A sphere of diameter $12\ cm$, 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 rised by $3\frac{5}{9}$. Find the diameter of the cylindrical vessel.

Given: Diameter of the sphere$=12\ cm$ rised water level in the cylindrical vessel$=3\frac{5}{9}$

To do: To Find the diameter of the cylindrical vessel.

Solution:

Radius of sphere, $r=6\ cm$

Volume of sphere$=\frac{4}{3} \pi r^{3}$

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

$=288\pi \ cm^{3}$

Let R be the radius of cylindrical vessel.

Rise in the water level of cylindrical vessel$=h=3\frac{5}{9} =\frac{32}{9} \ cm$

Increase in volume of cylindrical vessel$=\pi R^{2} h=\pi R^{2} \times \frac{32}{9}$

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

$\frac{32}{9} \pi R^{2} =288\pi$

$\therefore  R^{2} =\frac{288\times 9}{32}$

$\therefore \ R^{2} =81\ cm^{2}$

$\Rightarrow R=9\ cm$

$\therefore$ Diameter of the cylindrical vessel$=2\times \ R=2\ \times \ 9=18\ cm$


 

Updated on: 10-Oct-2022

84 Views

Related Articles

Kickstart Your Career

Get certified by completing the course

Get Started
Advertisements