A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 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. (Take = 3.14)
Step-by-Step Solution
Key Concept: Use the standard volume formulas from NCERT: \(V_{cone}=\frac13\pi r^{2}h\), \(V_{hemisphere}=\frac23\pi r^{3}\), and \(V_{cylinder}=\pi r^{2}h\). The radius of the base is half the given diameter. The height of the cylinder equals the total height of the toy (height of cone + radius of hemisphere).
1. Find the radius
The diameter of the base is 4 cm, so \(r = \frac{4}{2} = 2\) cm.
2. Volume of the cone
Height of cone, \(h_c = 2\) cm.
\[V_{cone}=\frac13\pi r^{2}h_c = \frac13\times \pi \times 2^{2}\times 2 = \frac{8}{3}\pi\]
3. Volume of the hemisphere
Height of a hemisphere = its radius = 2 cm.
\[V_{hemisphere}=\frac23\pi r^{3}=\frac23\times \pi \times 2^{3}=\frac{16}{3}\pi\]
4. Total volume of the toy
\[V_{toy}=V_{cone}+V_{hemisphere}=\frac{8}{3}\pi+\frac{16}{3}\pi=\frac{24}{3}\pi=8\pi\]
Substituting \(\pi = 3.14\):
\[V_{toy}=8\times 3.14 = 25.12\ \text{cm}^{3}\]
5. Dimensions of the circumscribing cylinder
- Radius = same as toy = 2 cm.
- Height = height of cone + height of hemisphere = 2 cm + 2 cm = 4 cm.
6. Volume of the cylinder
\[V_{cyl}=\pi r^{2}h = \pi \times 2^{2}\times 4 = 16\pi\]
With \(\pi = 3.14\):
\[V_{cyl}=16\times 3.14 = 50.24\ \text{cm}^{3}\]
7. Difference of volumes
\[\Delta V = V_{cyl}-V_{toy}=16\pi-8\pi = 8\pi\]
Numerically,
\[\Delta V = 8\times 3.14 = 25.12\ \text{cm}^{3}\]
Correct Answer: Volume of the toy = 25.12 cm³; Difference between cylinder and toy = 25.12 cm³.