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Surface Areas and Volumes
NCERT Exemplar Ch 11
CBSE_NCERT_EXEMPLAR_CH11
Grade 10

Question:

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. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Use $\pi = 3.14$)

Step-by-Step Solution

Key Concept: Radius $r = 2\text{ cm}$. Conical height $h = 2\text{ cm}$. Toy volume $= \text{Cone} + \text{Hemisphere}$. Cylinder radius $= 2\text{ cm}$, height $= 2 + 2 = 4\text{ cm}$.
Stepwise Solution:

Radius $r = 2\text{ cm}$, cone height $h = 2\text{ cm}$. [0.5 Mark]

Volume of cone $= \dfrac{1}{3} \pi r^2 h = \dfrac{1}{3} \times 3.14 \times 4 \times 2 = \dfrac{25.12}{3}\text{ cm}^3 = 8.37\text{ cm}^3$. [1.0 Mark]

Volume of hemisphere $= \dfrac{2}{3} \pi r^3 = \dfrac{2}{3} \times 3.14 \times 8 = \dfrac{50.24}{3}\text{ cm}^3 = 16.75\text{ cm}^3$.
Volume of toy $= \dfrac{25.12 + 50.24}{3} = \dfrac{75.36}{3} = 25.12\text{ cm}^3$. [1.5 Marks]

Cylinder circumscribing toy has radius $R = 2\text{ cm}$ and height $H = 2 + 2 = 4\text{ cm}$.
Volume of cylinder $= \pi R^2 H = 3.14 \times 4 \times 4 = 50.24\text{ cm}^3$. [1.0 Mark]

Difference of volumes $= 50.24 - 25.12 = 25.12\text{ cm}^3$. [1.0 Mark]

Marking Scheme:

• Calculating cone volume $= 8.37\text{ cm}^3$: 1.0 Mark
• Calculating hemisphere volume $= 16.75\text{ cm}^3$ and total toy volume $= 25.12\text{ cm}^3$: 1.5 Marks
• Calculating circumscribing cylinder volume $= 50.24\text{ cm}^3$: 1.5 Marks
• Difference in volumes $= 25.12\text{ cm}^3$: 1.0 Mark

Correct Answer:
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