The drift velocity of free electrons in a conductor is v, when a current i is flowing in it, Ifboth the radius and current are doubled, then the drift velocity will be :

Options

A. `(v_(d))/(8)`
B. `(v_(d))/(4)`
C. `(v_(d))/(2)`
D. `v_(d)`

Explanation

To solve the problem, we need to analyze how the drift velocity changes when both the radius of the conductor and the current flowing through it are doubled.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The relationship between current (I), charge density (n), charge of an electron (e), drift velocity (v_d), and the cross-sectional area (A) of the conductor is given by the formula:
\[ I = n \cdot e \cdot A \cdot v_d \]
where \( A = \pi r^2 \) (area of the circular cross-section of the conductor).

2. **Initial Conditions**:
Let the initial current be \( I \) and the initial radius be \( r \). Therefore, the initial area \( A \) is:
\[ A = \pi r^2 \]
Substituting this into the current equation gives:
\[ I = n \cdot e \cdot (\pi r^2) \cdot v_d \]
Rearranging for drift velocity \( v_d \):
\[ v_d = \frac{I}{n \cdot e \cdot \pi r^2} \]

3. **New Conditions**:
Now, both the current and the radius are doubled:
- New current \( I' = 2I \)
- New radius \( r' = 2r \)

The new area \( A' \) becomes:
\[ A' = \pi (2r)^2 = \pi \cdot 4r^2 = 4\pi r^2 \]

4. **Calculating New Drift Velocity**:
Substitute the new values into the drift velocity formula:
\[ v_d' = \frac{I'}{n \cdot e \cdot A'} \]
Substituting \( I' \) and \( A' \):
\[ v_d' = \frac{2I}{n \cdot e \cdot (4\pi r^2)} \]
Simplifying this gives:
\[ v_d' = \frac{2I}{4n \cdot e \cdot \pi r^2} = \frac{1}{2} \cdot \frac{I}{n \cdot e \cdot \pi r^2} \]
Therefore, we can express \( v_d' \) in terms of the original drift velocity \( v_d \):
\[ v_d' = \frac{1}{2} v_d \]

5. **Conclusion**:
Thus, the new drift velocity when both the radius and the current are doubled is:
\[ v_d' = \frac{v_d}{2} \]

### Final Answer:
The drift velocity will be \( \frac{v}{2} \).