CBSE Class 10 • 2024 • 4 Marks

Surface Areas and Volumes: Combination of Solids (Cone + Hemisphere)

Official examination question with verified M1/A1 mark scheme and step-by-step mathematical reasoning.

Problem Statement

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $2\text{ cm}$ and the diameter of the base is $4\text{ cm}$. Determine the volume of the toy. (Take $\pi = 3.14$)

Verified Solution & Marking Scheme

Identify dimensions
Diameter of base $d = 4\text{ cm} \implies$ radius $r = 2\text{ cm}$. Radius of hemisphere $r = 2\text{ cm}$. Height of cone $h = 2\text{ cm}$.
Formulate combined volume
$\text{Volume of toy } V = \text{Volume of hemisphere} + \text{Volume of cone}$ $V = \frac{2}{3}\pi r^3 + \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi r^2 (2r + h)$
Substitute numerical values
$V = \frac{1}{3} \times 3.14 \times (2)^2 \times [2(2) + 2]$ $V = \frac{1}{3} \times 3.14 \times 4 \times [4 + 2] = \frac{1}{3} \times 3.14 \times 4 \times 6 = 3.14 \times 8 = 25.12\text{ cm}^3$
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