Correct Answer - B
Frictional force between P and Q is `f = mu mg` which will retard P and accelerate Q.
Retardation of P is `a_(P)=-(f)/(m) =(-mumg)/(m)=-mug`
Acceleration of Q is `a_(Q)=(+f)/(M)=(mumg)/(M)`
Acceleration of Q relative to P is
`a_(QP)=a_(Q)-a_(P)=(mumg)/(M)-(-mug)`
`=mug[1+(m)/(M)]=0.3 xx 10[1+(1)/(6)]=3.5 m//s^(2)`