Two wires of the same material but of different diameters carry the same current `i`. If the ratio of their diameters is `2:1` , then the corresponding ratio of their mean drift velocities will be

Options

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

Answer

To solve the problem, we need to find the ratio of the mean drift velocities of two wires made of the same material but having different diameters, given that they carry the same current.

### Step-by-Step Solution:

1. **Understand the relationship between current and drift velocity**:
The current \( I \) in a wire is given by the formula:
\[ I = n \cdot e \cdot A \cdot v_d \]
where:
- \( n \) = number of free electrons per unit volume (constant for the same material),
- \( e \) = charge of an electron (constant),
- \( A \) = cross-sectional area of the wire,
- \( v_d \) = mean drift velocity of the charge carriers.

2. **Identify the given information**:
- The two wires carry the same current \( I \).
- The ratio of their diameters is \( d_1 : d_2 = 2 : 1 \).

3. **Relate the cross-sectional area to diameter**:
The cross-sectional area \( A \) of a wire is related to its diameter \( d \) by the formula:
\[ A = \frac{\pi d^2}{4} \]
Therefore, if we denote the diameters of the two wires as \( d_1 \) and \( d_2 \), their areas will be:
\[ A_1 = \frac{\pi d_1^2}{4}, \quad A_2 = \frac{\pi d_2^2}{4} \]

4. **Express the drift velocity in terms of area**:
Since the current \( I \) is the same for both wires, we can set up the following relationships:
\[ I = n \cdot e \cdot A_1 \cdot v_{d1} = n \cdot e \cdot A_2 \cdot v_{d2} \]
From this, we can derive that:
\[ v_{d1} = \frac{I}{n \cdot e \cdot A_1}, \quad v_{d2} = \frac{I}{n \cdot e \cdot A_2} \]

5. **Find the ratio of drift velocities**:
Taking the ratio of the drift velocities:
\[ \frac{v_{d1}}{v_{d2}} = \frac{A_2}{A_1} \]

6. **Substituting the areas**:
Using the areas derived from the diameters:
\[ \frac{A_1}{A_2} = \frac{\frac{\pi d_1^2}{4}}{\frac{\pi d_2^2}{4}} = \frac{d_1^2}{d_2^2} \]
Thus, we have:
\[ \frac{v_{d1}}{v_{d2}} = \frac{d_2^2}{d_1^2} \]

7. **Substituting the diameter ratio**:
Given \( d_1 : d_2 = 2 : 1 \), we can express this as:
\[ d_1 = 2d, \quad d_2 = d \]
Therefore:
\[ \frac{v_{d1}}{v_{d2}} = \frac{(d)^2}{(2d)^2} = \frac{d^2}{4d^2} = \frac{1}{4} \]

8. **Final ratio of drift velocities**:
This means:
\[ v_{d1} : v_{d2} = 1 : 4 \]

### Conclusion:
The ratio of the mean drift velocities of the two wires is \( 1 : 4 \).