## Subject:

**Problem**

A 523.6 cm^{3} solid spherical steel ball was melted and remolded into a hollow steel ball so that the hollow diameter is equal to the diameter of the original steel ball. Find the thickness of the hollow steel ball.

A. 1.3 cm | C. 1.2 cm |

B. 1.5 cm | D. 1.6 cm |

**Answer Key**

[ A ]

**Solution**

Radius of the original steel ball
$V = \frac{4}{3}\pi r^3$
$\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi r^3$
$t = R - r = 6.3 - 5$

$523.6 = \frac{4}{3}\pi r^3$

$r = 5 ~ \text{cm}$

Volume of the hollow steel ball is equal to the volume of the original steel ball.

Let *R* = outer radius of the hollow steel ball.

$\frac{4}{3}\pi R^3 = \frac{8}{3}\pi r^3$

$R^3 = 2r^3$

$R^3 = 2(5^3)$

$R = 6.3 ~ \text{cm}$

Thickness of the hollow steel ball

$t = 1.3 ~ \text{cm}$ ← Answer: [ A ]