Statement I: A wire of uniform cross-section and uniform resistivity is connected across an ideal cell. Now the length of the wire is doubled keeping volume of the wire constant. The drift velocity of electrons after stretching the wire becomes one-fouth of what it was before stretching the wire.
Statement II: If a wire (of uniform resistivity and uniform cross section) of length `l_0` is stretched to length `nl_0`, then its resistance becomes `n^2` times of what it was before stretching the wire (the volume of wire is kept constant in stretching process). Further at constant potential difference, current is inversely proportional to resistance. Finally, drift velocity of free electron is directly proportional to current and inversely proportional to cross-sectional area of current carrying wire.

Options

A. Statement I is true, Statement II is True, Statement II is a correct explanation for Statement I.
B. Statement I is True, Statement II is True, Statement II is not a correct explanation for Statement I.
C. Statement I is True, Statement II is False.
D. Statement I is False, Statement II is True.

Explanation

To solve the problem, we need to analyze both statements provided and determine their validity based on the principles of physics.

### Step 1: Analyze Statement I
Statement I claims that when a wire of uniform cross-section and uniform resistivity is stretched to double its length while keeping the volume constant, the drift velocity of electrons becomes one-fourth of what it was before stretching.

1. **Initial Parameters**:
- Let the initial length of the wire be \( L \).
- Let the initial cross-sectional area be \( A \).
- The volume of the wire is \( V = A \cdot L \).

2. **After Stretching**:
- The new length \( L' = 2L \).
- Since the volume remains constant, the new cross-sectional area \( A' \) can be calculated as:
\[ V = A \cdot L = A' \cdot L' \implies A' = \frac{A \cdot L}{2L} = \frac{A}{2} \]

3. **Resistance Calculation**:
- The resistance \( R \) of the wire is given by:
\[ R = \frac{\rho L}{A} \]
- The new resistance \( R' \) after stretching is:
\[ R' = \frac{\rho (2L)}{(A/2)} = \frac{4\rho L}{A} = 4R \]

4. **Current and Drift Velocity**:
- The current \( I \) through the wire can be expressed as:
\[ I = \frac{E}{R} \]
- Therefore, the new current \( I' \) is:
\[ I' = \frac{E}{R'} = \frac{E}{4R} = \frac{I}{4} \]

5. **Drift Velocity Relation**:
- The drift velocity \( V_d \) is related to current as:
\[ I = n \cdot A \cdot e \cdot V_d \]
- Thus, for the new wire:
\[ I' = n \cdot A' \cdot e \cdot V_d' \implies \frac{I}{4} = n \cdot \left(\frac{A}{2}\right) \cdot e \cdot V_d' \]
- Rearranging gives:
\[ V_d' = \frac{I}{4} \cdot \frac{2}{n \cdot A \cdot e} = \frac{1}{2} \cdot V_d \]

### Conclusion for Statement I
The drift velocity \( V_d' \) after stretching is actually half of the original drift velocity \( V_d \), not one-fourth. Therefore, **Statement I is false**.

### Step 2: Analyze Statement II
Statement II states that if a wire of length \( l_0 \) is stretched to length \( nl_0 \), then its resistance becomes \( n^2 \) times what it was before stretching.

1. **Resistance Before Stretching**:
- Initial resistance \( R_0 = \frac{\rho l_0}{A} \).

2. **Resistance After Stretching**:
- New length \( l' = nl_0 \) and new area \( A' = \frac{A}{n} \) (since volume is constant).
- New resistance \( R' \):
\[ R' = \frac{\rho (nl_0)}{(A/n)} = \frac{n^2 \rho l_0}{A} = n^2 R_0 \]

### Conclusion for Statement II
Statement II is true as the resistance indeed becomes \( n^2 \) times the original resistance.

### Final Conclusion
- **Statement I**: False
- **Statement II**: True