A disc of radius `R=2m` starts rotating wth constant angular acceleration `alpha=(2rad//s^(2))`. A block of mass `m=2kg` is kept at a distance `(R)/(2)` from the centre of disc. The coeffiecient of friction between disc and block is `mu_(s)=0.4,mu_(k)=0.3`. Acceleration of block when it just slips w.r.t. ground and w.r.t. disc are respectively`-`(`g=10ms^(-2)`)

Options

A. `3ms^(-2),1ms^(-2)`
B. `1ms^(-2),3ms^(-2)`
C. `4ms^(-2),1ms^(-2)`
D. `4ms^(-2),3ms^(-2)`

Answer

A

Explanation

Acceleration of block w.r.t. ground when it just slips `=mu_(k)g=(0.3)(10)=3ms^(-2)`

Acceleration of block w.r.t. disc when it just
slips`=(momega^(2)r-mu_(k)mg)/(m)`
`(mu_(s)mg-mu_(k)mg)/(m)`
`=(mu_(s)-mu_(k))g=(0.4-0.3)(10)=1 ms^(-2)`