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 $ \pi=22 / 7 $ ).

Given:

The surface area of a solid metallic sphere is \( 616 \mathrm{~cm}^{2} \). It is melted and recast into a cone of height \( 28 \mathrm{~cm} \).

To do:

We have to find the diameter of the base of the cone so formed.

Solution:

Let $r$ be the radius of the solid sphere.

Surface area of the solid sphere $=4 \pi r^{2}$

$=616 \mathrm{~cm}^{2}$

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

$=\frac{616 \times 7}{4 \times 22}$

$=7^{2}$

$\Rightarrow r=7 \mathrm{~cm}$

Let $R$ be the radius of the base of the cone.

Therefore,

Volume of the cone $=\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{H}$
Volume of the cone $=$ Volume of the solid sphere

$\frac{1}{3} \pi \mathrm{R}^{2} \mathrm{H}=\frac{4}{3} \pi r^{3}$

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

$\mathrm{R}^{2}=\frac{4 \times 7 \times 7^{2}}{28}$

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

$Rightarrow \mathrm{R}=7\ cm$
Diameter of the base of the cone $=2 \mathrm{R}$

$=2 \times 7$

$=14 \mathrm{~cm}$

The diameter of the base of the cone so formed is $14\ cm$.

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Updated on: 10-Oct-2022

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