A spherical ball of radius $3\ cm$ is melted and recast into three spherical balls. The radii of the two of the balls are $1.5\ cm$ and $2\ cm$ respectively. Determine the diameter of the third ball.

Given:

Radius of the spherical ball $=3\ cm$.

The radii of the two of the balls are $1.5\ cm$ and $2\ cm$ respectively.

To do:

We have to find the diameter of the third ball.

Solution:

Let $r_{3}$ be the radius of the third ball.

Volume of the spherical ball $=\frac{4}{3} \pi(\mathrm{R})^{3}$

$=\frac{4}{3} \pi(3)^{3}$

$=36 \pi \mathrm{cm}^{3}$

Radius of the first small ball recast $r_{1}=1.5 \mathrm{~cm}$

$=\frac{3}{2} \mathrm{cm}$

This implies,

Volume of the first small ball recast $=\frac{4}{3} \pi(r_{1})^{3}$

$=\frac{4}{3} \pi(\frac{3}{2})^{3}$

$=\frac{9 \pi}{2} \mathrm{~cm}^{3}$

Radius of the second ball recast $r_{2}=2 \mathrm{~cm}$

This implies,

Volume of the second ball $=\frac{4}{3} \pi(r_{2})^{3}$

$=\frac{4}{3} \pi \times(2)^{3}$

$=\frac{32}{3} \pi \mathrm{cm}^{3}$

Volume of the spherical ball $=$ Sum of the volumes of the three small balls

Therefore,

Volume of the third ball $=36 \pi-(\frac{9}{2} \pi+\frac{32}{3} \pi)$

$=36 \pi-(\frac{27+64}{6} \pi)$

$=\frac{216 \pi-91 \pi}{6}$

$=\frac{125}{6} \pi$

This implies,

$\frac{4}{3} \pi(r_{3})^{3}=\frac{125}{6} \pi$

$\Rightarrow r_{3}^{3}=\frac{125}{6} \pi \times \frac{3}{4 \pi}$

$=\frac{125}{8}$

$=(\frac{5}{2})^{3}$

$\Rightarrow r_{3}=\frac{5}{2} \mathrm{~cm}$

Diameter of the third ball $=2 r_{3}$

$=2 \times \frac{5}{2}$

$=5 \mathrm{~cm}$

The diameter of the third ball is $5\ cm$.

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Updated on: 10-Oct-2022

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