In the arrangement shown in the figure mass of the block `B` and `A` are `2m` , `8m` respectively. Surface between `B` floor is smooth. The block `B` is connected to block `C` by means of a pulley. If the whole system is released then the minimum value of mass of the block `C` so that the block `A` remains stationary with respect to `B` is: (Coefficient of friction between `A` and `B` is `mu` and pulley is ideal).

Options

A. `(m)/(mu)`
B. `(2m)/(mu+1)`
C. `(10m)/(1-mu)`
D. `(10m)/(mu-1)`

Answer

D

Explanation

FBD of `A`
If the acceleration of `'C'` is a
For block `'A'` `N=8ma` (1)
`8mg-muN=mu8ma` (2)
and acceleration a can be written by the equation of system `(A+B+C)`
`m_(1)g=(10m+m_(1))a` (3)
`8kmg=mu8m((m_(1)g)/(10m+m_(1)))`
`10m+m_(1)=mum_(1)`
`10m=m(mu-1)m_(1)`
implies `m_(1)=(10m)/(mu-1)` .