Question

A uniform solid right circular cone of base radius $R$ is joined to a uniform solid hemisphere of radius $R$ and of the same density, as shown. The center of mass of the composite solid lies at the center of base of the cone. The height of the cone is

• A. $1.5R$
• B. $\sqrt{3}R$
• C. $3R$
• D. $2\sqrt{3}R$

Answer 1

The correct option is B $\sqrt{3}R$

REF.Image.

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

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

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

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

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

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

or $p\times \frac{1}{3}\pi r^2[\frac{h^2}{4}-2r\times \frac{3r}{8}]=0$

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