A wire of length 5m and radius 1mm has a resistance of 1 ohm . What length of the wire of the same material at the same temperature and of radius 2mm will also have a resistance of 1 ohm

Options

A. `1.25m`
B. `2.5m`
C. `10m`
D. `20m`

Answer

To solve the problem, we need to find the length of a wire of the same material and temperature, but with a different radius, that has the same resistance as the original wire.

### Given Data:
1. Length of the original wire (L1) = 5 m
2. Radius of the original wire (r1) = 1 mm = 0.001 m
3. Resistance of the original wire (R1) = 1 ohm
4. Radius of the new wire (r2) = 2 mm = 0.002 m
5. Resistance of the new wire (R2) = 1 ohm (same as R1)

### Step-by-Step Solution:

**Step 1: Understand the formula for resistance.**
The resistance (R) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) = resistance
- \( \rho \) = resistivity of the material (constant for the same material)
- \( L \) = length of the wire
- \( A \) = cross-sectional area of the wire

**Step 2: Calculate the cross-sectional area of the original wire.**
The cross-sectional area (A) of a wire with radius \( r \) is given by:
\[ A = \pi r^2 \]
For the original wire:
\[ A_1 = \pi (0.001)^2 = \pi \times 10^{-6} \, \text{m}^2 \]

**Step 3: Calculate the cross-sectional area of the new wire.**
For the new wire with radius \( r_2 = 0.002 \, \text{m} \):
\[ A_2 = \pi (0.002)^2 = \pi \times 4 \times 10^{-6} \, \text{m}^2 \]

**Step 4: Set up the equation for resistance.**
Since the resistivity and resistance are the same for both wires, we can set up the following relationship:
\[ \frac{R_1}{R_2} = \frac{L_1 \cdot A_2}{L_2 \cdot A_1} \]

**Step 5: Substitute known values into the equation.**
Given that \( R_1 = R_2 = 1 \, \text{ohm} \):
\[ 1 = \frac{5 \cdot A_2}{L_2 \cdot A_1} \]

Substituting the areas:
\[ A_1 = \pi \times 10^{-6} \]
\[ A_2 = \pi \times 4 \times 10^{-6} \]

So the equation becomes:
\[ 1 = \frac{5 \cdot (\pi \times 4 \times 10^{-6})}{L_2 \cdot (\pi \times 10^{-6})} \]

**Step 6: Simplify the equation.**
The \( \pi \) terms cancel out:
\[ 1 = \frac{5 \cdot 4 \times 10^{-6}}{L_2 \cdot 10^{-6}} \]
\[ 1 = \frac{20}{L_2} \]

**Step 7: Solve for \( L_2 \).**
Rearranging gives:
\[ L_2 = 20 \, \text{m} \]

### Final Answer:
The length of the new wire (L2) that will also have a resistance of 1 ohm is **20 meters**.