An iron spherical ball has been melted and recast into smaller balls of equal size. If the radius of each of the smaller balls is \( 1 / 4 \) of the radius of the original ball, how many such balls are made? Compare the surface area, of all the smaller balls combined together with that of the original ball.

Given:

An iron spherical ball has been melted and recast into smaller balls of equal size.

The radius of each of the smaller balls is \( 1 / 4 \) of the radius of the original ball.

To do:

We have to find the number of balls made and compare the surface area of all the smaller balls combined together with that of the original ball.

Solution:

Let $R$ be the radius of the original ball.

This implies,

Radius of each smaller ball $r=\frac{1}{4} \mathrm{R}$

Volume of the original ball $V_1=\frac{4}{3} \pi \mathrm{R}^{3}$

Volume of each smaller ball $V_2=\frac{4}{3} \pi r^{3}$

$=\frac{4}{3} \pi(\frac{1}{4} \mathrm{R})^{3}$

$=\frac{1}{64} \times \frac{4}{3} \pi \mathrm{R}^{3}$

Number of smaller balls made $=V_1 \div V_2$

$=\frac{4}{3} \pi \mathrm{R}^{3} \div \frac{1}{64} \times \frac{4}{3} \pi \mathrm{R}^{3}$

$=\frac{4 \pi}{3} \mathrm{R}^{3} \times \frac{64 \times 3}{4 \pi \mathrm{R}^{3}}$

$=64$

Surface area of the original ball $=4 \pi \mathrm{R}^{2}$

Surface area of 64 smaller balls $=64 \times 4 \pi r^{2}$

$=256 \pi(\frac{1}{4} \mathrm{R})^{2}$

$=256 \pi \times \frac{1}{16} \mathrm{R}^{2}$

$=16 \pi \mathrm{R}^{2}$

Ratio of the surface areas $=16 \pi \mathrm{R}^{2}: 4 \pi \mathrm{R}^{2}$

$=16: 4$

$=4: 1$

Therefore, 64 such balls can be made.

Tutorialspoint

Updated on: 10-Oct-2022

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