If area of cross-section of a metalic wire becomes n times, its resistance becomes

Options

A. `((1)/(n ^(2)))` times
B. `n ^(2)` times
C. `n ^(4)` times
D. `((1)/(n ^(4)))` times

Answer

To solve the problem of how the resistance of a metallic wire changes when its cross-sectional area is increased by a factor of \( n \), we can follow these steps:

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

### Step 2: Define the initial conditions
Let:
- The initial cross-sectional area be \( A \).
- The initial length of the wire be \( L \).
- The initial resistance be \( R \).

Thus, we can write:
\[ R = \frac{\rho L}{A} \]

### Step 3: Define the new conditions after increasing the area
When the area is increased by \( n \) times, the new area \( A' \) becomes:
\[ A' = nA \]

### Step 4: Determine the new length of the wire
Since the volume of the wire remains constant, we can express the volume \( V \) as:
\[ V = A \times L \]
For the new conditions, the volume can also be expressed as:
\[ V = A' \times L' = nA \times L' \]
Setting these equal gives:
\[ A \times L = nA \times L' \]
From this, we can solve for the new length \( L' \):
\[ L' = \frac{L}{n} \]

### Step 5: Calculate the new resistance
Now, we can find the new resistance \( R' \) using the new area and the new length:
\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{n}\right)}{nA} \]
This simplifies to:
\[ R' = \frac{\rho L}{n^2 A} = \frac{1}{n^2} \cdot \frac{\rho L}{A} \]
Thus, we have:
\[ R' = \frac{R}{n^2} \]

### Conclusion
Therefore, when the area of cross-section of a metallic wire becomes \( n \) times, its resistance becomes:
\[ R' = \frac{R}{n^2} \]