Question

A hemisphere and a solid cone of same density have a common base. The centre of mass of the common structure coincides with the centre of the common base. If $R$ is the radius of hemisphere and $h$ is height of the cone, then :

• A. $\frac{h}{R}=\sqrt{3}$
• B. $\frac{h}{R}=\frac{1}{\sqrt{3}}$
• C. $\frac{h}{R}=3$
• D. $\frac{h}{R}=\frac{1}{3}$

Answer 1

The correct option is A $\frac{h}{R}=\sqrt{3}$

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

Mass of cone,$m_1=p\times \frac{1}{3}\pi r^{2}h$

mass of hemisphere,$m_2=p\times \frac{1}{2}\times \frac{4}{3}\pi r^3$

,$m_2=p\frac{2}{3}\pi r^3$

$y=\frac{m_1{y}_1+m_2{y}_2}{m_1+m_2}$

$0=\frac{p\times \frac{1}{3}\pi r^{2}h\times h/4+p\frac{2}{3}\pi r^3\times (-3r/8)}{p\times \frac{1}{3}\pi r^{2}h+p\frac{2}{3}\pi r^3}$

After calculation-

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

So,

$h=\sqrt{3}r$