Question

The inner radius of a hollow iron sphere of thickness 1 cm is 9 cm. Find the volume of iron used to make the ball.

• A. $1155.08{cm}^3$
• B. $1156.08{cm}^3$
• C. $1134.59{cm}^3$
• D. $1133.58{cm}^3$

Answer 1

The correct option is C

$1134.59{cm}^3$

$\text{Thickness of the iron = 1cm}$

$\text{Inner radius = 9cm}$
$\text{Therefore, inner volume of sphere}$
$=\frac{4}{3}$ $\pi$ $(r_{inner}{)}^3$
$=\frac{4}{3}$ $\times$ $\pi$ $\times$ $(9{)}^3$
$=3052.08{cm}^3$

$\text{Outer radius = 10 cm}$
$\text{Outer volume of sphere}$
$=\frac{4}{3}$ $\times$ $\pi$ $\times$ $(r_{outer}{)}^3$
$=\frac{4}{3}$ $\pi$ $(10{)}^3$
$=4186.67{cm}^3$

Then,
iron used
= Outer volume – Inner volume
$=1134.59{cm}^3$