Question

A building is in the form of a cylinder surmounted by a hemispherical dome.The base diameter of the dome is equal to $\frac{2}{3}$ of the total height of the building.Find the height of the building,if it contains 67$\frac{1}{21}{m}^3$ of air.

Answer 1

Let height of building = H
diameter of dome=diameter of cylinder= $\frac{2}{3}H$

$r=\frac{2}{3}H\times \frac{1}{2}=\frac{2}{6}H=\frac{1}{3}H$

height of cylinder = $H-\frac{1}{3}H=\frac{3H-H}{3}=\frac{2}{3}H$

volume = $67\times \frac{1}{21}=\frac{1408}{21}$ m

now volume of cylinder + volume of hemisphere = volume of building

$\frac{2}{3}\pi r^3+\pi r^{2}h=\frac{1408}{21}$

$\frac{2}{3}\times \frac{22}{7}\times \pi (\frac{H}{3}{)}^3+\frac{22}{7}\times (\frac{H}{3}{)}^2\times \frac{2H}{3}=\frac{1408}{21}$

$\frac{2\times 22}{21}\times \frac{H^3}{27}+\frac{22}{7}\times \frac{2H}{9}\times \frac{2H}{3}=\frac{1408}{21}$

$\frac{44}{21}\times \frac{H^3}{27}+\frac{22}{7}\times \frac{2H}{9}\times \frac{2H}{3}=\frac{1408}{21}$

$\frac{22}{7}\times \frac{H^3}{27}(\frac{2}{3}+2)=\frac{1408}{21}$

$\frac{H^3\times 8}{3}=64\times 9$

$H^3=\frac{64\times 9\times 3}{8}=8\times 27$

$H=\sqrt[3]{38\times 27}=2\times 3=6cm$\)

hence height of building = 6m