Question

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

Answer 1

Let R be the radius, and h be the height of the cylindrical portion.

Height of the building, R+h = 2R = D = internal diameter.

⇒ h = R.

Volume = Volume of cylindrical portion + Volume of hemispherical portion

⇒ $\pi R^{2}h+\frac{\frac{4}{3}\pi R^3}{2}=\frac{880}{21}$

⇒ $\frac{22}{7}(R^{2}h+\frac{2R^3}{3})=\frac{880}{21}$

⇒$R^2(h+\frac{2R}{3})=\frac{40}{3}$

⇒ $R^2(R+\frac{2R}{3})=\frac{40}{3}$

⇒$R^3\left(\frac{5}{3}\right)=\frac{40}{3}$

⇒$R^3=8$

∴R = 2 m

and Total Height of the building, = R+h = 2R=2(2) = 4 m