# The surface area of a solid metallic sphere is 616 cm2.

Question:

The surface area of a solid metallic sphere is $616 \mathrm{~cm}^{2}$. It is melted and recast into a cone of height $28 \mathrm{~cm}$. Find the diameter of the base of the cone so formed (Use $\left.\pi=22 / 7\right)$.

Solution:

The surface area of the metallic sphere is 616 square cm. Let the radius of the metallic sphere is r. Therefore, we have

$4 \pi r^{2}=616$

$\Rightarrow r^{2}=\frac{616 \times 7}{4 \times 22}$

$\Rightarrow r^{2}=7 \times 7$

$\Rightarrow r=7$

Therefore, the radius of the metallic sphere is 7 cm and the volume of the sphere is

$V_{1}=\frac{4}{3} \pi \times(7)^{3} \mathrm{~cm}^{3}$

The sphere is melted to recast a cone of height 28 cm. Let the radius of the cone is R cm. Therefore, the volume of the cone is

$V_{2}=\frac{1}{3} \pi \times(R)^{2} \times 28 \mathrm{~cm}^{3}$

Since, the volumes of the sphere and the cone are same; we have

$V_{1}=V_{2}$

$\Rightarrow \frac{4}{3} \pi \times(7)^{3}=\frac{1}{3} \pi \times(R)^{2} \times 28$'

$\Rightarrow R^{2}=\frac{4 \times(7)^{3}}{28}$

$\Rightarrow R^{2}=7^{2}$

$\Rightarrow R=7$

Hence, the diameter of the base of the cone so formed is two times its radius, which is $14 \mathrm{~cm}$.