Question

A cylinder of radius 12 cm contains water upto the height of 20 cm. A spherical iron ball is dropped into the cylinder and thus water level is raised by 6.75 cm. What is the radius of the ball?

Answer 1

Given:
Radius of the cylinder (R) = 12 cm
Level of water in the cylinder (H) = 20 cm
Now,
Let the radius of the iron ball be r cm.
When the iron ball is dropped into the cylinder, the water level is raised by 6.75 cm.
Thus, we have:
Volume of the iron ball = Volume of the water raised
$$\begin{gathered} \Rightarrow \frac{4}{3}\mathrm{\pi }{r}^{\mathit{3}}\mathit{=}\mathrm{\pi }{R}^{\mathit{2}}h \\ \mathit{\Rightarrow }\frac{\mathit{4}}{\mathit{3}}{r}^{\mathit{3}}\mathit{=}{12}^{\mathit{2}}\mathit{\times }6.75 \\ \mathit{\Rightarrow }{r}^{\mathit{3}}=\frac{144\times 6.75\times 3}{4}\mathit{} \\ \mathit{\Rightarrow }{r}^{\mathit{3}}\mathit{=}729 \\ \mathit{\Rightarrow }r\mathit{=}\sqrt[\mathit{3}]{\mathit{729}}=9 \end{gathered}$$

Hence, the radius of the iron ball is 9 cm.