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

Options

A. `(g)/(5)`
B. `6g`
C. `(g)/(2)`
D. `g`

Answer

A

Explanation

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 - T m_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 1/3 g= (5)/(3) a or a= (g)/(5)`