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A rigid rod AB moves on horizontal plane such that at the instant shown velocity of centre C of rod is $v_{0}$ at an angle of $45^{0}$ with the rod. While velocity of end A is $v_{0} / \sqrt{2}$ and has direction along the rod. Velocity of end B of the rod at the given instant is

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$2 v_{0}$

$v_{0} / \sqrt{2}$

$\sqrt{\frac{5}{2}} v_{0}$

$\sqrt{2} v_{0}$

answer is C.

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Detailed Solution

Angular velocity of rod is $ω = \frac{v_{0} \sin 45}{l}$
from figure $v_{B} = \sqrt{{(\frac{v_{0}}{\sqrt{2}})}^{2} + {(2 l ω)}^{2}} = (\sqrt{\frac{5}{2}}) v_{0}$
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