Length of a hollow tube is `5 m`, its outer diameter is `10 cm` and thickness of its wall is 5 mm. If resistivity of the material of the tube is `1.7 xx 10^(-8) Omega xx m` then resistance of tube will be

Options

A. `5.6 xx 10^(-5) Omega`
B. `2 xx 10^(-5)omega`
C. `4 xx 10^(-5) Omega`
D. None of these

Answer

To find the resistance of the hollow tube, we will follow these steps:

### Step 1: Identify the given values
- Length of the tube, \( L = 5 \, \text{m} \)
- Outer diameter, \( D_{outer} = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Thickness of the wall, \( t = 5 \, \text{mm} = 0.005 \, \text{m} \)
- Resistivity of the material, \( \rho = 1.7 \times 10^{-8} \, \Omega \cdot \text{m} \)

### Step 2: Calculate the outer radius
The outer radius \( r_{outer} \) is half of the outer diameter:
\[ r_{outer} = \frac{D_{outer}}{2} = \frac{0.1}{2} = 0.05 \, \text{m} \]

### Step 3: Calculate the inner radius
The inner radius \( r_{inner} \) can be calculated by subtracting the thickness from the outer radius:
\[ r_{inner} = r_{outer} - t = 0.05 - 0.005 = 0.045 \, \text{m} \]

### Step 4: Calculate the cross-sectional area of the hollow tube
The cross-sectional area \( A \) of the hollow tube can be calculated using the formula:
\[ A = \pi (r_{outer}^2 - r_{inner}^2) \]
Substituting the values:
\[ A = \pi \left((0.05)^2 - (0.045)^2\right) \]
Calculating the squares:
\[ A = \pi \left(0.0025 - 0.002025\right) = \pi \times 0.000475 \approx 0.00149 \, \text{m}^2 \]

### Step 5: Calculate the resistance using the formula
The resistance \( R \) of the tube can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
Substituting the values:
\[ R = \frac{1.7 \times 10^{-8} \times 5}{0.00149} \]
Calculating the resistance:
\[ R \approx \frac{8.5 \times 10^{-8}}{0.00149} \approx 5.7 \times 10^{-5} \, \Omega \]

### Final Answer
The resistance of the hollow tube is approximately:
\[ R \approx 5.7 \times 10^{-5} \, \Omega \]