A block A is kept on a rough inclined plane. Initially `theta = 0`. The co-efficient of kinetic friction between A and the inclined plane is `sqrt(3)` The angle `theta` is increased slowly. When `theta` becomes `60^(@)`, then :

Options

A. A slides down with constant velocity
B. A does not slide
C. A slides down with constant non-zero acceleration
D. A slides down with constant acceleration

Answer

B

Explanation

At `theta = 60^(@)`
`mg sin theta = mu mg cos theta`

`:.` Net force on the block is zero. Since initially block was at rest relative to inclined plane, it continues to remain in rest.