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The end B of the rod AB which makes angle `theta` with the floor is being pulled with a constant velocity `v_(v)` as shown. The length of the rod is `l`.
A. `(5v_(0))/(3l)`
B. `(3v_(0))/(5l)`
C. `(5v_(0))/(4l)`
D. `(4v_(0))/(5l)`

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Correct Answer - A
Differentiate `x=lcostheta` with respect to time
`(dx)/(dt)=lsintheta(d theta)/(dt)`
`(d theta)/(dt)=(V_(A))/(lsintheta)`
(negative sign indicates that `theta` is decreasing)
`omega=|(d theta)/(dt)|=(V_(A))/(lsin37^@)=(5v_(0))/(3l)`

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