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