An empty cone is of radius 3 cm and height 12 cm. Ice-cream is filled in it so that lower part of the cone which is $\left(\dfrac{1}{6}\right)^{th}$ of the volume of the cone is unfilled but hemisphere is formed on the top. Find volume of the ice-cream. (Take $\pi = 3.14$)
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer
Given: Cone radius $r = 3$ cm, height $h = 12$ cm.
Volume of cone:
$$V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3} \times 3.14 \times 9 \times 12 = 113.04 \text{ cm}^3$$
Volume of ice-cream in cone (cone is $\frac{1}{6}$ unfilled):
$$V_{cone\,filled} = \frac{5}{6} \times 113.04 = 94.2 \text{ cm}^3$$
Volume of hemisphere (radius = 3 cm):
$$V_{hemi} = \frac{2}{3}\pi r^3 = \frac{2}{3} \times 3.14 \times 27 = 56.52 \text{ cm}^3$$
Total volume of ice-cream:
$$= 94.2 + 56.52 = \boxed{150.72 \text{ cm}^3}$$
Source: Volume of a Combination of Solids, Chapter 12
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Explanation
- The cone is $\frac{1}{6}$ unfilled, so ice-cream fills $\frac{5}{6}$ of the cone's volume.
- The hemisphere on top has the same radius as the cone's open end (3 cm).
- Add the two volumes: filled cone part + hemisphere.
- Examiners expect clear formula, substitution, and final answer with units. Don't forget to multiply by $\frac{5}{6}$ — a common error is using the full cone volume.