Question

A uniform solid right circular cone of the base radius $r$ is joined to a uniform solid hemisphere of radius $r$ 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

• A. $2r$
• B. $\sqrt{3}r$
• C. $3r$
• D. $\sqrt{6}r$

Answer 1

Answer: (B)

Volume of cone $=\frac{1}{3}\pi r^{2}h$
Mass of cone, $m_1=\rho \times \frac{1}{3}\pi r^{2}h$
Mass of hemisphere, $m_2=\rho \times \frac{1}{2}\times \frac{4}{3}\pi r^3$
$=\rho \times \frac{2}{3}\pi r^3$

Now, $Y=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$
$\Rightarrow 0=\frac{\rho \times \frac{1}{3}\pi r^{2}h\times \frac{h}{4}+\frac{2}{3}\pi r^3(-\frac{3r}{8})}{\rho \times \frac{1}{3}\pi r^{2}h+\rho \times \frac{2}{3}\pi r^2}$
$\Rightarrow \rho \times \frac{1}{3}\pi r^2\left(\frac{h^2}{4}-2r\times \frac{3r}{8}\right)=0$
$\frac{h^2}{4}-\frac{3r^2}{4}=0$
or $h=\sqrt{3}r$