A metal pipe has a length of 8.9 cm. The internal and external diameters of the pipe are 4.2 cm and 5 cm respectively. (i) Find the volume of metal used in making the pipe. (ii) If the metal costs $8 per kg and the density of the metal is 2700 kg/m³, find the cost of the pipe.
The Correct Answer and Explanation is:
To solve the problem, we break it into two parts:
Given Data:
- Length of pipe, h=8.9 cmh = 8.9 \, \text{cm}h=8.9cm
- Internal diameter din=4.2 cmd_{\text{in}} = 4.2 \, \text{cm}din=4.2cm → radius rin=2.1 cmr_{\text{in}} = 2.1 \, \text{cm}rin=2.1cm
- External diameter dout=5.0 cmd_{\text{out}} = 5.0 \, \text{cm}dout=5.0cm → radius rout=2.5 cmr_{\text{out}} = 2.5 \, \text{cm}rout=2.5cm
- Density of metal = 2700 kg/m32700 \, \text{kg/m}^32700kg/m3
- Cost of metal = $8 per kg
Part (i): Volume of metal used
We calculate the volume of the pipe as the difference between the volume of the outer cylinder and the inner hollow part:V=πh(rout2−rin2)V = \pi h (r_{\text{out}}^2 – r_{\text{in}}^2)V=πh(rout2−rin2)
Substitute the values (in cm):V=π×8.9×(2.52−2.12)V = \pi \times 8.9 \times (2.5^2 – 2.1^2)V=π×8.9×(2.52−2.12)V=π×8.9×(6.25−4.41)=π×8.9×1.84V = \pi \times 8.9 \times (6.25 – 4.41) = \pi \times 8.9 \times 1.84V=π×8.9×(6.25−4.41)=π×8.9×1.84V≈3.1416×8.9×1.84≈51.4 cm3V \approx 3.1416 \times 8.9 \times 1.84 \approx 51.4 \, \text{cm}^3V≈3.1416×8.9×1.84≈51.4cm3
So, the volume of metal used is 51.4 cm³.
Part (ii): Cost of the pipe
Convert volume to cubic meters:51.4 cm3=51.4×10−6 m3=5.14×10−5 m351.4 \, \text{cm}^3 = 51.4 \times 10^{-6} \, \text{m}^3 = 5.14 \times 10^{-5} \, \text{m}^351.4cm3=51.4×10−6m3=5.14×10−5m3
Now calculate the mass:Mass=Density×Volume=2700×5.14×10−5=0.13878 kg\text{Mass} = \text{Density} \times \text{Volume} = 2700 \times 5.14 \times 10^{-5} = 0.13878 \, \text{kg}Mass=Density×Volume=2700×5.14×10−5=0.13878kg
Now calculate cost:Cost=0.13878×8=1.11024\text{Cost} = 0.13878 \times 8 = 1.11024Cost=0.13878×8=1.11024
So, the cost of the pipe is approximately $1.11.
Explanation
This problem involves geometric and physical calculations. In the first part, we are asked to find the volume of metal used to make a pipe. A pipe is a hollow cylinder, which means it has both an outer and inner cylindrical surface. To get the volume of metal, we subtract the volume of the inner cylinder (the hole) from the volume of the outer cylinder (the full pipe).
We use the formula for the volume of a cylinder: V=πr2hV = \pi r^2 hV=πr2h. Instead of calculating two volumes separately, we use a single expression: V=πh(rout2−rin2)V = \pi h (r_{\text{out}}^2 – r_{\text{in}}^2)V=πh(rout2−rin2). This formula saves time and avoids extra steps. After calculating, we find that the volume is 51.4 cm³.
In the second part, we convert this volume into cubic meters because the density is given in kilograms per cubic meter. One cubic meter is equal to 1,000,000 cubic centimeters. Therefore, to convert cm³ to m³, we multiply by 10−610^{-6}10−6.
Using the converted volume and the density, we calculate the mass using the formula Mass=Density×Volume\text{Mass} = \text{Density} \times \text{Volume}Mass=Density×Volume. The result is about 0.139 kg. Finally, since the cost is given per kilogram, we multiply the mass by the cost per kg to find the total cost of the pipe. The final cost comes out to be approximately $1.11.
This question combines knowledge from geometry, unit conversion, and basic physics, making it a good example of practical math application.
