Two objects of mass 10 kg and 20 kg respectively are connected to the two ends of a rigid rod of length 10 m with negligible mass. The distance of the center of mass of the system from the 10 kg mass is :

Given, 
M₁ = 10 kg, M₂ = 20 kg, Length of the rod = 10 m

Let M₁ be at origin, then x₁ and x₂ is 0 m and 10 m, respectively.
Position of centre of mass from M₁ is
 $\mathrm{x}=\frac{({\mathrm{M}}_1{\mathrm{x}}_1+{\mathrm{M}}_2{\mathrm{x}}_2)}{{\mathrm{M}}_1+{\mathrm{M}}_2}$

Putting the value in the above equation, 
 $\mathrm{x}=\frac{10\times 0 + 20\times 10}{10+20}$
$\mathrm{x}=\frac{200}{30}=\frac{20}{3}$

The distance of the centre of mass of system from 10 kg mass is 20/3 m.