In the figure shown the block `B` moves with , velocity `10 ms^(-1)`. The velocity of `A` in the position shown is

Options

A. `12.5 ms^(-1)`
B. `25 ms^(-1)`
C. `8 ms^(-1)`
D. `16 ms^(-1)`

Answer

D

Explanation

`x + x + sqrt(y^(2) + c^(2)) = l =` length of string
Differenting w.r.t. time, we get

`2 (dx)/(dt) = (y)/sqrt(y^(2) + c^(2)) ((- dy)/(dt))`
or `2 nu_(B) = (1)/( cos theta) nu_(A)`
`:. nu_(A) = 2 nu_(B) cos theta`
`= (2) (10) (0.8)`
`= 16 m//s`