A spherical ball of radius $ 3 \mathrm{~cm} $ is melted and recast into three spherical balls. The radii of two of the balls are $ 1.5 \mathrm{~cm} $ and $ 2 \mathrm{~cm} $. Find the diameter of the third ball.

Given:

A spherical ball of radius \( 3 \mathrm{~cm} \) is melted and recast into three spherical balls.

The radii of two of the balls are \( 1.5 \mathrm{~cm} \) and \( 2 \mathrm{~cm} \).

To do:

We have to find the diameter of the third ball.

Solution:

Radius of the original spherical ball $R = 3\ cm$

Volume of the spherical ball $=4 \pi R^3$

$=4 \pi (3)^3$
$=4 \pi (27)$

$=104 \pi$

Radius of the first small ball $r_1=1.5\ cm$

Volume of the ball of radius $(r_1)=4 \pi (1.5)^3$

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

$=\frac{4}{3} \pi \times \frac{27}{8}$

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

Radius of the second ball $r_2=2\ cm$

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

$=\frac{4}{3} \pi \times 8$

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

Let the radius of the third small ball be $r_3$.

Therefore,

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

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

$=36 \pi-\frac{91}{6} \pi$

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

$=\frac{125}{6} \pi \mathrm{cm}^{3}$

Thi implies,

$\frac{4}{3} \times \pi(r_3)^{3}=\frac{125}{6} \pi \mathrm{cm}^{3}$

$(r_3)^{3}=\frac{3}{4 \pi} \times \frac{125}{6} \pi$

$r_3=\sqrt[3]{\frac{125}{6} \pi \times \frac{3}{4 \pi}}$

$r_3=\sqrt[3]{\frac{125}{8}}$

$r_3=\sqrt[3]{(\frac{5}{2})^{3}}$

$r_3=\frac{5}{2} \mathrm{~cm}$

$r_3=2.5 \mathrm{~cm}$

This implies,

Diameter of the third small ball $=2 \times r_3$

$=2 \times 2.5$

$=5 \mathrm{~cm}$

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

Tutorialspoint

Updated on: 10-Oct-2022

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