Question

A conical vessel, with a base radius of $5cm$ and a height $24cm$, is full of water. This water is emptied into a cylindrical vessel of base radius $10cm$. Find the height to which the water will rise in the cylindrical vessel.

Answer 1

We know that when one thing is emptied into the other thing then the volume is same for both the shapes. So

Step 1 - Finding the volume of the conical vessel

$$\begin{gathered} \mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{the}\mathrm{cone}=\frac{1}{3}{\mathrm{\pi r}}^2\mathrm{h} \\ \mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{cone}=\frac{1}{3}\times \frac{22}{7}\times 5\times 5\times 24 \\ \mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{cone}=\frac{1760}{7}{\mathrm{cm}}^2 \end{gathered}$$

Since, the volume of the water will be same and the water will take the shape of the cylinder whose radius is given. So, we can find the height of the water.

Step 2-

$\begin{array}{rcl}Volumeofwaterincylindricalvessel& =& Volumeofwaterinconicalvessel\\ {\mathrm{\pi r}}^2\mathrm{h}& =& \frac{1760}{7}\\ \frac{22}{7}\times 10\times 10\times \mathrm{h}& =& \frac{1760}{7}\\ \mathrm{h}& =& \frac{200}{100}\\ \mathrm{h}& =& 2\mathrm{cm}\end{array}$

Therefore, the height of water in the cylindrical vessel will be $2cm$