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 cetain `theta`.

Answer

B

Explanation

For the block to slide, the angle of inclination should be equal to the angle of repose, i.e.,
`tan^-1mu=tan^-1sqrt3=60^@`.
Therefore, option (a) is wrong.
For the block to topple, the condition of the block will be as shown in the figure.

In `DeltaPOM`, `tantheta=(PM)/(OM)=(5 cm)/(7.5 cm)=2/3`
For this, `thetalt60^@`. From this we can conclude that the block will topple at lesser angle of inclination. Thus the block will remain at rest on the plane up to a certain agnle `theta` and then it will topple.