If a uniform wire of resistance R is uniformly strethced to n times the orginal length , then new resistance of the wire becomes

Options

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

Explanation

To find the new resistance of a uniformly stretched wire, we can follow these steps:

### Step 1: Understand the relationship between resistance, 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: Define the initial conditions
Let the initial length of the wire be \( L \) and the initial resistance be \( R \). Thus, we have:
\[ R = \frac{\rho L}{A} \]

### Step 3: Determine the new length after stretching
If the wire is stretched to \( n \) times its original length, the new length \( L' \) becomes:
\[ L' = nL \]

### Step 4: Determine the new cross-sectional area
When the wire is stretched, its volume remains constant. The volume \( V \) of the wire can be expressed as:
\[ V = L \cdot A \]
After stretching, the volume can also be expressed as:
\[ V' = L' \cdot A' = nL \cdot A' \]
Setting the two expressions for volume equal gives:
\[ L \cdot A = nL \cdot A' \]
From this, we can solve for the new area \( A' \):
\[ A' = \frac{A}{n} \]

### Step 5: Substitute the new length and area into the resistance formula
Now we can find the new resistance \( R' \):
\[ R' = \frac{\rho L'}{A'} = \frac{\rho (nL)}{(A/n)} = \frac{n \rho L}{A/n} = n^2 \frac{\rho L}{A} \]
Thus, we can express the new resistance in terms of the original resistance \( R \):
\[ R' = n^2 R \]

### Conclusion
The new resistance of the wire after being stretched to \( n \) times its original length is:
\[ R' = n^2 R \]