When the diameter of a wire is doubled, its resistance becomes:

Options

A. double
B. four times
C. one-half
D. one-fourth

Answer

To determine how the resistance of a wire changes when its diameter is doubled, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Resistance**:
The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
where:
- \( R \) is the resistance,
- \( \rho \) is the resistivity of the material,
- \( L \) is the length of the wire,
- \( A \) is the cross-sectional area of the wire.

2. **Determine the Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire with diameter \( D \) can be calculated using the formula for the area of a circle:
\[ A = \pi \left(\frac{D}{2}\right)^2 = \frac{\pi D^2}{4} \]

3. **Calculate Initial Resistance**:
Substituting the expression for area into the resistance formula, we get:
\[ R = \frac{\rho L}{\frac{\pi D^2}{4}} = \frac{4\rho L}{\pi D^2} \]

4. **Consider the New Diameter**:
If the diameter is doubled, the new diameter \( D' \) becomes:
\[ D' = 2D \]
The new cross-sectional area \( A' \) will be:
\[ A' = \pi \left(\frac{D'}{2}\right)^2 = \pi \left(\frac{2D}{2}\right)^2 = \pi D^2 \]

5. **Calculate New Resistance**:
The new resistance \( R' \) with the new diameter is:
\[ R' = \frac{\rho L}{A'} = \frac{\rho L}{\pi D^2} \]

6. **Relate New Resistance to Original Resistance**:
Now, we can relate the new resistance \( R' \) to the original resistance \( R \):
\[ R' = \frac{\rho L}{\pi D^2} = \frac{1}{4} \left(\frac{4\rho L}{\pi D^2}\right) = \frac{1}{4} R \]

7. **Conclusion**:
Therefore, when the diameter of the wire is doubled, the resistance becomes:
\[ R' = \frac{R}{4} \]
This means the resistance is reduced to one-fourth of the original resistance.

### Final Answer:
When the diameter of a wire is doubled, its resistance becomes one-fourth of the original resistance.

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