A building is in the form of a cylinder surmounted by a hemisphere valted dome and contains $41\dfrac{19}{21} m^3$ of air. If the internal diameter of dome is equal to the total height of the building, find the height of the building.

The below figure shows the building in the form of a cylinder surmounted by a hemisphere valted dome:

Let the radius of the dome be r.

∴ Internal diameter = 2r.

Given, internal diameter is equal to height.

∴ Height of building (h) = 2r.

Height of hemispherical area = r.

So, height of cylindrical area, h₁ = 2r - r = r.

Volume of building (V) = Volume of cylindrical area + Volume of hemispherical area.

$\therefore V = πr^2h_1 + \dfrac{2}{3}πr^3 \\[1em] \Rightarrow V = πr^2.r + \dfrac{2}{3}πr^3 \\[1em] \Rightarrow V = πr^3 + \dfrac{2}{3}πr^3 \\[1em] \Rightarrow V = \dfrac{5}{3}πr^3.$

Given, V = $41\dfrac{19}{21} = \dfrac{880}{21} \text{ m}^3$

$\therefore \dfrac{5}{3} \times \dfrac{22}{7} \times r^3 = \dfrac{880}{21} \\[1em] \Rightarrow r^3 = \dfrac{880 \times 3 \times 7}{5 \times 22 \times 21} \\[1em] \Rightarrow r^3 = \dfrac{18480}{2310} \\[1em] \Rightarrow r^3 = 8 \\[1em] \Rightarrow r = 2 \text{ m}.$

h = 2r = 2(2) = 4 m.

Hence, the height of the building is 4 m.