Question

A solid is in the form of a cone of vertical height $h$ mounted on the top base of a right circular cylinder of height $\frac{1}{3}h$. The circumference of the base of the cone and that of the cylinder are both equal to $C$. If $V$ be the volume of the solid, then prove that $C=\sqrt{\frac{6\pi V}{h}}$.

Answer 1

The volume of the cone is given by, $V_{cone}=\frac{1}{3}\pi r^{2}h$, where '$r$' is the base radius of the cylinder and cone.

The volume of the cylinder is given by, $V_{cyl}=\frac{1}{3}\pi r^{2}h$

The circumference of the base is given by, $C=2\pi r$ or $r=\frac{C}{2\pi }$

The total volume of the solid is $V=\frac{1}{3}\pi r^{2}h+\frac{1}{3}\pi r^{2}h=\frac{2}{3}\pi r^{2}h$
Substituting for $r$ in $V$,
$V=\frac{2}{3}\pi {\left(\frac{C}{2\pi }\right)}^{2}h\Rightarrow V=\frac{2}{3}\pi h\times \frac{C^2}{4{\pi }^2}\Rightarrow V=\frac{C^2}{6\pi }h\Rightarrow C=\sqrt{\frac{6\pi V}{h}}$.