Q1. [2] § 12.2 Surface Area of a Combination of Solids § 12.4 Summary
A solid metallic sphere of radius $10.5$ cm is melted and recast into a number of smaller cones, each of radius $3.5$ cm and height $3$ cm. Find the number of cones so formed.
Previously asked in CBSE board exam
2022 30/1/1 Q2
Generated by claude-sonnet-4-6 · 2026-06-15 10:34 · grounding rag
Model Answer
Volume of sphere = $\dfrac{4}{3}\pi r^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times (10.5)^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 1157.625 = 4851 \text{ cm}^3$
Volume of one cone = $\dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times (3.5)^2 \times 3 = \dfrac{1}{3} \times \dfrac{22}{7} \times 12.25 \times 3 = 38.5 \text{ cm}^3$
Number of cones = $\dfrac{4851}{38.5} = \boxed{126}$
Source: Chapter 12, Section 12.3
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Explanation
- The key principle: volume of sphere = total volume of all cones formed (no material is lost in recasting).
- Use $r = 10.5$ cm for sphere; $r = 3.5$ cm, $h = 3$ cm for each cone.
- Examiner expects the formula, substitution, and final answer clearly shown — all three steps earn the 2 marks.
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