Answer:2 m right
Explanation:
Given
First child is sitting 1 m left from Pivot point
Mass of second child is half of first
Let mass of first child be m
Second child [tex]\frac{m}{2}[/tex]
In order to balance see saw torque must be zero about pivot
[tex]m\times 1=\frac{m}{2}\times x[/tex]
x=2 m right from pivot point