Two masses `m_(1)` and `m_(2)` are attached to a spring balance `S` as shown in figure. `m_(1) gt m_(2)` then the reading of spring balance will be .
.

Options

A. `(m_(1)-m_(2))`
B. `(m_(1)+m_(2))`
C. `(2m_(1)m_(2))/(m_(1)+m_(2))`
D. `(m_(1)m_(2))/(m_(1)+m_(2))`

Answer

C

Explanation

`F_("net") =ma`,From `FBD` of `m_(1), m_(1)g -T_(1) =m_(1)a`
From `FBD` of `m_(2),T_(2) -m_(2)g =m_(2)a`
take `T_(1) =T_(2)` solving the above eqs we get 'a' .