A hemisphere of lead of radius $7\ cm$ is cast into a right circular cone of height $49\ cm$. Find the radius of the base.

 Given:

A hemisphere of lead of radius $7\ cm$ is cast into a right circular cone of height $49\ cm$. 

To do:

We have to find the radius of the base.

Solution:

Radius of the hemisphere $(r) = 7\ cm$

This implies,

Volume of the hemisphere $=\frac{2}{3} \pi r^{3}$

$=\frac{2}{3} \times \pi \times 7 \times 7 \times 7$

$=\frac{686}{3} \pi \mathrm{cm}^{3}$

Therefore,

Volume of the cone $=\frac{686}{3} \pi \mathrm{cm}^{3}$

Height of the cone $=49 \mathrm{~cm}$

This implies,

Radius of the cone $=\sqrt{\frac{\text { Volume } \times 3}{\pi h}}$

$=\sqrt{\frac{686 \pi \times 3}{3 \times \pi \times 49}}$

$=\sqrt{\frac{686}{49}}$

$=\sqrt{14} \mathrm{~cm}$

$=3.74 \mathrm{~cm}$

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Updated on: 10-Oct-2022

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