Correct Answer - A::B
`y_(1)+(y_(1)-y_(0))+x=1_(1)`….(1)
`y_(2)+(y_(2)-y_(1))=1_(2)`....(2) `v_(B)=4v_(A)`
`a_(B)=4a_(A) m_(A)g-T=m_(a)a_(A)`
`(T)/(4)-muM_(B)g=M_(B)a_(B) , M_(A)g-T=M_(a)a_(A)`
`T-4muM_(B)g=M_(B)4a_(B)=M_(B)16 a_(a)`
(1) `a_(A)=[(M_(A)-4muM_(B))/(16M_(B)+M_(A))]=(5-4xx0.1xx10)/((16xx10)+5)x10`
`-(10)/(165)=0.06ms^(-2)`
(2) `a_(B)=4a_(A)=0.2ms^(-2)`
(3) `T=m_(A)(g-a_(A))`
(Between `A` and pulley `1`)`= 5[10-0.06]=50N`
(4) `(T)/(4)=12.5N` ( Between pulley `3` and block`B`)