A material 'B' has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter of a wire made of 'A'. Then for the two wires to have the same resistance, the ratio `l_(B)//l_(A)` of their respective lengths must be

Options

A. 1
B. `1/2`
C. `1/4`
D. 2

Explanation

To solve the problem, we need to find the ratio of the lengths of two wires made of different materials, A and B, such that they have the same resistance. Let's denote the specific resistance (resistivity) of material A as ρ and that of material B as 2ρ (since it is given that B has twice the specific resistance of A).

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Specific resistance of material A: ρ
- Specific resistance of material B: 2ρ
- Diameter of wire B: 2D (where D is the diameter of wire A)

2. **Calculate the Cross-Sectional Area:**
- The cross-sectional area \( A_A \) of wire A can be calculated using the formula for the area of a circle:
\[ A_A = \frac{\pi D^2}{4} \]
- The diameter of wire B is twice that of wire A, so the diameter of wire B is \( 2D \). Therefore, the cross-sectional area \( A_B \) of wire B is:
\[ A_B = \frac{\pi (2D)^2}{4} = \frac{\pi \cdot 4D^2}{4} = \pi D^2 \]

3. **Write the Formula for Resistance:**
- The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho l}{A} \]
- For wire A:
\[ R_A = \frac{\rho L_A}{A_A} = \frac{\rho L_A}{\frac{\pi D^2}{4}} = \frac{4\rho L_A}{\pi D^2} \]
- For wire B:
\[ R_B = \frac{2\rho L_B}{A_B} = \frac{2\rho L_B}{\pi D^2} \]

4. **Set the Resistances Equal:**
- Since we want the resistances to be equal, we set \( R_A = R_B \):
\[ \frac{4\rho L_A}{\pi D^2} = \frac{2\rho L_B}{\pi D^2} \]

5. **Cancel Common Terms:**
- We can cancel \( \rho \) and \( \frac{\pi D^2}{4} \) from both sides:
\[ 4L_A = 2L_B \]

6. **Rearrange to Find the Length Ratio:**
- Rearranging gives:
\[ \frac{L_B}{L_A} = \frac{4}{2} = 2 \]

### Final Answer:
The ratio of the lengths of wire B to wire A is:
\[ \frac{L_B}{L_A} = 2 \]