The velocity of point A on the rod is `2ms^(-1)` (leftwards) at the instant shown in fig. The velocity of the point B on the rod at this instant is

Options

A. `(2)/(sqrt(3)) ms^(-1)`
B. `1 ms^(-1)`
C. `(1)/(2sqrt(3)) ms^(-1)`
D. `(sqrt(3))/(2) ms^(-1)`

Answer

B

Explanation

Here y is constant `(dl)/(dt)=v_(B)`
`l^(2)=y^(2)+x^(2)`
`implies 2l(dl)/(dt)= 2x(dx)/(dt)`
`implies v_(B)=(x)/(l) v_(A)=v_(A) cos 60^(@)=1ms^(-1)`