Question:
A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical iron ball is dropped into the tub and thus the level of water is raised by 6.75 cm. What is the radius of the ball?
Solution:
Suppose that the radius of the ball is r cm.
Radius of the cylindrical tub =12 cm
Depth of the tub= 20 cm
Now, volume of the ball = volume of water raised in the cylinder
$\Rightarrow \frac{4}{3} \pi r^{3}=\pi \times 12^{2} \times 6.75$
$\Rightarrow r^{3}=\frac{144 \times 6.75 \times 3}{4}=36 \times 6.5 \times 3=729$
$\Rightarrow r=9 \mathrm{~cm}$
∴ The radius of the ball is 9 cm.