A rod AB of length 10 cm, has its mass per unit length varying uniformly with distance from 1kg/m (at end A) to 2Kg/m (at end B). The distance of centre of mass from end B is:

$\text{ Given : Mass per unit length, }\lambda =\frac{\left({\lambda }_2-{\lambda }_1\right)}{L}x+{\lambda }_1$

$\lambda =10x+1(\text{ in }\mathrm{kg}/\mathrm{m})$

$\text{ Center of mass is given by, }$

$x_{\mathrm{cm}}=\frac{\int \mathrm{xdm}}{\int \mathrm{dm}}=\frac{\int \mathrm{x}\left(\lambda \mathrm{dx}\right)}{\int \lambda \mathrm{dx}}=\frac{{\int }_0^{\mathrm{L}}\left(10{\mathrm{x}}^2+\mathrm{x}\right)\mathrm{dx}}{{\int }_0^{\mathrm{L}}(10\mathrm{x}+1)\mathrm{dx}}$

$\Rightarrow {\mathrm{x}}_{\mathrm{cm}}=\frac{10\frac{{\mathrm{L}}^3}{3}+\frac{{\mathrm{L}}^2}{2}}{10\frac{{\mathrm{L}}^2}{2}+\mathrm{L}}=\frac{1}{18} \mathrm{m}=\frac{50}{9} \mathrm{cm}$

$\text{ here }\mathrm{L}=0.1 \mathrm{m}$

So this will be at a distance of $\frac{40}{9}\mathrm{cm}$from end B