The diameters of two wires of same material is `n:1`. The length of wires is 4 m each. On applying the same load, the increase in length of this wire will be

Options

A. `n^(2)` times
B. n times
C. 2n times
D. none

Explanation

To solve the problem, we need to determine the increase in length of two wires made of the same material, given that their diameters are in the ratio of \( n:1 \) and their lengths are both 4 m. We will use the formula for elongation in a wire, which is given by:

\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

where:
- \( \Delta L \) = increase in length
- \( F \) = applied load
- \( L \) = original length of the wire
- \( A \) = cross-sectional area of the wire
- \( Y \) = Young's modulus of the material

### Step-by-Step Solution:

1. **Identify the Cross-Sectional Area**:
The cross-sectional area \( A \) of a wire with diameter \( d \) is given by:
\[ A = \frac{\pi d^2}{4} \]
For two wires with diameters \( d_1 \) and \( d_2 \) (where \( d_1 = n \cdot d_2 \)), we can express their areas as:
\[ A_1 = \frac{\pi d_1^2}{4} = \frac{\pi (n \cdot d_2)^2}{4} = \frac{\pi n^2 d_2^2}{4} \]
\[ A_2 = \frac{\pi d_2^2}{4} \]

2. **Use the Formula for Increase in Length**:
Since both wires have the same length \( L = 4 \, \text{m} \) and are subjected to the same load \( F \), we can write the equations for the increase in length for both wires:
\[ \Delta L_1 = \frac{F \cdot L}{A_1 \cdot Y} = \frac{F \cdot 4}{\frac{\pi n^2 d_2^2}{4} \cdot Y} = \frac{16F}{\pi n^2 d_2^2 Y} \]
\[ \Delta L_2 = \frac{F \cdot L}{A_2 \cdot Y} = \frac{F \cdot 4}{\frac{\pi d_2^2}{4} \cdot Y} = \frac{16F}{\pi d_2^2 Y} \]

3. **Find the Ratio of Increases in Length**:
Now, we can find the ratio of the increases in length of the two wires:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{\frac{16F}{\pi n^2 d_2^2 Y}}{\frac{16F}{\pi d_2^2 Y}} = \frac{1}{n^2} \]
This implies:
\[ \Delta L_1 = \frac{1}{n^2} \Delta L_2 \]

4. **Express the Increase in Length**:
If we denote the increase in length of the second wire as \( \Delta L_2 \), then the increase in length of the first wire can be expressed as:
\[ \Delta L_1 = \frac{\Delta L_2}{n^2} \]

### Conclusion:
Thus, the increase in length of the first wire is \( \Delta L_1 = \frac{\Delta L_2}{n^2} \).