Two identical wires of iron and copper with their Young's modulus in the ratio `3:1` are suspended at same level. They are to be loaded so as to have the same extension and hence level. Ratio of the weight is

Options

A. `1:3`
B. `2:1`
C. `3:1`
D. `4:1`

Explanation

To solve the problem of finding the ratio of the weights of iron and copper wires that will have the same extension, we can follow these steps:

### Step 1: Understand the relationship between Young's modulus, stress, and strain.
Young's modulus (Y) is defined as the ratio of stress to strain. It can be expressed mathematically as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
where:
- \( F \) is the force applied (weight in this case),
- \( A \) is the cross-sectional area,
- \( \Delta L \) is the extension,
- \( L \) is the original length.

### Step 2: Write the expression for extension in terms of Young's modulus.
The extension \( \Delta L \) can be expressed as:
\[ \Delta L = \frac{F L}{A Y} \]
This means that the extension in a wire is directly proportional to the force applied and the original length, and inversely proportional to the cross-sectional area and Young's modulus.

### Step 3: Set the extensions of both wires equal to each other.
Since the extensions of both wires (iron and copper) are the same, we can write:
\[ \Delta L_{\text{iron}} = \Delta L_{\text{copper}} \]
Substituting the expression for extension, we get:
\[ \frac{F_{\text{iron}} L}{A Y_{\text{iron}}} = \frac{F_{\text{copper}} L}{A Y_{\text{copper}}} \]

### Step 4: Simplify the equation.
Since the lengths \( L \) and cross-sectional areas \( A \) are the same for both wires, they can be canceled out from the equation:
\[ \frac{F_{\text{iron}}}{Y_{\text{iron}}} = \frac{F_{\text{copper}}}{Y_{\text{copper}}} \]

### Step 5: Rearrange to find the ratio of the weights.
Rearranging gives us:
\[ \frac{F_{\text{iron}}}{F_{\text{copper}}} = \frac{Y_{\text{iron}}}{Y_{\text{copper}}} \]

### Step 6: Substitute the given ratio of Young's moduli.
We are given that the ratio of Young's moduli of iron to copper is \( 3:1 \):
\[ \frac{Y_{\text{iron}}}{Y_{\text{copper}}} = \frac{3}{1} \]
So we can substitute this into our equation:
\[ \frac{F_{\text{iron}}}{F_{\text{copper}}} = \frac{3}{1} \]

### Step 7: Conclude the ratio of weights.
Thus, the ratio of the weights (which is the same as the forces since weight \( F = mg \)) is:
\[ \frac{W_{\text{iron}}}{W_{\text{copper}}} = \frac{3}{1} \]

### Final Answer:
The ratio of the weights of iron to copper is \( 3:1 \).

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