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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 from `0^@`. Then

A

at `theta = 30^@`, the block will starts 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`.

Text Solution

Verified by Experts

The correct Answer is:
B
The block will start sliding down (if it does not topple) at angle of repose.
ie, `mu = tan theta` :. `theta = 60^@`
The block will start toppling (if it does not slide at angle `theta` id
`mg sin theta((15)/(2)) = mg cos theta ((10)/(2)) rArr theta_(t), = tan^(-1) ((2)/(3))`
.
:. As `theta_(t) gt theta_(s)` block will first topple before it slides.
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