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 will follow these steps:

### Step 1: Understand the relationship between stress, strain, and Young's modulus
Young's modulus (Y) is defined as the ratio of stress to strain. Mathematically, it can be expressed as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Where:
- \( F \) = Force applied
- \( A \) = Cross-sectional area of the wire
- \( \Delta L \) = Change in length
- \( L \) = Original length

### Step 2: Express the change in length (\( \Delta L \))
From the definition of Young's modulus, we can rearrange the equation to express the change in length (\( \Delta L \)):
\[ \Delta L = \frac{F L}{A Y} \]

### Step 3: Determine the cross-sectional area of the wires
The cross-sectional area \( A \) of a wire can be expressed in terms of its diameter \( d \):
\[ A = \frac{\pi}{4} d^2 \]
Thus, for two wires (thick wire A and thin wire B), we have:
- For wire A (thick): \( A_A = \frac{\pi}{4} d_A^2 \)
- For wire B (thin): \( A_B = \frac{\pi}{4} d_B^2 \)

### Step 4: Set up the ratio of the increase in lengths
Since both wires are made of the same material, have the same length, and are subjected to the same load, we can compare the increases in length of the two wires:
\[ \frac{\Delta L_B}{\Delta L_A} = \frac{A_A}{A_B} \]

Substituting the expressions for the areas:
\[ \frac{\Delta L_B}{\Delta L_A} = \frac{\frac{\pi}{4} d_A^2}{\frac{\pi}{4} d_B^2} = \frac{d_A^2}{d_B^2} \]

### Step 5: Use the given ratio of diameters
We are given the ratio of diameters of the two wires as \( n:1 \), which means:
\[ \frac{d_A}{d_B} = n \]
Thus, we can express the ratio of the squares of the diameters:
\[ \frac{d_A^2}{d_B^2} = n^2 \]

### Step 6: Relate the increases in length
Now substituting this back into our ratio of increases in length:
\[ \frac{\Delta L_B}{\Delta L_A} = n^2 \]
This implies:
\[ \Delta L_B = n^2 \Delta L_A \]

### Conclusion
The increase in the length of the thin wire (wire B) is \( n^2 \) times the increase in the length of the thick wire (wire A).

### Final Answer
Thus, the increase in the length of the thin wire will be:
\[ \Delta L_B = n^2 \Delta L_A \]

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