A force F is applied on a block of mass `sqrt(3) kg` which rests on a horizontal surface with a coefficient of friction `1/(2sqrt3)`. The maximum value of F for which the block doesn't move, is [Take `g = 10 ms^(-2)]`

Options

A. 20N
B. 10N
C. 12N
D. 15N

Answer

A

Explanation

From acting on block are shown in adjoining figure
As the block does not move hence
`F cos `
`60^(@)=f=muN=mu(Mg+Fsin60^(@))`

`:.F1/2=1/(2sqrt(3))(sqrt(3)xx10+F .(sqrt(3))/2)`
On simplification we get F=20N