If the coefficient of friction between `A` and `B` is `mu`, the maximum acceleration of the wedge `A` for which `B` will remain at rest with respect to the wedge is

Options

A. `mug`
B. `g((1+mu)/(1-mu))`
C. `(g)/(mu)`
D. `g((1-mu)/(1+mu))`

Answer

B

Explanation

FBD of block `B w.r.t.` wedge `A`,for maximum `'a':`
Perpendicular to wedge `:`
`Sigmaf_(y)=(mg cos theta+ma sin theta-N)=0`
and `Sigmaf_(x)=mgsintheta+muN-macos theta=0(` for maximum `a)`

`rArr mgsin theta+mu(mg cos theta +ma sin theta)-ma cos theta =0`
`rArr a=((g sin theta+mu g cos theta))/(cos theta-musintheta)`
for `theta=45^(@)`
`a=g((tan 45^(@)+mu)/(cot 45^(@)-mu)),a=g((1+mu)/(1-mu))Ans.`