Answer:
Option C
Explanation:
Lert x1 and x2 be the position of masses m1 and m2, respectively
The position of centre of mass is
$X_{CM}$=$\frac{x_{1}m_{1}+x_{2}m_{2}}{m_{1}+m_{2}}$
If $\triangle x_{1}$ and $\triangle x_{2}$ be the changes in positions , then change in the position of the centre of mass,
$\triangle X_{CM}=\frac{\triangle x_{1}m_{1}+\triangle x_{2}m_{2}}{m_{1}+m_{2}}$
Given that , the centre of mass remains unchannged i.e, $\triangle$XCM =0 and $\triangle x_{1}$ =d
$\Rightarrow 0=\frac{dm_{1}+m_{2}\triangle x_{2}}{m_{1}+m_{2}}$
or $\triangle x_{2}=-\frac{m_{1}}{m_{2}} d$
