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A block of base 10 cm xx 10 cm and heigh...

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

A

At `theta=30^(@)`, the block will start sliding down the plane

B

The block will remain at rest on the plane up to certaiin `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`

Text Solution

Verified by Experts

The correct Answer is:
B

For sliding :
`tan theta ge sqrt(3)~~1.732`
For toppling ,
`tantheta ge (2)/(3)~~0.67`
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