A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains $ 41 \frac{19}{21} \mathrm{~m}^{3} $ of air. If the internal diameter of dome is equal to its total height above the floor, find the height of the building?

Given:

A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains \( 41 \frac{19}{21} \mathrm{~m}^{3} \) of air.

The internal diameter of dome is equal to its total height above the floor

To do:

We have to find the height of the building.

Solution:

Let the total height of the building be $2r$.

This implies,

Internal diameter of the dome $= 2r$

Radius of the dome $=\frac{2r}{2}$

$ = r$

Height of the cylindrical part $= 2r-r$

$= r$

Therefore,

Volume of the cylindrical part $=\pi r^{2}(r)$

$=\pi r^{3} \mathrm{~m}^{3}$

Volume of the hemispherical dome $=\frac{2}{3} \pi r^{3} \mathrm{~m}^{3}$

Total volume of the building $=$ Volume of the cylindrical part $+$ Volume of hemispherical dome

$=\pi r^{3}+\frac{2}{3} \pi r^{3}$

$=\frac{5}{3} \pi r^{3} \mathrm{~m}^{3}$

Volume of the building $=$ Volume of the air

$\Rightarrow \frac{5}{3} \pi r^{3}=41 \frac{19}{21}$

$\Rightarrow \frac{5}{3} \pi r^{3}=\frac{880}{21}$

$\Rightarrow r^{3}=\frac{880 \times 7 \times 3}{21 \times 22 \times 5}$

$\Rightarrow r^{3}=\frac{40 \times 21}{21 \times 5}$

$\Rightarrow r^{3}=8$

$\Rightarrow r^{3}=8$

$\Rightarrow r=2$

This implies,

Height of the building $=2 r$

$=2 \times 2$

$=4 \mathrm{~m}$

The height of the building is $4\ m$.

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Updated on: 10-Oct-2022

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