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

A monkey climbs up and another monkey climbs down a rope hanging from

1.	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
Answer» `(G)/(5)` `6g` `(g)/(2)` `g` Solution :Let T be the tension in the ROPE. When MONKEY climbs up with acceleration a, `T-m_1 g = m_1 a ` When another monkey climbs down with same acceleration a, ` m_2 g - Tm_2 a ` Adding (i) and (ii), we get `(m_2-m_1 )g=(m_1 +m_2) a` `or(1- (m_1)/(m_2)) g=[(m_1)/(m_2)+1] a or (1- (2)/(3)) g= [(2)/(3) +1]a` ` or 1/3g= (5)/(3)a ora= (g)/(5)`