From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm2.
Step-by-Step Solution
Key Concept: The total surface area of the solid after removing the conical cavity consists of (i) the curved surface area of the original cylinder, (ii) the curved surface area of the cone (which becomes an interior surface), and (iii) the area of the one circular base of the cylinder that remains untouched. The base of the cone is not counted because it is removed, and the opposite base of the cylinder becomes a point (vertex of the cone).
1. Given data
- Height of cylinder (and cone) \(h = 2.4\,\text{cm}\)
- Diameter = 1.4 cm \(\Rightarrow\) radius \(r = \frac{1.4}{2}=0.7\,\text{cm}\)
2. Slant height of the cone
\[ l = \sqrt{r^{2}+h^{2}} = \sqrt{0.7^{2}+2.4^{2}} = \sqrt{0.49+5.76}=\sqrt{6.25}=2.5\,\text{cm} \]
3. Curved surface area of the cylinder
\[ \text{CSA}_{\text{cyl}} = 2\pi r h = 2\pi(0.7)(2.4)=3.36\pi\ \text{cm}^{2} \]
4. Curved surface area of the cone (interior surface)
\[ \text{CSA}_{\text{cone}} = \pi r l = \pi(0.7)(2.5)=1.75\pi\ \text{cm}^{2} \]
5. Area of the remaining circular base of the cylinder
\[ \text{Base area}=\pi r^{2}=\pi(0.7)^{2}=0.49\pi\ \text{cm}^{2} \]
6. Total surface area of the remaining solid
\[ \text{TSA}=\text{CSA}_{\text{cyl}}+\text{CSA}_{\text{cone}}+\text{Base area}
=(3.36\pi+1.75\pi+0.49\pi)\ \text{cm}^{2}
=5.60\pi\ \text{cm}^{2} \]
7. Numerical value
\[ 5.60\pi \approx 5.60 \times 3.1416 = 17.59\ \text{cm}^{2} \]
Rounded to the nearest square centimetre, \(\boxed{18\ \text{cm}^{2}}\).
Hence, the total surface area of the solid after the conical cavity is removed is 18 cm² (nearest integer).
Correct Answer: 18 cm² (nearest integer)