A monkey climbs up and another monkey climbs down a rope hanging from a tree with same uniform acceleration separately.  If the respective masses of monkeys are in the ratio 2 : 3, the common acceleration must be

Let T is the tension of the rope.  When monkey climbs up with acceleration a, then T−m1g=m1a ……….(i)  When another monkey goes, down with same acceleration a, then m2g−T=m2a                   …… (ii)Hence, (m2−m1)g−T=(m1+m2)a (or) (1−m1m2)g=(m1m2+1)a (or)(1−23)g=(23+1)a (or) a=g5