A block is placed on an inclined plane. The block is moving towards right horizontal with an acceleration `a_(0)=g`. The length of the inclined plane (AC) is equal to 1 m. Whole the situation are shown in the figure. Assume that all the surfaces are frictionless. The time taken by the block to reach from C to A is (Take,`g=10m//s^(2)`)

Options

A. 0.74 s
B. 0.9 s
C. 0.52 s
D. 1.24 s

Answer

A

Explanation

The forces on smaller block is given as

For the motion of the block along the incline plane in upward direction.
Net force on the block = mass acceleration of the block
`rArr mgcos30^(@)-mgsin30^(@)=ma " " (thereforea_(0)=g)`
`rArr a =((sqrt(3)-1)/(2))g=3.66m//s^(2)`
Now, from equation of motion `s = (1)/(2) at^(2)`
`rArr t = sqrt((2s)/(a))=sqrt((2xx1)/(3.66))=0.74s`