A block of `3 kg` rests in limiting equilibrium on an inclined plane of inclination `30^(@)`. If the plane is raised to an inclination `60^(@)` , the force along the plane required to support it is

Options

A. `(10)/(sqrt(3)) N`
B. `10 sqrt(3) N`
C. `10 N`
D. `(10)/(3) N`

Answer

B

Explanation

`mg sin 30^(@) = mu mg cos 30^(@)`
`rArr mu = tan 30^(@) = (1)/(sqrt(3))`
`mg sin 60 = F + mu mg cos 60^(@)`
`3 xx 10 xx (sqrt(3))/(2) = F + (1)/(sqrt(3)) xx 3 xx 10 xx (1)/(2)`
`F = 15 sqrt(3) - 5 sqrt(3) = 10 sqrt(3) N`