Two blocks `m_(1)` and `m_(2)` are allowed to move without friction Block `m_(1)` is on block `m_(2)` and `m_(2)` slides on smooth fixed incline as shown The angle of inclination of inclined plane is `theta`
.
Normal reaction on `m_(1)` is .

Options

A. `m_(1)g`
B. `(m_(1)+m_(2))g`
C. `(m_(1)m_(2)g cos^(2)theta)/(m_(2)+m_(1)sin^(2)theta)`
D. `(m_(1)g[1-(m_(1)+m_(2))sin ^(2)theta])/(m_(1)+m_(2)sin ^(2)theta)`

Answer

C

Explanation

`m_(1)g-N_(1)=m_(1)a_(1)` ….(i)
`(N_(1)+m_(2)g)sin theta=m_(2)a_(2)` ….(ii)
From (i) and (ii)
`[(m_(1)g-m_(1)a_(1))+m_(2)g]sin theta=m_(2)a_(2)`
`(m_(1)+m_(2))g sin theta-m_(1)a_(1)sin theta=m_(2)a_(2)` ....(iii)
Block `m_(1)` can move only in vertical direction as there is no force in horizontal direction. Since acceleration along normal at point of contact of the block and wedge should be equal.
So, `a_(1)=a_(2) sin theta` ...(iv)