Question

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

Answer 1

Given, diameter of the vessel $=12cm$

So, radius of the vessel $=\frac{12}{2}=6cm$

We know that, volume of the sphere $=\frac{4}{3}$${\mathrm{\pi r}}^3$

On substituting the known values we get$=\frac{4}{3}$$\mathrm{\pi }{6}^3$

$=\frac{4}{3}\mathrm{\pi }\times 6\times 6\times 6$

$$\begin{gathered} =4\times \mathrm{\pi }\times 6\times 6\times 2 \\ =288\mathrm{\pi } \end{gathered}$$

Now, water displaced is equal to the volume of the sphere

Given, level of the water increased $h=3\frac{5}{9}$

$=$$\frac{32}{9}$

The shape of the given to be cylindrical.

Let $r$ is the radius of the cylinder.

Now, volume of the water displaced$=288\mathrm{\pi }$

$$\begin{gathered} \Rightarrow {\mathrm{\pi r}}^2\mathrm{h}=288\mathrm{\pi } \\ \Rightarrow {\mathrm{r}}^2\frac{32}{9}=288 \end{gathered}$$

$$\begin{gathered} \Rightarrow r^2=\frac{288\times 9}{32} \\ \Rightarrow r^2=81 \\ \Rightarrow r=9 \end{gathered}$$

Hence, diameter of the cylinder $=2r=2\times 9=18cm$