When a piece of aliminium wire of finite length is drawn through a series of dies to reduce its diameter to half its original value, its resistance will become

Options

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

Answer

To solve the problem of how the resistance of an aluminum wire changes when its diameter is reduced to half its original value, we can follow these steps:

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

### Step 2: Determine the relationship between area and diameter
The cross-sectional area \( A \) of a wire with diameter \( D \) is given by:
\[ A = \frac{\pi D^2}{4} \]
If the diameter is reduced to half, then the new diameter \( D' \) is:
\[ D' = \frac{D}{2} \]
The new area \( A' \) can be calculated as:
\[ A' = \frac{\pi (D')^2}{4} = \frac{\pi \left(\frac{D}{2}\right)^2}{4} = \frac{\pi \frac{D^2}{4}}{4} = \frac{\pi D^2}{16} \]

### Step 3: Relate the original and new areas
The original area \( A \) is:
\[ A = \frac{\pi D^2}{4} \]
The new area \( A' \) is:
\[ A' = \frac{\pi D^2}{16} \]
Thus, we can see that:
\[ A' = \frac{A}{4} \]

### Step 4: Analyze the effect on resistance
Since the volume of the wire remains constant when it is drawn through the dies, the length \( L' \) of the wire will increase when the diameter decreases. The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
Since the volume remains constant:
\[ A \cdot L = A' \cdot L' \]
Substituting for \( A' \):
\[ A \cdot L = \frac{A}{4} \cdot L' \]
From this, we can find \( L' \):
\[ L' = 4L \]

### Step 5: Calculate the new resistance
Now, substituting the new values into the resistance formula:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (4L)}{\frac{A}{4}} = \frac{16 \rho L}{A} = 16R \]
Thus, the new resistance \( R' \) is 16 times the original resistance \( R \).

### Conclusion
Therefore, when the diameter of the aluminum wire is reduced to half its original value, its resistance will become:
\[ R' = 16R \]