A room is in the form of cylinder surmounted by a hemi-spherical dome. The base radius of hemisphere is one-half the height of cylindrical part. Find total height of the room if it contains $\left(\dfrac{1408}{21}\right)$ m³ of air. (Take $\pi = \dfrac{22}{7}$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Let the base radius of hemisphere = $r$, then height of cylinder $h = 2r$ (given $r = \frac{h}{2}$).
Total volume of air = Volume of cylinder + Volume of hemisphere
$$\frac{1408}{21} = \pi r^2 h + \frac{2}{3}\pi r^3$$
$$\frac{1408}{21} = \pi r^2(2r) + \frac{2}{3}\pi r^3 = 2\pi r^3 + \frac{2}{3}\pi r^3 = \frac{8}{3}\pi r^3$$
$$\frac{1408}{21} = \frac{8}{3} \times \frac{22}{7} \times r^3$$
$$r^3 = \frac{1408}{21} \times \frac{3 \times 7}{8 \times 22} = \frac{1408 \times 21}{21 \times 176} = \frac{1408}{176} = 8$$
$$r = 2 \text{ m}$$
Total height = $h + r = 2r + r = 3r = 3 \times 2 = \boxed{6 \text{ m}}$
Source: Chapter 12, Section 12.3 – Volume of a Combination of Solids
---
Explanation
- The key condition is: radius of hemisphere = half the height of cylinder, i.e., $r = h/2$, so $h = 2r$.
- Total volume = cylinder + hemisphere (not half-sphere surface — full volumes add up).
- Simplify to a single variable $r$, substitute $\pi = 22/7$, and solve for $r$.
- Total height = height of cylinder + radius of hemisphere (since the dome sits on top) = $2r + r = 3r$.
- Examiners expect the setup equation, substitution, and final answer clearly labelled.