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 $T-m_{1}g=m_{1}a$ ……….(i)

 When another monkey goes, down with same acceleration a, then $m_{2}g-T=m_{2}a$                   …… (ii)

Hence, $\left(m_2-m_1\right)g-T=\left(m_1+m_2\right)a$ (or) $\left(1-\frac{m_1}{m_2}\right)g=\left(\frac{m_1}{m_2}+1\right)a$ (or)

$\left(1-\frac{2}{3}\right)g=\left(\frac{2}{3}+1\right)a$ (or) $a=\frac{g}{5}$