Two metal wires P and Q of same length and material are stretched by same load. Yheir masses are in the ratio `m_(1):m_(2)`. The ratio of elongation of wire P to that of Q is

Options

A. `m_1^2:m_2^2`
B. `m_2^2:m_1^2`
C. `m_2:m_1`
D. `m_1 : m_2`

Answer

To solve the problem, we need to find the ratio of elongation of two metal wires P and Q, given that they have the same length, material, and are subjected to the same load. Their masses are in the ratio \( m_1 : m_2 \).

### Step-by-Step Solution:

1. **Understand the relationship between stress, strain, and Young's modulus**:
- Young's modulus \( Y \) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
- Rearranging this gives:
\[ \Delta L = \frac{F L}{A Y} \]

2. **Express the area of cross-section in terms of mass**:
- The mass \( m \) of a wire can be expressed as:
\[ m = \rho V = \rho A L \]
- Rearranging gives:
\[ A = \frac{m}{\rho L} \]

3. **Substitute the area into the elongation formula**:
- Substituting the expression for area \( A \) into the elongation formula:
\[ \Delta L = \frac{F L}{(\frac{m}{\rho L}) Y} = \frac{F \rho L^2}{m Y} \]

4. **Determine the ratio of elongation for wires P and Q**:
- Let \( \Delta L_P \) be the elongation of wire P and \( \Delta L_Q \) be the elongation of wire Q.
- The elongations can be expressed as:
\[ \Delta L_P = \frac{F \rho L^2}{m_1 Y} \quad \text{and} \quad \Delta L_Q = \frac{F \rho L^2}{m_2 Y} \]
- Now, taking the ratio of elongations:
\[ \frac{\Delta L_P}{\Delta L_Q} = \frac{\frac{F \rho L^2}{m_1 Y}}{\frac{F \rho L^2}{m_2 Y}} = \frac{m_2}{m_1} \]

5. **Final Result**:
- Therefore, the ratio of elongation of wire P to that of wire Q is:
\[ \frac{\Delta L_P}{\Delta L_Q} = \frac{m_2}{m_1} \]

### Conclusion:
The ratio of elongation of wire P to that of wire Q is \( \frac{m_2}{m_1} \).

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