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

Explanation

To find the new resistance of a wire when it is uniformly stretched to n times its original length, we can follow these steps:

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

### Step 2: Set up the initial conditions
Let:
- The original length of the wire be \( L_1 \)
- The original cross-sectional area be \( A_1 \)
- The original resistance be \( R_1 = R \)

### Step 3: Apply the condition of constant volume
When the wire is stretched uniformly to n times its original length, the new length \( L_2 \) becomes:
\[ L_2 = n L_1 \]
Since the volume of the wire remains constant during stretching, we have:
\[ A_1 L_1 = A_2 L_2 \]
Substituting \( L_2 \):
\[ A_1 L_1 = A_2 (n L_1) \]

### Step 4: Solve for the new cross-sectional area
Rearranging the equation gives:
\[ A_2 = \frac{A_1}{n} \]

### Step 5: Calculate the new resistance
Now, we can find the new resistance \( R_2 \) using the new dimensions:
\[ R_2 = \rho \frac{L_2}{A_2} \]
Substituting \( L_2 \) and \( A_2 \):
\[ R_2 = \rho \frac{n L_1}{\frac{A_1}{n}} = \rho \frac{n^2 L_1}{A_1} \]
Since \( R_1 = \rho \frac{L_1}{A_1} \), we can express \( R_2 \) in terms of \( R_1 \):
\[ R_2 = n^2 R_1 \]

### Step 6: Substitute the original resistance
Since \( R_1 = R \), we have:
\[ R_2 = n^2 R \]

### Conclusion
Thus, the new resistance of the wire after being stretched to n times its original length is:
\[ \boxed{n^2 R} \]

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