Radius of the original steel ball
$V = \frac{4}{3}\pi r^3$
$523.6 = \frac{4}{3}\pi r^3$
$r = 5 ~ \text{cm}$
The amount of material in the hollow steel ball is equal to the amount of material in the the original solid steel ball.
Let R = outer radius of the hollow steel ball.
$\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi r^3$
$\frac{4}{3}\pi R^3 = \frac{8}{3}\pi r^3$
$R^3 = 2r^3$
$R^3 = 2(5^3)$
$R = 6.3 ~ \text{cm}$
Thickness of the hollow steel ball
$t = R - r = 6.3 - 5$
$t = 1.3 ~ \text{cm}$ ← Answer: [ A ]
MATHalinoEngineering Math Review