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

Explanation

To find the resistance of the hollow tube, we will use the formula for resistance:

\[ R = \frac{\rho \cdot L}{A} \]

where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the tube,
- \( A \) is the cross-sectional area of the hollow part of the tube.

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

### Step 2: Calculate the outer radius \( R_2 \)
The outer radius \( R_2 \) is half of the outer diameter:
\[ R_2 = \frac{D_{outer}}{2} = \frac{0.1 \, m}{2} = 0.05 \, m \]

### Step 3: Calculate the inner radius \( R_1 \)
The inner radius \( R_1 \) can be calculated by subtracting the thickness of the wall from the outer radius:
\[ R_1 = R_2 - t = 0.05 \, m - 0.005 \, m = 0.045 \, m \]

### Step 4: Calculate the cross-sectional area \( A \)
The cross-sectional area \( A \) of the hollow tube is given by the difference between the area of the outer circle and the area of the inner circle:
\[ A = \pi (R_2^2 - R_1^2) \]
Substituting the values:
\[ A = \pi \left((0.05)^2 - (0.045)^2\right) \]
Calculating the squares:
\[ A = \pi \left(0.0025 - 0.002025\right) = \pi \times 0.000475 \]
Calculating \( A \):
\[ A \approx 0.00149 \, m^2 \, (\text{using } \pi \approx 3.14) \]

### Step 5: Substitute values into the resistance formula
Now, substituting \( \rho \), \( L \), and \( A \) into the resistance formula:
\[ R = \frac{1.7 \times 10^{-8} \cdot 5}{0.00149} \]
Calculating the numerator:
\[ 1.7 \times 10^{-8} \cdot 5 = 8.5 \times 10^{-8} \]
Now calculating \( R \):
\[ R \approx \frac{8.5 \times 10^{-8}}{0.00149} \approx 5.7 \times 10^{-5} \, \Omega \]

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