A given length of a wire is doubled on itself and this process is repeated once again. By what factor does the resistance of the wire change?

Answer

To solve the problem step by step, we need to analyze how the resistance of the wire changes when it is doubled on itself twice.

### Step-by-Step Solution:

1. **Initial Length and Area**:
- Let the initial length of the wire be \( L \).
- Let the initial cross-sectional area of the wire be \( A \).

2. **First Doubling**:
- When the wire is doubled on itself, the effective length of the wire becomes \( \frac{L}{2} \) (since it is folded).
- The cross-sectional area doubles, so the new area becomes \( 2A \).

3. **Resistance Calculation after First Doubling**:
- The resistance \( R_1 \) of the wire can be calculated using the formula:
\[ R_1 = \frac{\rho L_1}{A_1} \]
- Substituting the values:
\[ R_1 = \frac{\rho \left(\frac{L}{2}\right)}{2A} = \frac{\rho L}{4A} \]

4. **Second Doubling**:
- Now, we double the wire again. The effective length now becomes \( \frac{L}{4} \) (since we are folding the already halved length).
- The cross-sectional area doubles again, so the new area becomes \( 4A \).

5. **Resistance Calculation after Second Doubling**:
- The new resistance \( R_2 \) is given by:
\[ R_2 = \frac{\rho L_2}{A_2} \]
- Substituting the new values:
\[ R_2 = \frac{\rho \left(\frac{L}{4}\right)}{4A} = \frac{\rho L}{16A} \]

6. **Finding the Factor of Change in Resistance**:
- To find the factor by which the resistance changes, we take the ratio of the new resistance \( R_2 \) to the original resistance \( R_0 \):
\[ R_0 = \frac{\rho L}{A} \]
- Thus, the factor of change is:
\[ \frac{R_2}{R_0} = \frac{\frac{\rho L}{16A}}{\frac{\rho L}{A}} = \frac{1}{16} \]

### Conclusion:
The resistance of the wire changes by a factor of \( \frac{1}{16} \).