Two conductors are made of the same material and have the same length. Conductor A is a solid wire of radius 1.0 mm. Conductor B is a hollow tube of outside radius 2.2 mm and inside radius 1.0 mm. What is the resistance rate `R_A//R_B` measured between their ends?

Answer

To solve the problem of finding the resistance ratio \( R_A / R_B \) between two conductors, we will follow these steps:

### Step 1: Understand the given data
- Conductor A is a solid wire with a radius \( r_A = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \).
- Conductor B is a hollow tube with an outer radius \( r_{B_{\text{outer}}} = 2.2 \, \text{mm} = 2.2 \times 10^{-3} \, \text{m} \) and an inner radius \( r_{B_{\text{inner}}} = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \).
- Both conductors have the same length \( L \) and are made of the same material, implying they have the same resistivity \( \rho \).

### Step 2: Write the formula for resistance
The resistance \( R \) of a conductor is given by the formula:
\[ R = \frac{\rho L}{A} \]
where \( A \) is the cross-sectional area.

### Step 3: Calculate the cross-sectional area of Conductor A
For Conductor A, which is a solid wire, the cross-sectional area \( A_A \) is given by:
\[ A_A = \pi r_A^2 = \pi (1.0 \times 10^{-3})^2 = \pi \times 10^{-6} \, \text{m}^2 \]

### Step 4: Calculate the cross-sectional area of Conductor B
For Conductor B, which is a hollow tube, the cross-sectional area \( A_B \) is given by the difference between the outer and inner areas:
\[ A_B = \pi r_{B_{\text{outer}}}^2 - \pi r_{B_{\text{inner}}}^2 = \pi \left( (2.2 \times 10^{-3})^2 - (1.0 \times 10^{-3})^2 \right) \]
Calculating the areas:
\[ A_B = \pi \left( (4.84 \times 10^{-6}) - (1.0 \times 10^{-6}) \right) = \pi \times 3.84 \times 10^{-6} \, \text{m}^2 \]

### Step 5: Write the expressions for resistance
Now we can write the expressions for the resistances of both conductors:
\[ R_A = \frac{\rho L}{A_A} = \frac{\rho L}{\pi \times 10^{-6}} \]
\[ R_B = \frac{\rho L}{A_B} = \frac{\rho L}{\pi \times 3.84 \times 10^{-6}} \]

### Step 6: Find the ratio of the resistances
To find the ratio \( \frac{R_A}{R_B} \):
\[ \frac{R_A}{R_B} = \frac{\frac{\rho L}{A_A}}{\frac{\rho L}{A_B}} = \frac{A_B}{A_A} \]
Substituting the areas:
\[ \frac{R_A}{R_B} = \frac{\pi \times 3.84 \times 10^{-6}}{\pi \times 10^{-6}} = \frac{3.84}{1} = 3.84 \]

### Final Answer
Thus, the resistance ratio \( \frac{R_A}{R_B} \) is:
\[ \frac{R_A}{R_B} = 3.84 \]