Correct Answer - C
Let ` AB = X, then ` AC= x cos theta`
:. ` (d(AC)/(dt) = d/(dt) (x cos theta) = (dx)/(dt) cos theta + x (- sin theta) (d theta)/(dt)`
But `(d(AC)/(dt) = 0 `, so ` 0= (dx)/(dt) cos theta + x ( - sin theta) ( d theat)/(dt)`
or ` v cos theta = x sin theta = x sin theta (d theta)/(dt)` or ` (dtheta)/(dt) = ( v cos theta)/(x sin theta)`
![image](https://learnqa.s3.ap-south-1.amazonaws.com/images/161078191214376847201610781912.png)
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Also , ` CB = x sin theta`.
Velocity of block will be
` u = sin theta + x cos theta xx (v cos theta)/( x sin theta)`
` (v (sin^@ theta + cos^2 theta)/( sin theta) = v/(sin theta)`.