​ A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains 41 19/21 m^3 of air. ​

Question

A building is in the form of a cylinder surmounted by a hemispherical vaulted dome and contains \(41\frac{19}{21}\) m³ of air. If the internal diameter of the dome is equal to its total height above the floor, find the height of the building?

(a) 2 m 

(b) 6 m 

(c) 4 m 

(d) 8 m

Answer 1

Answer :(c) = 4 m

Let the height of the building 

= internal diameter of the dome = 2r m. 

∴  Radius of the building = radius of dome = \(\frac{2r}{2}\) = r m.

Height of cylindrical portion = 2r – r = r m.

Volume of the cylinder = πr²(r) = πr³  m³

Volume of hemispherical dome = \(\frac{2}{3} \pi r^3 \, m^3\) 

∴ Total volume of the building 

= \(\pi r^3 +\frac{2}{3}\pi r^3 = \frac{5}{3} \pi r^3 \,m^3\)  

Given,

\(\frac{5}{3}\pi r^3 = 41\frac{19}{21} = \frac{880}{21} \implies r^3 = \frac{880\times 7 \times 3}{5\times 22\times21} =8\)  

∴ r = \(\sqrt[3]{8}\) 

= 2 m

Hence, height of the building = 2r = 4 m.