Two wires have the same material and length, but their masses are in the ratio of `4:3`. If they are stretched by the same force, their elongations will be in the ratio of

Options

A. `2:3`
B. `3:4`
C. `4:3`
D. `9:16`

Explanation

To solve the problem, we need to find the ratio of elongations of two wires made of the same material and having the same length, but with masses in the ratio of 4:3 when stretched by the same force.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- Let the masses of the two wires be \( m_1 \) and \( m_2 \).
- Given that \( \frac{m_1}{m_2} = \frac{4}{3} \).
- Both wires have the same length \( L \) and are made of the same material, meaning they have the same Young's modulus \( Y \).

2. **Young's Modulus Relation:**
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Rearranging gives:
\[ \Delta L = \frac{F L}{A Y} \]

3. **Finding the Area of Cross-section:**
- The mass of a wire can be expressed as:
\[ m = \rho V = \rho A L \]
- Since both wires have the same material and length, we can express the masses in terms of their areas:
\[ m_1 = \rho A_1 L \quad \text{and} \quad m_2 = \rho A_2 L \]
- Taking the ratio of the masses:
\[ \frac{m_1}{m_2} = \frac{\rho A_1 L}{\rho A_2 L} = \frac{A_1}{A_2} \]
- Therefore, we have:
\[ \frac{A_1}{A_2} = \frac{4}{3} \implies \frac{A_2}{A_1} = \frac{3}{4} \]

4. **Finding the Ratio of Elongations:**
- From the elongation formula, we know:
\[ \Delta L_1 = \frac{F L}{A_1 Y} \quad \text{and} \quad \Delta L_2 = \frac{F L}{A_2 Y} \]
- Taking the ratio of the elongations:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{A_2}{A_1} \]
- Substituting the area ratio:
\[ \frac{\Delta L_1}{\Delta L_2} = \frac{3}{4} \]

5. **Final Result:**
- The ratio of elongations \( \Delta L_1 : \Delta L_2 \) is therefore:
\[ \Delta L_1 : \Delta L_2 = 3 : 4 \]

### Conclusion:
The elongations of the two wires when stretched by the same force are in the ratio \( 3:4 \).