In the system shown in fig. `m_(1) gt m_(2)` . The system is held at rest by thread BC. Now thread BC is burnt. Answer the following:

Just after burning the thread, what is the acceleration of `m_(2)`?

Options

A. `m_(1)g`
B. `m_(2)g`
C. Zero
D. Cannot say

Answer

A

Explanation

Before Burning BC, the free-body diagram are shown in the fig.

`T_(2)=T_(1)+m_(2)g` …(i)
`kx=T_(2)=m_(1)g`..(ii)
where x is extension in the spring. Just after burning, `T_(1)` will become zero. but `T_(2)` will remain same.
`T_(2)-m_(2)g=m_(2)g implies a = ((m_(1)-m_(2)g))/(m_(2))`
As `T_(2)` remains same, acceleration of block A will still remain zero.