NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.4
Question 15

Question. 15

A building is a cylinder surmounted by a hemispherical dome and contains \(41\dfrac{19}{21}\,\text{m}^3\) of air. If the internal diameter of the dome equals the total height above the floor, find the height of the building.

Answer:

4 m

Detailed Answer with Explanation:

Step 1: Convert the given volume into an improper fraction.

\(41\dfrac{19}{21} = \dfrac{41 \times 21 + 19}{21} = \dfrac{861 + 19}{21} = \dfrac{880}{21}\,\text{m}^3.\)

Step 2: Let the radius of the hemispherical dome be \(r\,\text{m}.\)

The internal diameter of the dome = total height of the building.

So, \(2r = h + r \;\Rightarrow\; h = r.\)

This means the cylindrical part and the radius of the dome are equal in height.

Step 3: Write the formula for total volume of the building.

Total volume = Volume of cylinder + Volume of hemisphere

\(= \pi r^2h + \dfrac{2}{3}\pi r^3\)

Step 4: Substitute \(h = r\).

Volume \(= \pi r^2(r) + \dfrac{2}{3}\pi r^3\)

\(= \pi r^3 + \dfrac{2}{3}\pi r^3\)

\(= \dfrac{5}{3}\pi r^3.\)

Step 5: Equate with given volume.

\(\dfrac{5}{3}\pi r^3 = \dfrac{880}{21}\)

Step 6: Take \(\pi = \dfrac{22}{7}.\)

\(\dfrac{5}{3} \times \dfrac{22}{7} r^3 = \dfrac{880}{21}\)

\(\dfrac{110}{21} r^3 = \dfrac{880}{21}\)

Step 7: Simplify.

\(110 r^3 = 880\)

\(r^3 = 8\)

\(r = 2\,\text{m}.\)

Step 8: Find the height of the cylindrical part.

Since \(h = r = 2\,\text{m},\) the cylinder’s height = 2 m.

Step 9: Find the total height of the building.

Total height = Height of cylinder + Radius of dome

= \(2 + 2 = 4\,\text{m}.\)

Final Answer: The height of the building = 4 m.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 12: Surface Areas & Volumes – Exercise 12.4 | Detailed Answers