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Question
A building is in the form of a cylinder surmounted by a hemi-spherical vaulted dome and contains \[41\frac{19}{21} m^3\] of air. If the internal diameter of dome is equal to its total height above the floor , find the height of the building ?
Solution
let the total height of the building be H m.
let the radius of the base be r m. Therefore the radius of the hemispherical dome is r m.
Now given that internal diameter = total height
\[\Rightarrow 2r = H\]
Total height of the building = height of the cylinder +radius of the dome
⇒ H = h + r
⇒ 2r = h + r
⇒ r = h
Volume of the air inside the building = volume of the cylinder+ volume of the hemisphere
\[\Rightarrow 41\frac{19}{21} = \pi r^2 h + \frac{2}{3} \pi r^3 \]
\[ \Rightarrow \frac{880}{21} = \pi h^2 h + \frac{2}{3} \pi h^3 \]
\[ \Rightarrow \frac{880}{21} = \pi h^3 \left( 1 + \frac{2}{3} \right)\]
\[ \Rightarrow \frac{880}{21} = \pi h^3 \left( \frac{5}{3} \right)\]
\[ \Rightarrow h = 2 m\]
Hence, height of the building H = 2 × 2 = 4m
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