The diameter of aluminium wire is reduced to half its original value, then its resistance will become

Options

A. four times
B. two times
C. eight times
D. sixteen times

Explanation

To solve the problem of how the resistance of an aluminum wire changes when its diameter is reduced to half, 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 \( \rho \) is the resistivity of the material, \( L \) is the length of the wire, and \( A \) is the cross-sectional area.

2. **Determine the Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire with diameter \( d \) is given by:
\[ A = \frac{\pi d^2}{4} \]

3. **Substitute the Area into the Resistance Formula**:
Substituting the expression for \( A \) into the resistance formula, we get:
\[ R = \frac{\rho L}{\frac{\pi d^2}{4}} = \frac{4\rho L}{\pi d^2} \]

4. **Effect of Reducing the Diameter**:
If the diameter is reduced to half its original value, the new diameter \( d' \) becomes:
\[ d' = \frac{d}{2} \]
The new area \( A' \) becomes:
\[ A' = \frac{\pi (d')^2}{4} = \frac{\pi \left(\frac{d}{2}\right)^2}{4} = \frac{\pi \frac{d^2}{4}}{4} = \frac{\pi d^2}{16} \]

5. **Calculate the New Resistance**:
Now, substituting \( A' \) into the resistance formula gives:
\[ R' = \frac{\rho L}{A'} = \frac{\rho L}{\frac{\pi d^2}{16}} = \frac{16\rho L}{\pi d^2} \]

6. **Relate the New Resistance to the Original Resistance**:
From the original resistance \( R = \frac{4\rho L}{\pi d^2} \), we can see that:
\[ R' = 4 \times R \]
Thus:
\[ R' = 16R \]

### Conclusion:
When the diameter of the aluminum wire is reduced to half its original value, the resistance becomes **16 times** the original resistance.