A carpenter has constructed a toy as shown in the figure. If the density of the material of the sphere is 12 times that of cone, the position of the centre of mass of the toy is given by:

Let density of cone is $\mathrm{\rho }$, then density of sphere is 12$\mathrm{\rho }$.
Mass of cone:

$m_1 = \frac{1}{3}\mathrm{\pi }{\left(2\mathrm{R}\right)}^2\left(4\mathrm{R}\right)\mathrm{\rho } = \frac{16{\mathrm{\pi \rho R}}^3}{3}$

mass of sphere: $m_2 = \frac{4}{3}{\mathrm{\pi R}}^3\left(12\mathrm{\rho }\right) = \frac{48{\mathrm{\pi \rho R}}^3}{3}$

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

       = $\frac{{\displaystyle \frac{16{\mathrm{\pi \rho R}}^3}{3}}\left({\displaystyle \frac{4\mathrm{R}}{4}}\right)+{\displaystyle \frac{48}{3}}{\mathrm{\pi \rho R}}^3\left(5\mathrm{R}\right)}{{\displaystyle \frac{16{\mathrm{\pi \rho R}}^3}{3}}+{\displaystyle \frac{48{\mathrm{\pi \rho R}}^3}{3}}} = 4\mathrm{R}$