The ratio of diameters of two wires of same material is `n:1`. The length of each wire is `4 m`. On applying the same load, the increases in the length of the thin wire will be `(n gt 1)`

Options

A. `n^(2)`times
B. `n` times
C. `2n` times
D. `(2n+1)` times

Explanation

To solve the problem, we need to analyze the relationship between the elongation of two wires made of the same material, with a given ratio of diameters and equal lengths.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- We have two wires made of the same material.
- The ratio of their diameters is given as \( n:1 \).
- The length of each wire is \( L = 4 \, \text{m} \).
- The same load is applied to both wires.

2. **Using the Formula for Elongation**:
- The formula for elongation (\( \Delta L \)) of a wire under a load is given by:
\[ \Delta L = \frac{F L}{A Y} \]
where \( F \) is the force applied, \( L \) is the original length, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus of the material.

3. **Cross-Sectional Area**:
- The cross-sectional area \( A \) of a wire with diameter \( d \) is given by:
\[ A = \frac{\pi d^2}{4} \]
- For the two wires, let \( d_1 = n \) (thicker wire) and \( d_2 = 1 \) (thinner wire).

4. **Calculating Areas**:
- The area of the thicker wire (\( A_1 \)):
\[ A_1 = \frac{\pi (n)^2}{4} \]
- The area of the thinner wire (\( A_2 \)):
\[ A_2 = \frac{\pi (1)^2}{4} = \frac{\pi}{4} \]

5. **Finding the Ratio of Elongations**:
- Since both wires are subjected to the same load \( F \) and have the same length \( L \), we can set up the ratio of elongations:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{A_1}{A_2} \]
- Substituting the areas:
\[ \frac{\Delta L_2}{\Delta L_1} = \frac{\frac{\pi (n)^2}{4}}{\frac{\pi}{4}} = n^2 \]

6. **Expressing the Elongation of the Thinner Wire**:
- Rearranging gives:
\[ \Delta L_2 = n^2 \Delta L_1 \]
- This indicates that the increase in length of the thinner wire (\( \Delta L_2 \)) is \( n^2 \) times the increase in length of the thicker wire (\( \Delta L_1 \)).

7. **Conclusion**:
- Therefore, the increase in the length of the thin wire will be \( n^2 \) times the increase in length of the thicker wire.

### Final Answer:
The increase in the length of the thin wire will be \( n^2 \) times the increase in length of the thicker wire.

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