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. Let's denote the diameters of the two wires as \(D_1\) and \(D_2\) where \(D_1:D_2 = 2:1\).

### Step-by-step Solution:

1. **Understanding the relationship between current and drift velocity**:
The current \(I\) in a wire can be expressed using the formula:
\[ I = n A e v_d \]
where:
- \(n\) = number density of charge carriers (constant for the same material)
- \(A\) = cross-sectional area of the wire
- \(e\) = charge of an electron (constant)
- \(v_d\) = mean drift velocity of the charge carriers

2. **Cross-sectional area of the wires**:
The cross-sectional area \(A\) of a wire with diameter \(D\) is given by:
\[ A = \frac{\pi D^2}{4} \]
Therefore, for the two wires:
- For wire 1: \(A_1 = \frac{\pi D_1^2}{4}\)
- For wire 2: \(A_2 = \frac{\pi D_2^2}{4}\)

3. **Setting up the equations for both wires**:
Since both wires carry the same current \(I\), we can write:
\[ I = n A_1 e v_{d1} = n A_2 e v_{d2} \]
This implies:
\[ n A_1 e v_{d1} = n A_2 e v_{d2} \]
Here, \(n\) and \(e\) cancel out, leading to:
\[ A_1 v_{d1} = A_2 v_{d2} \]

4. **Expressing the areas in terms of diameters**:
Substituting the expressions for areas:
\[ \frac{\pi D_1^2}{4} v_{d1} = \frac{\pi D_2^2}{4} v_{d2} \]
The \(\frac{\pi}{4}\) cancels out:
\[ D_1^2 v_{d1} = D_2^2 v_{d2} \]

5. **Finding the ratio of drift velocities**:
Rearranging gives us:
\[ \frac{v_{d1}}{v_{d2}} = \frac{D_2^2}{D_1^2} \]

6. **Substituting the ratio of diameters**:
Given that \(D_1:D_2 = 2:1\), we can express this as:
\[ \frac{D_1}{D_2} = 2 \implies D_1 = 2D_2 \]
Therefore:
\[ \frac{D_2^2}{D_1^2} = \frac{D_2^2}{(2D_2)^2} = \frac{D_2^2}{4D_2^2} = \frac{1}{4} \]

7. **Final ratio of drift velocities**:
Hence, we have:
\[ \frac{v_{d1}}{v_{d2}} = \frac{1}{4} \implies v_{d1} : v_{d2} = 1 : 4 \]

### Conclusion:
The ratio of the mean drift velocities \(v_{d1} : v_{d2}\) is \(1 : 4\).