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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