A uniform wire of resistance R is uniformly compressed along its length, until its radius becomes n times the original radius. Now resistance of the wire becomes

Options

A. `(R)/(n^(4))`
B. `(R)/(n^(2))`
C. `(R)/(n)`
D. `nR`

Answer

To solve the problem of finding the new resistance of a uniformly compressed wire, we will follow these steps:

### Step 1: Understand the relationship between resistance, resistivity, length, and area
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.

### Step 2: Determine the original dimensions of the wire
Let the original radius of the wire be \( r \) and the original length be \( L \). The cross-sectional area \( A \) of the wire can be expressed as:
\[ A = \pi r^2 \]
Thus, the original resistance \( R \) can be written as:
\[ R = \frac{\rho L}{\pi r^2} \]

### Step 3: Analyze the effect of compression on the wire
When the wire is uniformly compressed, its radius becomes \( n \) times the original radius. Therefore, the new radius \( r' \) is:
\[ r' = n \cdot r \]

### Step 4: Calculate the new cross-sectional area
The new cross-sectional area \( A' \) can be calculated as:
\[ A' = \pi (r')^2 = \pi (n \cdot r)^2 = \pi n^2 r^2 \]

### Step 5: Determine the new length of the wire
Since the volume of the wire remains constant during compression, we can set the initial volume equal to the final volume:
\[ \text{Initial Volume} = \text{Final Volume} \]
\[ \pi r^2 L = \pi n^2 r^2 L' \]
Cancelling \( \pi r^2 \) from both sides gives:
\[ L = n^2 L' \]
This implies:
\[ L' = \frac{L}{n^2} \]

### Step 6: Calculate the new resistance
Now we can find the new resistance \( R' \) using the new length \( L' \) and the new area \( A' \):
\[ R' = \frac{\rho L'}{A'} = \frac{\rho \left(\frac{L}{n^2}\right)}{\pi n^2 r^2} \]
Substituting \( A = \pi r^2 \) into the equation gives:
\[ R' = \frac{\rho \left(\frac{L}{n^2}\right)}{\pi n^2 r^2} = \frac{1}{n^2} \cdot \frac{\rho L}{\pi r^2} \cdot \frac{1}{n^2} = \frac{R}{n^4} \]

### Final Result
Thus, the new resistance \( R' \) of the wire after compression is:
\[ R' = \frac{R}{n^4} \]