Two copper wires A and B of length l and 2l respectively, have the same area of cross-section. The ratio of the resistivity of wire A to the resistivity of wire B is

Options

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

Explanation

To solve the problem of finding the ratio of the resistivity of wire A to wire B, we can follow these steps:

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

### Step 2: Identify the lengths and areas of the wires
Let:
- Length of wire A, \( L_A = l \)
- Length of wire B, \( L_B = 2l \)
- Area of cross-section for both wires, \( A_A = A_B = A \)

### Step 3: Write the resistance for both wires
Using the formula for resistance:
- Resistance of wire A, \( R_A \):
\[ R_A = \rho_A \frac{L_A}{A} = \rho_A \frac{l}{A} \]
- Resistance of wire B, \( R_B \):
\[ R_B = \rho_B \frac{L_B}{A} = \rho_B \frac{2l}{A} \]

### Step 4: Find the ratio of the resistances
To find the ratio of the resistivities, we can express the ratio of the resistances:
\[ \frac{R_A}{R_B} = \frac{\rho_A \frac{l}{A}}{\rho_B \frac{2l}{A}} = \frac{\rho_A}{\rho_B} \cdot \frac{l}{2l} = \frac{\rho_A}{\rho_B} \cdot \frac{1}{2} \]

### Step 5: Rearranging to find the ratio of resistivities
From the above equation, we can rearrange to find the ratio of the resistivities:
\[ \frac{\rho_A}{\rho_B} = 2 \cdot \frac{R_A}{R_B} \]

### Step 6: Conclusion
Since the problem does not provide specific values for the resistances \( R_A \) and \( R_B \), we conclude that the ratio of the resistivity of wire A to wire B is:
\[ \frac{\rho_A}{\rho_B} = 2 \]