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

Answer

B

Explanation

FBD of m w.r.t wedge
Relative acceleration along the inclined plane
`a_(r ) = (m a cos theta - m g sin theta)/(m)`
`a cos theta - g sin theta`
`= (1.0 sqrt(3)) ((sqt(3))/(2)) - (10) ((1)/(2)) = 10 m//s^(2)`
`t = sqrt((2s)/(a_(r ))) = (sqrt(2 xx 1))/(10) = (1)/(sqrt(5)) s`