A juice seller was serving his customers using glasses as shown in Fig. 12.13. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use = 3.14.)
Step-by-Step Solution
Key Concept: Use the formula for volume of a cylinder \(V_{cyl}=\pi r^{2}h\) to obtain the apparent capacity. The bottom contains a hemispherical protrusion, whose volume is half of a sphere: \(V_{hem}=\frac{2}{3}\pi r^{3}\). The actual capacity is the cylindrical volume minus the hemispherical volume.
1. Given data
- Inner diameter of glass = 5 cm \(\Rightarrow\) radius \(r = \frac{5}{2}=2.5\) cm.
- Height of glass (cylindrical part) = 10 cm.
- \(\pi = 3.14\).
2. Apparent capacity (volume of cylinder)
\[V_{\text{cyl}} = \pi r^{2} h = 3.14 \times (2.5)^{2} \times 10\]
\[(2.5)^{2}=6.25\]
\[V_{\text{cyl}} = 3.14 \times 6.25 \times 10 = 3.14 \times 62.5 = 196.25 \text{ cm}^{3}\]
Hence, apparent capacity = 196.25 cm³.
3. Volume of the hemispherical raised portion
Volume of a sphere = \(\frac{4}{3}\pi r^{3}\).
Therefore, volume of a hemisphere = \(\frac{1}{2}\times \frac{4}{3}\pi r^{3}=\frac{2}{3}\pi r^{3}\).
\[V_{\text{hem}} = \frac{2}{3} \times 3.14 \times (2.5)^{3}\]
\[(2.5)^{3}=15.625\]
\[V_{\text{hem}} = \frac{2}{3} \times 3.14 \times 15.625 = \frac{2 \times 49.0625}{3}= \frac{98.125}{3}= 32.7083 \text{ cm}^{3}\]
Approximate to two decimal places: 32.71 cm³.
4. Actual capacity
\[V_{\text{actual}} = V_{\text{cyl}} - V_{\text{hem}} = 196.25 - 32.71 = 163.54 \text{ cm}^{3}\]
Hence, actual capacity = 163.54 cm³ (approximately).
5. Result
- Apparent capacity = 196.25 cm³
- Actual capacity = 163.54 cm³
Correct Answer: Apparent capacity = 196.25 cm³; Actual capacity \(\approx 163.54\) cm³.