In the figure the wedge is pushed with an acceleration of `sqrt3 m//s ^(2)`. It is seen that the block start climbing up on the smooth inclined face of wedge . What will be the time taken by the block to reach the top?

Options

A. `(2)/(sqrt5) s`
B. `(1)/(sqrt5) s`
C. `sqrt(5) s`
D. `(sqrt5)/(2) s`

Answer

B

Explanation

`FBD` of `m` w.r.t. wedge
Relative acceleration along the inclined plane
`a_(r) = (ma cos theta - mg sin theta)/(m)`
`= a cos theta - g sin theta`
`= (10 sqrt3) ((sqrt3)/(2)) - (10) ((1)/(2))`
`= 10 m//s^(2)`
`t = sqrt((2 s)/(a_(r))) = sqrt((2 xx 1)/(10)) = (1)/(sqrt5) s`