The coefficient of friction between the block `A` of mass `m` & block `B` of mass `2m` is `mu`. There is no friction between `A & B` is released from rest & there is no slipping between `A & B` then:

Options

A. `2 theta le sin ^(-1)(2mu)`
B. `theta le tan^(-1)(mu)`
C. `2 theta le cos^(-1)(2mu)`
D. `2 theta le tan^(-1)(mu//2)`

Answer

B

Explanation

The acceleration of blocks down the incline will be `g sin theta`. Horizontal component of this acceleration is `a_(H)=a cos theta` and vertical components `a_(v)=a sin theta`
`a_(H)= a cos theta = g sin theta cos theta` and `a_(v)= g sin^(2)theta`
For body `A: mg-N=ma_(v)`, or,
`N=mg-mg sin^(2) theta=mg cos^(2)theta`
And, `muN ge ma_(H)`
`mu mg cos^(2) theta ge mg sin theta=mg cos^(2)theta`
`mu getan theta` or `theta le tan^(-1)(mu)`