In the given figure, the wedge is acted upon by a constant horizontal force F. The wedge is moving on a smooth horizontal surface. A ball of mass 'm' is at rest relative to the wedge. The ratio of forces exerted on 'm' by the wedge when F is acting, and when F is withdrawn, assuming no friction between the wedge and the ball, the ratio is equal to

Options

A. `sec^(2)theta`
B. `cos^(2)theta`
C. 1
D. none of these.

Answer

A

Explanation

The ball moves along x-axis with an acceleration, say a, whereas it is in equilibrium along y axis
`rArr Sigma F_(x)=0 rArr R sin theta=ma`
and `SigmaF_(y)=0rArr R cos theta=mg`
`rArr R =mg sec theta " " ...(1)`

When `F-0`, Next the ball moves down the plane, along X axis, say with an acceleration a.
`rArr mg sin theta=ma.`
The ball does not move perpendicular to the plane. It is in equilibriumalong y - axis
`rArr mg cos theta-R.=0`
`rArr R.=mg cos theta" " ...(2)`
Dividing (1) by (2) `rArr (R)/(R.)=sec^(2)theta`