Two uniform rods A and B of lengths 5m and 3m are placed end to end .If their linear Densities are 3 kg/m and 2 kg/m, the position of their center of mass from their interface is

Length of the rod A = 5 m

Length of the rod B = 3 m

${\mathrm{\lambda }}_1=\frac{{\mathrm{m}}_1}{\mathrm{l}}\Rightarrow {\mathrm{m}}_1={\mathrm{\lambda }}_1\times l$

$3\times 5={\mathrm{m}}_1$

$m_1=15$

$\mathrm{x}=-2.5$

${\mathrm{\lambda }}_2=\frac{{\mathrm{m}}_2}{l}\Rightarrow {\mathrm{m}}_2={\mathrm{\lambda }}_2\times l=2\times 3=6$

${\mathrm{m}}_2=6 {\mathrm{x}}_2=1.5$

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{m}}_1{\mathrm{x}}_1+{\mathrm{m}}_2{\mathrm{x}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

$=\frac{15\left(-2.5\right)+6\left(1.5\right)}{15+6}$

$=\frac{-37.5+9}{2}=\frac{-19}{14}=\frac{19}{14}\mathrm{m}$

On the side of heavier rod