A building is in the form of a cylinder surmounted by a hemispherical dome. The base diameter of the dome is equal to $ \frac{2}{3} $ of the total height of the building. Find the height of the building, if it contains $ 67 \frac{1}{21} \mathrm{~m}^{3} $ of air.

Given:

A building is in the form of a cylinder surmounted by a hemispherical dome.

The base diameter of the dome is equal to \( \frac{2}{3} \) of the total height of the building.

It contains \( 67 \frac{1}{21} \mathrm{~m}^{3} \) of air. 

To do:

We have to find the height of the building.

Solution:

Let the radius of the hemispherical dome be $r$ and the total height of the building be $h$.

The base diameter of the dome $=\frac{2}{3} \times$ The total height

This implies,

$2r= \frac{2}{3}h$

$r= \frac{h}{3}$

Let the height of the cylindrical portion be $H$.

Therefore,

$H = h - \frac{h}{3}$

$=\frac{2}{3}h$ 

Volume of the air inside the building $=$ Volume of the air inside the dome $+$ Volume of the air inside the cylindrical part

$=\frac{2}{3} \pi r^3 + \pi r^2 H$

$=\frac{2}{3} \pi (\frac{h}{3})^{3}+\pi(\frac{h}{3})^{2}(\frac{2}{3} h)$

$=\frac{8}{81} \pi h^{3}$

Volume of the air inside the building $= 67 \frac{1}{21} \mathrm{~m}^{3}$

Therefore,

$\frac{8}{81} \pi h^{3}=\frac{1408}{21}$

$\Rightarrow h^{3}=\frac{1408 \times 81 \times 7}{21 \times 8 \times 22}$

$\Rightarrow h^{3}=216$

$\Rightarrow h^{3}=(6)^{3}$

$\Rightarrow h=6 \mathrm{~m}$

The height of the building is $6\ m$.

Tutorialspoint

Updated on: 10-Oct-2022

447 Views

Kickstart Your Career

Get certified by completing the course

Get Started