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Surface Areas and Volumes
RD Sharma
CBSE
Grade 10

Question:

A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is $2\text{ cm}$ and diameter of 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.

Step-by-Step Solution

Key Concept: $r = 2\text{ cm}, h_{cone} = 2\text{ cm}$. Toy Vol $= \dfrac{1}{3} \pi (4)(2) + \dfrac{2}{3} \pi (8) = \dfrac{8\pi}{3} + \dfrac{16\pi}{3} = 8 \pi = 25.12\text{ cm}^3$.<br>Cylinder height $= 2 + 2 = 4\text{ cm}$. Cylinder Vol $= \pi (2)^2 (4) = 16 \pi = 50.24\text{ cm}^3$.<br>Difference $= 16 \pi - 8 \pi = 8 \pi = 25.12\text{ cm}^3$.
Toy Vol $= \dfrac{1}{3}\pi(4)(2) + \dfrac{2}{3}\pi(8) = 8\pi = 25.12\text{ cm}^3$. [1.5 Marks]
Cylinder Vol $= \pi(4)(4) = 16\pi = 50.24\text{ cm}^3$. [1.0 Mark]
Difference $= 16\pi - 8\pi = 8\pi = 25.12\text{ cm}^3$. [0.5 Mark]

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🎯 Official CBSE Marking Scheme:
Calculating toy volume $= 25.12\text{ cm}^3$: 1.5 Marks
Calculating circumscribing cylinder volume $= 50.24\text{ cm}^3$: 1.0 Mark
Evaluating volume difference $= 25.12\text{ cm}^3$: 0.5 Mark

Correct Answer:
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