Question

A wedge `B` of mass 2 m is placed on a rough horizontal suface. The coefficient of friction between wedge and the horizontal surface is `mu_(1)`. A block of mass m is placed on wedge as shown in the figure. The coefficient of friction between block and wedge is `mu_(2)`. The block and wedge are released from rest.
Q. Suppose the inclined surface of the wedge is at `theta=37^(@)` from angle from horizontal and `mu_(2)=0` are wedge will remain in equilibrium if
A. `mu_(1)=0.1`
B. `mu_(1)=0.2`
C. `mu_(1)=0.3`
D. `mu_(1)=0.4`

Answer 1

Correct Answer - B::C::D

`FBD` of wedge
`Nsin37^(@)=mu_(1)(2mg+Ncos37^(@))`
`mgcos37^(@)sin37^(@)=mu(2mg+mgcos^(2)37^(@))`
`(12)/(25)=mu_(1)(1+(16)/(25))`
`mu_(1)=(12)/(66)=(2)/(11)`