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The velocity of point A on the rod is 2 m s^-1 (leftwards) at the instant shown in figure. The velocity of the point B on the rod at this instant is

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$\frac{2}{\sqrt{3}} m s^{- 1}$

$1 {m s}^{- 1}$

$\frac{1}{2 \sqrt{3}} m s^{- 1}$

$\frac{\sqrt{3}}{2} m s^{- 1}$

answer is B.

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Detailed Solution

Here y is constant $\frac{dl}{dt} = v_{B}$

$\begin{array}{l} l^{2} = y^{2} + x^{2} \\ \Rightarrow 2 l \frac{dl}{dt} = 2 x \frac{dx}{dt} \\ \Rightarrow v_{B} = \frac{x}{l} v_{A} = v_{A} \cos 60^{\circ} = 1 {ms}^{- 1} \end{array}$

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