If a wire of resistacne R is fold n times so that its length becomes `((1)/(n )) ^(th)` of its initial length, then its new resistance becomes

Options

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

Answer

To find the new resistance of a wire that has been folded n times, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- Let the initial resistance of the wire be \( R \).
- Let the initial length of the wire be \( L \).
- Let the initial cross-sectional area of the wire be \( A \).

2. **Determine the New Length**:
- When the wire is folded n times, its new length becomes \( \frac{L}{n} \). This is because folding the wire reduces its length by a factor of n.

3. **Determine the New Cross-Sectional Area**:
- When the wire is folded, its cross-sectional area increases. The new cross-sectional area becomes \( nA \) because the area increases proportionally to the number of folds.

4. **Use the Resistance Formula**:
- The resistance \( R \) of a wire is given by the formula:
\[ R = \frac{\rho L}{A} \]
- Where \( \rho \) is the resistivity of the material.

5. **Calculate the New Resistance**:
- Substitute the new length and new cross-sectional area into the resistance formula:
\[ R' = \frac{\rho \left(\frac{L}{n}\right)}{nA} \]
- Simplifying this gives:
\[ R' = \frac{\rho L}{n^2 A} \]

6. **Relate New Resistance to Initial Resistance**:
- Since the initial resistance \( R \) is given by \( R = \frac{\rho L}{A} \), we can express the new resistance in terms of the initial resistance:
\[ R' = \frac{R}{n^2} \]

### Final Answer:
Thus, the new resistance after folding the wire n times is:
\[ R' = \frac{R}{n^2} \]