A uniform solid right circular cone of base radius r is joined to a uniform solid hemisphere of radius r and…

A uniform solid right circular cone of base radius $\mathrm{r}$ is joined to a uniform solid hemisphere of radius $\mathrm{r}$ and of the same density, so as to have a common face. The centre of mass of the composite solid lies on the common face. The height of the cone is

1. $2\mathrm{r}$
2. $\sqrt{3}\mathrm{r}$
3. $3\mathrm{r}$
4. $\sqrt{6}\mathrm{r}$

Solution

Volume of cone$=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

Mass of cone, ${\mathrm{m}}_1=\mathrm{\rho }\times \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}$

Mass of hemisphere, ${\mathrm{m}}_2=\mathrm{\rho }\times \frac{1}{2}\times \frac{4}{3}{\mathrm{\pi r}}^3$

                                        $=\mathrm{\rho }\times \frac{2}{3}{\mathrm{\pi r}}^3$



Now, $\mathrm{Y}=\frac{{\mathrm{m}}_1{\mathrm{y}}_1+{\mathrm{m}}_2{\mathrm{y}}_2}{{\mathrm{m}}_1+{\mathrm{m}}_2}$

$\Rightarrow 0=\frac{\mathrm{\rho }\times \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}\times \frac{\mathrm{h}}{4}+\frac{2}{3}{\mathrm{\pi r}}^3(-\frac{3\mathrm{r}}{8})}{\mathrm{\rho }\times \frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h}+\mathrm{\rho }\times \frac{2}{3}{\mathrm{\pi r}}^2}$
$\Rightarrow \mathrm{\rho }\times \frac{1}{3}{\mathrm{\pi r}}^3\left(\frac{{\mathrm{h}}^2}{4}-2\mathrm{r}\times \frac{3\mathrm{r}}{8}\right)=0$

$\frac{{\mathrm{h}}^2}{4}-\frac{3{\mathrm{r}}^2}{4}=0$ or  $\mathrm{h}=\sqrt{3}\mathrm{r}$

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