Question

A hemisphere of radius 21 cm, 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 hemisphere to that of the cone?

• 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,
The radius of the hemisphere = 21 cm
The height of the cone is = 21 cm
Let the radius of the cone be r.

Thus, the volume of the hemisphere
$=\frac{2}{3}\pi r^3=\frac{2}{3}\pi \times (21{)}^3=6174\pi c{m}^3$

And, curved surface area of the hemisphere is
$=3\pi r^2=3\pi \times (21{)}^2=1323\pi c{m}^2$

When the hemisphere is melted and moulded to form a cone, the volume of both the solids remains same.

Now,
$\text{Volume of cone = Volume of hemisphere}$
$\Rightarrow \frac{1}{3}\pi r^{2}h=6174\pi$
$\Rightarrow \frac{1}{3}\pi \times r^2\times 21=6174\pi$
$\Rightarrow 7r^2=6174\Rightarrow r=21\sqrt{2}cm$

Thus, the radius of the cone $=21\sqrt{2}cm$

Curved surface of cone $=\pi rl$
where, l = lateral height of the cone
$and{l}^2=r^2+h^2$
Solving it by putting values of r and h of the cone,
we get, lateral height of cone
$l=21\sqrt{3}cm$

Thus, curved surface area of cone
$=\pi rl=\pi \times 21\sqrt{2}\times 21\sqrt{3}=441\sqrt{6}\pi c{m}^2$

Now, the ratio of curved surface area of hemisphere and cone is
$=\frac{1323\pi c{m}^2}{441\sqrt{6}\pi c{m}^2}=\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{3}:\sqrt{2}$