A force `vec(F)` is applied to block `(m = 6 kg)` at rest on an inclined plane of inclination `30^(@)`. The force `vec(F)` is horizontal and parallel to surface of inclined plane. Maximum value of `F` so that block remains at rest is `40 N`. The coefficient of friction of the surface is.

Options

A. `(2)/(sqrt(3))`
B. `sqrt(3)`
C. `(5)/(3 sqrt(3))`
D. `(5 sqrt(3))/(2)`

Answer

C

Explanation

`mu mg cos theta = sqrt(F^(2)+(mg sin theta)^(2))`
`m 60 (sqrt(3))/(2) = sqrt((40)^(2)+(30)^(2))`
`mu 30 sqrt(3) = 50 rArr mu = (5)/(3 sqrt(3))`.