In the figure shown 'p' is a plate on which a wedge `B` is placed and on `B` a block `A` of mass m `m` is placed. The plate is suddenly removed and system of `B` and `A` is allowed to fall under gravity. Neglecting any force due to air on `A` and `B` , the normal force on `A` due to `B` is

Options

A. `(mg)/(costheta)`
B. `mg costheta`
C. zero
D. `(2mg)/(costheta)`

Answer

C

Explanation

F.B.D. of block `A`
Applying Newton's second law for block `A` in vertical direction
`m_(A)g-Ncostheta=m_(A)g`
where theta is the angle of the wedge
implies `Ncostheta=0`
as `thetalt90^(@)`
:. `N=0`