A uniform ring of mass `m` is placed on a rough horizontal fixed surface as shown in the figure. The coefficient of friction between the left part of the ring and left part of the horizontal surface is `mu_(1)=0.6pi` and between right half and the surface is `mu_(2)=0.2pi`. At the instant shown, now the ring has been imparted an angular velocity in clockwise sense in the figure shown. At this moment magnitude of acceleration of centre `O` of the ring (in `m//s^(2)`) is (take `g=10m//s^(2)`)

Answer

4

Explanation

`2[int_(0)^(pi//2)(mu_(1)-mu_(2))Rglamdacosthetad theta]=2(mu_(1)-mu_(2))Rlamdag`
`:.a=(2(mu_(1)-mu_(2))Rlamdag)/(2piRlamda)=4`