Question

The radius of a solid metallic sphere is $r$. A solid metallic cone of height $h$ has base radius $r$. The two are melted together and recast into a solid right circular cone with base radius $r$. Prove that the height of the resulting cone is $4r+h.$

Answer 1

Volume of resulting cone $=$ Volume of sphere $+$ Volume of cone which are melted.
Volume of a sphere of radius ${}^{\prime }{r}^{\prime }=\frac{4}{3}\pi r^3$
Volume of a cone $=\frac{1}{3}\pi r^{2}h$

where, $r$ is the radius of the base of the cone and $h$ is the height.
Let the height of the resulting cone be $H$.
Hence, $\frac{1}{3}\pi r^{2}H=\frac{4}{3}\pi r^3+\frac{1}{3}\pi r^{2}h$
$=>H=4r+h$
Hence, height of the resulting cone is $4r+h$