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

Given: Base radius of the conical vessel $r_{1}=5\ cm$, height of the conical vessel, $h_{1}=24\ cm$, base radius of the cylinderical vessel $r_{2}=10\ cm$.

To do: To find the height of the rise of the water in cylinderical vessel, when the water is emptied in the in the cylinderical vessel.

Solution: Let the water will rise upto the height of the $h_{2}=24\ cm$.

$\because$ the water is emptied from conical vessel to the cylinderical vessel.

$\therefore$ Volume of the water in conical vessel$=$Volume of the water in cylinderical vessel 

As known volume of the cone$=\frac{1}{3} \pi r^{2} h$

and volume of the cone$=\pi r^{2} h$

On using the formulas to the given question,

We have,

$\frac{1}{3} \pi r^{2}_{1} h_{1} =\pi r^{2}_{2} h_{2} \ $

$\Rightarrow r^{2}_{1} h_{1} =3r^{2}_{2} h_{2}$

$\Rightarrow 5\times 5\times 24=3\times 10\times 10\times h_{2}$

$\Rightarrow 300h_{2} =600$

$\Rightarrow h_{2} =\frac{600}{300} =2\ cm$

Thus the water will rise upto the height of $2\ cm$ in the cylinderical vessel.
 

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Updated on: 10-Oct-2022

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