Question

The inner radius of a hollow iron sphere is 9 cm. If iron is 1 cm thick, volume of iron used to make the ball =

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

Answer 1

The correct option is C 1134.59 cm${}^3$

Given, Inner radius, $r_i=9\text{cm}$
Thichnes $=1\text{cm}$
Outer radius, $r_o=9+1=10\text{cm}$

$\therefore$ Inner volume of sphere$=\frac{4}{3}\pi (r_i{)}^3=\frac{4}{3}\times 3.14\times 9^3=3052.08{\text{cm}}^3$

Outer volume $=\frac{4}{3}\times \pi \times (r_o{)}^3=\frac{4}{3}\times 3.14\times {10}^3=4186.67{\text{cm}}^3$

Volume of Iron used $=\text{Outer volume}–\text{Inner volume}$
$=1134.59{\text{cm}}^3$