A block of base `10 cm xx 10 cm` and height `15 cm` is kept on an inclined plane. The corfficient of friction between them is `sqrt(3)`. The inclination `theta` of this inclined plane from the horizontal plane is gradually increased frm `0^@`. Then

Options

A. at `theta = 30^(@)`, the block will start sliding down the plane
B. the block will remain 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`

Answer

B

Explanation

The maxima angle when the block will not slide should be equal to the angle of repose.

`phi = tan^(-1) mu = tan^(-1) sqrt(3) = 60^(@)`
And for toppling, maximum angle of inclination is given by (about pont A) `mg sin theta h/2 le m g cos theta b/2`
`rArr tan theta le b/a rArr theta_("max") = tan^(-1) 2/3 therefore theta_("max") lt 60^(@)`