus takes the shape of a square is

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$\frac{V}{\sqrt{2}}$

$\frac{V}{4}$

$\frac{V}{2}$

answer is B.

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Detailed Solution

$AC = x, BE = Y {BE}^{2} + {AE}^{2} = l^{2} Y^{2} + {(\frac{x}{2})}^{2} = l^{2} 2 Y \frac{dy}{dt} + \frac{x}{2} \frac{dx}{dt} = 0 \frac{- dy}{dt} = \frac{1}{2} (\frac{x}{2 y}) \frac{dx}{dt}$

x=2y, when rhombus is a

Square. Hence $V_{B} = \frac{1}{2} V_{C} = \frac{V}{2}$

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