Correct Answer - B::C
`a=(mgsintheta-mu_k mg cos theta)/(m)`
`:.` `a_A=g sin theta-mu_(k,A)g cos theta` …(i)
and `a_B= g sin theta -mu_(k,B) g cos theta` …(ii)
Putting values we get
`a_A=4sqrt2m//s^2` and `a_B=3.5sqrt2m//s^2`
Let `a_(AB)` is relative acceleration of A w.r.t. B. Then
`a_(AB)=a_A-a_B`
`L=sqrt2m`
[where L is the relative distance between A and B]
Then `L=(1)/(2)a_(AB^(t^2))`
or `t^2=(2L)/(a_(AB))=(2L)/(a_A-a_B)`
Putting values we get, `t^2=4` or `t=2s`.
Distantce moved by B during that time is given by
`S=(1)/(2)a_(B^(t^2))=1/2xx3.5sqrt2xx4=7sqrt2m`
Similarly for `A=8sqrt2m`.