In the given diagram , find the relation between acceleration of blocks `m_(1) and m_(2)` . (`m_(2)` remains horizontal ).

Explanation

For bpulley of type II, we can write
` a_(2) = (a_(1) - a_(2))/2`
`rArr a_(1) = 3a_(2)`
Now , for `m_(2)`
` m_(2)g - 3T = m_(2)a_(2)`
or ` m_(2)g - 3T = (m_(2)a_(1))/3 " "` ….(i)
and for `m_(1)`
`T- m_(1)g = m_(1)a_(1) " "`...(ii)
By (i) `+ 3 xx `(ii),
` m _(2)g - 3m_(1)g = 3(m_(1) + m_(2)/9)a_(1)`
` rArr (a_(1) = 3(m_(2) - 3m_(1))g)/(9m_(1)+m_(2)),a_(1) = 3a_(2)`