Question:
A cylindrical tub of radius 12 cm contains water to a depth of 20 cm. A spherical form 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:
Radius of cylindrical tub = 12 cm
Depth = 20 cm
Let r be the radius of the ball
Then
Volume of the ball = Volume of water raised
$\frac{4}{3} \pi r^{3}=\pi r^{2} h$
$r^{3}=\frac{3.14 \times(12)^{2} \times 6.75 \times 3}{4}$
$\mathrm{r}^{3}=729$
$r=\sqrt[3]{729}$
r = 9 cm
Therefore radius of the ball = 9 cm