A block of base `10cmxx10cm` and height 15cm is kept on an inclined plane. The coefficient of friction between them is `sqrt3`. The inclination `theta` of this inclined plane from the horizontal plane is gradually increased from `0^@` Then

Options

A. At `theta=30^(@)`, the block will start sliding down the plane
B. The block will remian at rest on the plane up to certain `theta` and then it will topple
C. At `theta=60^(@)`, the block will start sliding down the plane and continue to do so at higher angles
D. At `theta=60^(@)`, the block will start sliding down the plane and on further increasing `theta`, it will topple at certain `theta`

Explanation

Condition for sliding:

`mg sin theta=f=mu mg costheta`
`tan theta=muimpliestheta=tan^(-1)(mu)=tan^(-1)(sqrt3)`
`theta=60^(@)`
Condition for toppling:

Torque about `C`
`Nxx5=fxx(15)/(2)`
`mg cos thetaxx5=mg sinthetaxx(15)/(2)`
`tan theta=(2)/(3)`
`theta` for toppling`lt theta` for sliding