A block of mass m starting from rest slides down a smooth inclined plane of gradient e, fixed in a lift moving upwards with an acceleration an as shown in figure. If the base of the inclined plane has length L, the time taken by the block to slide from top to bottom of the inclined plane, will be

Options

A. `( L)/( ( g _ a_0) sin 2 theta )`
B. `[( 4L )/((g +a_0 ) sin 2 theta)]^(1//2)`
C. `((g +a_0)sin 2 theta)/(2L)`
D. `[(2 L)/((g+a_0) sin theta)]^2`

Answer

B

Explanation

Acceleration along the inclined plane,
`a=(g + a_0 )sin theta `
Slope length `=(L )/( cos theta )`
using ` s- ut +1/2 at^2 , ` we get `(L )/( cos theta ) = 0 xx t + 1/2 (g +a_0 ) sin theta t^2 `
` or t= sqrt((2L )/((g+a_0) sin theta cos theta)) or t= sqrt((4L)/((g+a_0) sin theta))`