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. `sqrt5s`
D. `(sqrt5)/(2)s`

Answer

B

Explanation

`10sqrt3 cos 30^(@) = 10 sqrt3 (sqrt3)/(2) = 15m//s^(2)`
`:. A = 15 -5 =10m//s^(2) :. S = (1)/(2) at^(2)`
`1 = (1)/(2) (10)t^(2) or t = (1)/(sqrt5)S`
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