Question

A right circular cone is divided by a plane parallel to its base in two equal volumes. In what ratio will the plane divide the axis of the cone

Answer 1

Let VAB be a cone of height $h$ and base radius $r$. Suppose it is cut by a plane parallel to the base of the cone at point $O^{\prime }$. Let $O^{\prime }{A}^{\prime }=r_1$ and $V{O}^{\prime }=h_1$

Clearly,
$\therefore \frac{VO}{V{O}^{\prime }}=\frac{OA}{O^{\prime }{A}^{\prime }}\Rightarrow \frac{h}{h_1}=\frac{r}{r_1}$

It is given that
Volume of cone VA'B' = volume of the frustum ABB'A'

$\frac{1}{3}\pi r_1^2{h}_1=\frac{1}{3}\pi (r^2+r_1^2+r{r}_1)(h-h_1)$

$\Rightarrow r_1^2{h}_1=(r^2+r_1^2+r{r}_1)(h-h_1)$

$\Rightarrow 1=\{{\left(\frac{r}{r_1}\right)}^2+1+\frac{r}{r_1}\}(\frac{h}{h_1}-1)\Rightarrow 1={\left(\frac{h}{h_1}\right)}^3-1^3\Rightarrow {\left(\frac{h}{h_1}\right)}^3=2\Rightarrow \frac{h}{h_1}=2^{1/3}$

Hence the ratio $=\frac{h_1}{h-h_1}=\frac{1}{(\frac{h}{h_1}-1)}=\frac{1}{2^{1/3}-1}$