Question

A hemisphere of radius 21cm, made up of iron is melted and moulded to form a cone of height 21 cm. What is the ratio of the curved surface area of the cone to that of the hemisphere?

• A. $1:2$
• B. $2\sqrt{3}:\sqrt{5}$
• C. $2\sqrt{2}:3$
• D. $\sqrt{3}:\sqrt{2}$

Answer 1

The correct option is D

$\sqrt{3}:\sqrt{2}$

Given, radius of hemisphere and height of cone = 21 cm
Let $r$ be the radius of the cone.

Volume of the hemisphere $=\frac{2}{3}\pi (21{)}^3=6174\pi c{m}^3$

The hemisphere is molded into a cone, hence, the volume of cone will be equal to the volume of the hemisphere.

$\frac{1}{3}\pi \times r^2\times 21=6174\pi r=21\sqrt{2}cm$

Slant height of the cone, $l^2=h^2+r^2={21}^2+\left(2\right){.21}^2\Rightarrow l=21\sqrt{3}cm$

Curved surface area of the cone $=\pi rl=\pi \left(21\sqrt{2}\right)\left(21\sqrt{3}\right)=\pi (21{)}^2\sqrt{2}.\sqrt{3}$

Curved surface area of the hemisphere $=2\pi r^2=2\pi (21{)}^2$

Ratio of the curved surface areas of cone to hemisphere = $\frac{\pi (21{)}^2\sqrt{2}.\sqrt{3}}{2\pi (21{)}^2}$

$=\sqrt{3}:\sqrt{2}$