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A spherical ball is placed between a ver...

A spherical ball is placed between a vertical wall and a movable inclined plane as shown. There is no friction between the wall and the ball, and no friction between the ball and the inclined plane. The coefficient of friction between the inclined plane and the ground is The system is in equilibrium.

The minimum value of `mu` is :

A

`(m tan theta)/(M+m)`

B

`(m cos theta)/(M+m)`

C

`(m tan theta)/(M+m sin theta infty cos theta)`

D

`(m sin theta)/(M+m cos^(2) theta)`

Text Solution

Verified by Experts

The correct Answer is:
a
Friction acts on the incline to balance the horizontal component of Balancing forces on incline,
`R_(0)=Mg+N cos theta rarr R_(0)=(M+m)g`
`N sin theta f rarr f=N sin theta mg tan theta`
Now put `f=muR_(0)` to get
`mu=(m tan theta )/(M+m)`
This is the minimum value of `mu`
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