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, we can follow these steps:

### Step 1: Understand the relationship between resistance, length, and cross-sectional area
The resistance \( R \) of a wire is given by the formula:
\[ R = \rho \frac{L}{A} \]
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.

### Step 2: Determine the effect of reducing the diameter
When the diameter of the wire is reduced to half, the radius \( r \) of the wire also reduces to half. The cross-sectional area \( A \) of the wire is given by:
\[ A = \pi r^2 \]
If the diameter is halved, the new radius \( r' \) becomes:
\[ r' = \frac{r}{2} \]
Thus, the new area \( A' \) becomes:
\[ A' = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} = \frac{A}{4} \]

### Step 3: Substitute the new area into the resistance formula
Now, substituting the new area \( A' \) into the resistance formula, we have:
\[ R' = \rho \frac{L}{A'} = \rho \frac{L}{\frac{A}{4}} = 4 \rho \frac{L}{A} = 4R \]
This means that the resistance increases by a factor of 4 due to the reduction in diameter.

### Step 4: Consider the effect of length
However, when the wire is drawn through the dies, its length \( L \) also increases. If we assume that the volume of the wire remains constant during this process, we can express the relationship between the original and new dimensions.

The volume \( V \) of the wire is given by:
\[ V = A \cdot L \]
Since the area decreases to \( \frac{A}{4} \), if the length increases to \( L' \), we have:
\[ A \cdot L = \frac{A}{4} \cdot L' \]
This implies:
\[ L' = 4L \]

### Step 5: Calculate the new resistance with the new length
Now substituting \( L' \) back into the resistance formula:
\[ R' = \rho \frac{L'}{A'} = \rho \frac{4L}{\frac{A}{4}} = 16 \rho \frac{L}{A} = 16R \]

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

**Final Answer: The resistance will become 16 times the original resistance.**

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