A hollow copper tube of 5 m length has got external diameter equal to 10 cm and the walls are 5 mm thick . If specific resistance of copper is `1.7xx10^(-8)` ohm meter. Calculate the resistance of the tube.

Options

A. `5.77xx10^(5)Omega`
B. `5.77xx10^(-5)Omega`
C. `5.77xx10^(-7)Omega`
D. `5.77xx10^(7)Omega`

Answer

To calculate the resistance of a hollow copper tube, we can follow these steps:

### Step 1: Identify the given parameters
- Length of the tube (L) = 5 m
- External diameter = 10 cm = 0.1 m
- Thickness of the wall = 5 mm = 0.005 m
- Specific resistance of copper (ρ) = \(1.7 \times 10^{-8}\) ohm meter

### Step 2: Calculate the external radius (R2)
The external radius (R2) can be calculated from the external diameter:
\[ R2 = \frac{\text{External diameter}}{2} = \frac{10 \text{ cm}}{2} = 5 \text{ cm} = 0.05 \text{ m} \]

### Step 3: Calculate the internal radius (R1)
The internal radius (R1) can be calculated by subtracting the thickness of the wall from the external radius:
\[ R1 = R2 - \text{thickness} = 0.05 \text{ m} - 0.005 \text{ m} = 0.045 \text{ m} \]

### Step 4: Calculate the cross-sectional area (A)
The cross-sectional area (A) of the hollow tube can be calculated using the formula for the area of a circle:
\[ A = \pi (R2^2 - R1^2) \]
Substituting the values:
\[ A = \pi \left((0.05)^2 - (0.045)^2\right) \]
Calculating the squares:
\[ A = \pi \left(0.0025 - 0.002025\right) = \pi \times 0.000475 \text{ m}^2 \]

### Step 5: Calculate the resistance (R)
Using the formula for resistance:
\[ R = \frac{\rho L}{A} \]
Substituting the known values:
\[ R = \frac{1.7 \times 10^{-8} \times 5}{\pi \times 0.000475} \]
Calculating the numerator:
\[ 1.7 \times 10^{-8} \times 5 = 8.5 \times 10^{-8} \]
Now substituting this back into the resistance formula:
\[ R = \frac{8.5 \times 10^{-8}}{\pi \times 0.000475} \]
Calculating the denominator:
\[ \pi \times 0.000475 \approx 0.001492 \]
Now calculate R:
\[ R \approx \frac{8.5 \times 10^{-8}}{0.001492} \approx 5.69 \times 10^{-5} \text{ ohms} \]

### Final Answer
The resistance of the hollow copper tube is approximately:
\[ R \approx 5.69 \times 10^{-5} \text{ ohms} \]

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