Mathematics — Standard (041) — CBSE Class 10 board question

Let the height of the cylindrical part = $h$ m.

Then, base radius of hemisphere = $r = \dfrac{h}{2}$ m.

Total volume of room = Volume of cylinder + Volume of hemisphere

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

$$= \pi \left(\frac{h}{2}\right)^2 h + \frac{2}{3}\pi \left(\frac{h}{2}\right)^3$$

$$= \frac{\pi h^3}{4} + \frac{2\pi h^3}{24} = \frac{\pi h^3}{4} + \frac{\pi h^3}{12} = \frac{3\pi h^3 + \pi h^3}{12} = \frac{4\pi h^3}{12} = \frac{\pi h^3}{3}$$

Given volume $= \dfrac{1408}{21}$ m³:

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

$$\frac{22}{7} \times \frac{h^3}{3} = \frac{1408}{21}$$

$$h^3 = \frac{1408}{21} \times \frac{21}{22} = \frac{1408}{22} = 64$$

$$h = 4 \text{ m}$$

The height of the cylindrical part is 4 m.

Source: Chapter 12, Section 12.3 — Volume of a Combination of Solids

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• Key setup: since the base radius of the hemisphere equals the radius of the cylinder, and $r = h/2$, express everything in terms of $h$.
• Volume = $\pi r^2 h + \frac{2}{3}\pi r^3$ (cylinder + hemisphere). Combine carefully to get $\frac{\pi h^3}{3}$.
• Substituting $\pi = \frac{22}{7}$ and the given volume simplifies to $h^3 = 64$, so $h = 4$ m.
• Show all algebraic steps clearly — 3-mark questions reward working, not just the final answer.