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. (a)g/2 upwards, g downwards
B. (b)g upwards, g/2 downwards
C. (c)g upwards, 2g downwards
D. (d)2g upwards, g downwards

Answer

A

Explanation

For equilibrium of mass m : T = mg
Now, the string is cut. Therefore, force is decreased on mass m upwards and downwards on mass 2m. Hence, net force on m is now mg in downward direction. Hence, the acceleration of ‘m’
`:. A_(m) = (mg)/(2m) = (g)/(2)` (downwords)
For ‘2m’ block net force will be mg in upward direction. Hence, acceleration of the block ‘2m will be 2m will be
`:. a_(2m) = (mg)/(2m) = (g)/(2)` (upwards)