A small block of mass of 0.1 kg lies on a fixed inclined plane PQ which makes an angle `theta` with the horizontal. A horizontal force of 1N acts on the block through its centre of mass as shown in figure.

The block remains stationary if (take `g=10m//s^2`)

Options

A. `theta = 45^(@)`
B. `theta gt 45^(@)`and a frictional force acts on the block towards `p`
C. `theta gt 45^(@)` and a frictional force acts on the block towards `Q`
D. `theta lt 45^(@)` and a frictional force acts on the block towards `Q`

Answer

a,c

Explanation

Herer `m = 0.1 kg, F = 1 N, g = 10 m//s`
We have resolved mg into two rectangular compenents : `mg sin theta` along `QP` Again two rectangular com ponents of `F` are `F cos theta` along `QP` Again two rectangular com ponents of `F` are `Fcos theta` along `PQ` and `F sin theta bot PQ`
The block will remain stationary when forces along `PQ` have zero resultant
`mg sin theta = F cos theta`
`0.1 xx 10 sin theta= 1 cos theta`
or `tan theta = 1` or `theta =45^(@)`
Choice (a) is correct
`If theta gt 45^(@), mg sin theta gt F cos theta`
the block tends to move down the incline `QP` Frictional force must act on the block towards `Q` Choice (c) is correct With the same argument we can show that choices (b) and (d) are incorrect
.