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

Answer

B

Explanation

Let `a =` maximum acceleration of `A`.
Under no slip condition acceleration of `B` is also `a`
`FBD of A`w.r.t. ground

`sum F_(y) = 0`
` :. (N)/(sqrt2) = mg + (mu N)/(sqrt2)` …(i)
`sum F_(x) = ma`
`:. (N)/(sqrt2) + (mu N)/(sqrt2) = ma` ...(ii)
Solving these two equations, we get
`a = g ((1 + mu)/(1 - mu))`