# 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?

Let r be the radius of the iron ball.

Volume of iron ball = Volume of water raised in the tub

$\begin{array}{l} \Rightarrow \frac{4}{3} \pi r^{3}=\pi r^{2} h \\ \Rightarrow \frac{4}{3} r^{3}=(16)^{2} \times 9 \\ \Rightarrow r^{3}=\frac{27 \times 16 \times 16}{4} \\ \Rightarrow r^{3}=1728 \\ \Rightarrow r=12 \mathrm{~cm} \end{array}$