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A Building is in the Form of a Cylinder Surmounted by a Hemi-spherical Vaulted Dome and Contains 41 19 21 M 3 of Air - Mathematics

Answer in Brief

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

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Solution

let the total height of the building be H m.

let the radius of the base be r m. Therefore the radius of the hemispherical dome is r m.

 Now given that internal diameter = total height

\[\Rightarrow 2r = H\]

Total height of the building = height of the cylinder +radius of the dome
⇒ H = h + r
⇒ 2r = h + r
⇒ r = h

Volume of the air inside the building = volume of the cylinder+ volume of the hemisphere

\[\Rightarrow 41\frac{19}{21} = \pi r^2 h + \frac{2}{3} \pi r^3 \]

\[ \Rightarrow \frac{880}{21} = \pi h^2 h + \frac{2}{3} \pi h^3 \]

\[ \Rightarrow \frac{880}{21} = \pi h^3 \left( 1 + \frac{2}{3} \right)\]

\[ \Rightarrow \frac{880}{21} = \pi h^3 \left( \frac{5}{3} \right)\]

\[ \Rightarrow h = 2 m\]

Hence, height of the building H = 2 × 2 = 4m

  Is there an error in this question or solution?
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APPEARS IN

RD Sharma Class 10 Maths
Chapter 14 Surface Areas and Volumes
Exercise 14.2 | Q 34 | Page 62
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