Answer :
Sure, let's solve the problem step by step.
### Step-by-step Solution:
#### Given:
1. Total volume of water required to fill the whole tube is [tex]\(2.82 \, \text{cm}^3\)[/tex].
2. Volume of water required to fill the tube to a level 1 cm below the top is [tex]\(262 \, \text{cm}^3\)[/tex].
Recall that the Lesitube consists of a cylinder and a hemisphere.
#### Definitions:
- The volume [tex]\(V_\text{cylinder}\)[/tex] of a cylinder with radius [tex]\(r\)[/tex] and height [tex]\(h\)[/tex] is given by the formula:
[tex]\[ V_\text{cylinder} = \pi r^2 h \][/tex]
- The volume [tex]\(V_\text{hemisphere}\)[/tex] of a hemisphere with radius [tex]\(r\)[/tex] is given by the formula:
[tex]\[ V_\text{hemisphere} = \frac{2}{3} \pi r^3 \][/tex]
#### Total volume of the tube:
The total volume [tex]\(V_\text{total}\)[/tex] is the sum of the volumes of the cylinder and hemisphere. Hence:
[tex]\[ V_\text{total} = V_\text{cylinder} + V_\text{hemisphere} \][/tex]
[tex]\[ V_\text{total} = \pi r^2 h + \frac{2}{3} \pi r^3 \][/tex]
Given [tex]\(V_\text{total} = 2.82 \, \text{cm}^3\)[/tex]:
[tex]\[ 2.82 = \pi r^2 h + \frac{2}{3} \pi r^3 \][/tex]
#### Volume when filled 1 cm below the top:
When the tube is filled to 1 cm below the top, the height of the water in the cylinder part is [tex]\(h - 1\)[/tex]. Therefore, the volume of water in the tube is:
[tex]\[ V_\text{below\_top} = \pi r^2 (h - 1) + \frac{2}{3} \pi r^3 \][/tex]
Given [tex]\(V_\text{below\_top} = 262 \, \text{cm}^3\)[/tex]:
[tex]\[ 262 = \pi r^2 (h - 1) + \frac{2}{3} \pi r^3 \][/tex]
#### Equating and solving the equations:
We have the system of equations:
1. [tex]\(2.82 = \pi r^2 h + \frac{2}{3} \pi r^3\)[/tex]
2. [tex]\(262 = \pi r^2 (h - 1) + \frac{2}{3} \pi r^3\)[/tex]
By solving these equations simultaneously, we find:
- The radius [tex]\(r\)[/tex] of the tube:
[tex]\[ r = -9.08292663743988i \quad (\text{imaginary solution}) \][/tex]
- The height [tex]\(h\)[/tex] of the cylindrical part of the tube:
[tex]\[ h = -0.0108804691720039 + 6.05528442495992i \quad (\text{imaginary solution}) \][/tex]
- The volume [tex]\(V_\text{hemisphere}\)[/tex] of the hemisphere:
[tex]\[ V_\text{hemisphere} = 1569.40861726111i \quad (\text{imaginary solution}) \][/tex]
### Conclusion:
(a) The radius of the tube is [tex]\(-9.08292663743988i\)[/tex] (imaginary number).
(b) The volume of the hemisphere is [tex]\(1569.40861726111i\)[/tex] (imaginary number).
The purely imaginary solutions indicate potentially complex physical conditions or constraints that may have been overlooked or a mismatch in expected units or values in the problem statement. For practical or real-world applications, the expectations would be to have real positive values, implying further review or reinterpretation may be necessary.
### Step-by-step Solution:
#### Given:
1. Total volume of water required to fill the whole tube is [tex]\(2.82 \, \text{cm}^3\)[/tex].
2. Volume of water required to fill the tube to a level 1 cm below the top is [tex]\(262 \, \text{cm}^3\)[/tex].
Recall that the Lesitube consists of a cylinder and a hemisphere.
#### Definitions:
- The volume [tex]\(V_\text{cylinder}\)[/tex] of a cylinder with radius [tex]\(r\)[/tex] and height [tex]\(h\)[/tex] is given by the formula:
[tex]\[ V_\text{cylinder} = \pi r^2 h \][/tex]
- The volume [tex]\(V_\text{hemisphere}\)[/tex] of a hemisphere with radius [tex]\(r\)[/tex] is given by the formula:
[tex]\[ V_\text{hemisphere} = \frac{2}{3} \pi r^3 \][/tex]
#### Total volume of the tube:
The total volume [tex]\(V_\text{total}\)[/tex] is the sum of the volumes of the cylinder and hemisphere. Hence:
[tex]\[ V_\text{total} = V_\text{cylinder} + V_\text{hemisphere} \][/tex]
[tex]\[ V_\text{total} = \pi r^2 h + \frac{2}{3} \pi r^3 \][/tex]
Given [tex]\(V_\text{total} = 2.82 \, \text{cm}^3\)[/tex]:
[tex]\[ 2.82 = \pi r^2 h + \frac{2}{3} \pi r^3 \][/tex]
#### Volume when filled 1 cm below the top:
When the tube is filled to 1 cm below the top, the height of the water in the cylinder part is [tex]\(h - 1\)[/tex]. Therefore, the volume of water in the tube is:
[tex]\[ V_\text{below\_top} = \pi r^2 (h - 1) + \frac{2}{3} \pi r^3 \][/tex]
Given [tex]\(V_\text{below\_top} = 262 \, \text{cm}^3\)[/tex]:
[tex]\[ 262 = \pi r^2 (h - 1) + \frac{2}{3} \pi r^3 \][/tex]
#### Equating and solving the equations:
We have the system of equations:
1. [tex]\(2.82 = \pi r^2 h + \frac{2}{3} \pi r^3\)[/tex]
2. [tex]\(262 = \pi r^2 (h - 1) + \frac{2}{3} \pi r^3\)[/tex]
By solving these equations simultaneously, we find:
- The radius [tex]\(r\)[/tex] of the tube:
[tex]\[ r = -9.08292663743988i \quad (\text{imaginary solution}) \][/tex]
- The height [tex]\(h\)[/tex] of the cylindrical part of the tube:
[tex]\[ h = -0.0108804691720039 + 6.05528442495992i \quad (\text{imaginary solution}) \][/tex]
- The volume [tex]\(V_\text{hemisphere}\)[/tex] of the hemisphere:
[tex]\[ V_\text{hemisphere} = 1569.40861726111i \quad (\text{imaginary solution}) \][/tex]
### Conclusion:
(a) The radius of the tube is [tex]\(-9.08292663743988i\)[/tex] (imaginary number).
(b) The volume of the hemisphere is [tex]\(1569.40861726111i\)[/tex] (imaginary number).
The purely imaginary solutions indicate potentially complex physical conditions or constraints that may have been overlooked or a mismatch in expected units or values in the problem statement. For practical or real-world applications, the expectations would be to have real positive values, implying further review or reinterpretation may be necessary.
