System shown in fig is in equilibrium and at rest. The spring and string are massless now the string is cut. The acceleration of mass 2m and m just after the string is cut will be:

Options

A. `g//2` upwards , g downwards
B. g upwards, `g//2` down wards
C. g upwards, `2g` downwards
D. `2g` upwards, g downwards

Answer

A

Explanation

After string is cut, FBD of m
`a=(mg)/(m)=g darr`

FBD of 2m (when string is cut tension in the spring takes finite time to becomes zero. How ever tension in the string immediately become zero.)

`a=(3mg-2mg)/(2m)=g/2 uarr`