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(C) Let the total height of the building be equal to the internal diameter of the dome,which is $2r \, m$.Therefore,the radius of the building (or dom

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$4$

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(C) Let the total height of the building be equal to the internal diameter of the dome,which is $2r \, m$.Therefore,the radius of the building (or dome) is $\frac{2r}{2} = r \, m$.The height of the cylindrical part is $2r - r = r \, m$.Volume of the cylinder $= \pi r^2 h = \pi r^2 (r) = \pi r^3 \, m^3$.Volume of the hemispherical dome $= \frac{2}{3} \pi r^3 \, m^3$.Total volume of the building $=$ Volume of the cylinder $+$ Volume of the hemispherical dome.Total volume $= \pi r^3 + \frac{2}{3} \pi r^3 = \frac{5}{3} \pi r^3 \, m^3$.Given that the volume of air inside is $41 \frac{19}{21} \, m^3 = \frac{880}{21} \, m^3$.Equating the volumes: $\frac{5}{3} \pi r^3 = \frac{880}{21}$.Using $\pi = \frac{22}{7}$,we get $\frac{5}{3} 	imes \frac{22}{7} 	imes r^3 = \frac{880}{21}$.$\frac{110}{21} r^3 = \frac{880}{21}$.$r^3 = \frac{880}{110} = 8$.$r = \sqrt[3]{8} = 2 \, m$.Total height of the building $= 2r = 2 	imes 2 = 4 \, m$.

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

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