Question:
A cylindrical tub of radius 16 cm contains water to a depth of 30 cm. A spherical iron ball is dropped into the tub and thus level of water is raised by 9 cm. What is the radius of the ball?
Solution:
Let r be the radius of the iron ball
Radius of the cylinder = 16 cm
Then,
Volume of iron ball = Volume of water raised in the hub
$\frac{4}{3} \pi r^{3}=\pi r^{2} h$
$\frac{4}{3} r^{3}=(16)^{2} \times 9$
$\mathrm{r}^{3}=\frac{27 \times 16 \times 16}{4}$
$r^{3}=1728$
r = 12 cm
Therefore radius of the ball = 12 cm.