Question

A sphere of diameter $12cm,$ is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by $3\frac{5}{9}cm.$. Find the diameter of the cylindrical vessel.

Answer 1

Step 1: Finding the volume of the sphere

The diameter of the sphere$=12cm$

Therefore, the radius of the sphere$=r=\frac{12}{2}cm=6cm$

The volume of the sphere$=\frac{4}{3}{\mathrm{\pi r}}^3$

$=\frac{4}{3}\mathrm{\pi }{\left(6\right)}^3$

Therefore, the volume of the sphere is $\frac{4}{3}\mathrm{\pi }{\left(6\right)}^3$.

Step 2: Finding the volume of the water rise in the cylindrical vessel

The height of water rise in the cylindrical vessel$=3\frac{5}{9}=\frac{32}{9}cm$.

The volume of the water rise in the cylindrical vessel$={\mathrm{\pi r}}^2\mathrm{h}$

$={\mathrm{\pi r}}^2\left(\frac{32}{9}\right)$

Step 3: Finding the diameter of the cylindrical vessel

If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by $3\frac{5}{9}cm.$

The volume of the sphere$=$Volume the rise of water in the cylindrical vessel

$$\begin{gathered} \Rightarrow \frac{4}{3}\cancel{\mathrm{\pi }}{\left(6\right)}^3=\cancel{\mathrm{\pi }}{\mathrm{r}}^2\left(\frac{32}{9}\right) \\ \Rightarrow \frac{4}{3}{\left(6\right)}^3={\mathrm{r}}^2\left(\frac{32}{9}\right) \\ \Rightarrow {\mathrm{r}}^2=\frac{4\times 6\times 6\times 6\times 9}{3\times 32} \\ \Rightarrow {\mathrm{r}}^2=81 \\ \Rightarrow \mathrm{r}=\sqrt{81} \\ \Rightarrow \mathrm{r}=9 \end{gathered}$$

$\Rightarrow$The radius of the cylindrical vessel is $9cm$.

We know that the diameter of the cylindrical vessel is twice its radius .

$\therefore$Diameter of the cylindrical vessel$=2\times r$

$$\begin{gathered} =2\times 9 \\ =18cm \end{gathered}$$

Therefore, the diameter of the cylindrical vessel is $18cm$.