In the pulley-block arrangement shown in figure , find the relation between acceleration of blocks `A` and `B`

Explanation

Let `m_(2)` goes downward with acceleration `a_(2) and m_(1)` goes upward with acceleration `a_(1)`
From the diagram , we have ` a_(2) = (a_(1)+0)/2`

` 2a_(2) = a_(1) " " `….(i)
Now, `m_(2)g-2T = m_(2)a_(2)" "` ....(ii)
and ` T- m_(1)g = m_(1)a_(1)" "` ....(iii)
By (ii)n `+ 2 xx `(iii), we get
` m _(2)g - 2m_(1)g = m_(2)(a_(2))+ 4m_(1)a_(2) " " ( :' a_(1) = 2a_(2))`
` a_(2) ((m_(2)-2m_(1))/(m_(2)+4m_(1)))g`
From (i)
` a_(1) = (2(m_(2) - 2m_(1)))/((m_(1)+4m_(1)))g`
Note : For pulley II, downward force is 2T. This is so because the pulley is light and the net force on it ,must be zero .