Two wires of the same material and same mass are stretched by the same force. Their lengths are in the ratio 2:3. Their elongation are in the ratio

Options

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

Explanation

To solve the problem of finding the ratio of elongations of two wires of the same material and mass, stretched by the same force, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Two wires are made of the same material, which means they have the same Young's modulus (Y).
- The lengths of the wires are in the ratio \( l_1 : l_2 = 2 : 3 \).
- The wires have the same mass and are stretched by the same force.

2. **Express Mass in Terms of Density and Volume:**
- The mass (m) of a wire can be expressed as:
\[ m = \text{Density} \times \text{Volume} \]
- The volume (V) of a wire can be expressed as:
\[ V = \text{Area} \times \text{Length} \quad (V = A \cdot l) \]
- Therefore, we can write:
\[ m = \text{Density} \times A \cdot l \]

3. **Set Up the Equation for Both Wires:**
- For wire 1:
\[ m = \rho \cdot A_1 \cdot l_1 \]
- For wire 2:
\[ m = \rho \cdot A_2 \cdot l_2 \]
- Since the mass and density are the same for both wires, we can equate:
\[ A_1 \cdot l_1 = A_2 \cdot l_2 \]

4. **Find the Ratio of Areas:**
- Rearranging gives:
\[ \frac{A_1}{A_2} = \frac{l_2}{l_1} \]
- Substituting the ratio of lengths:
\[ \frac{A_1}{A_2} = \frac{3}{2} \quad \text{(since } l_1 : l_2 = 2 : 3\text{)} \]

5. **Use Young's Modulus to Relate Elongation:**
- The elongation (Δl) is given by:
\[ \Delta l = \frac{F l}{A Y} \]
- Since both wires are subjected to the same force (F) and have the same Young's modulus (Y), we can write:
\[ \frac{\Delta l_1}{\Delta l_2} = \frac{F l_1 / A_1}{F l_2 / A_2} = \frac{l_1 A_2}{l_2 A_1} \]

6. **Substitute the Ratios:**
- Substitute \( l_1 = 2k \) and \( l_2 = 3k \) for some constant \( k \):
\[ \frac{\Delta l_1}{\Delta l_2} = \frac{2k \cdot A_2}{3k \cdot A_1} = \frac{2 \cdot A_2}{3 \cdot A_1} \]
- Now substitute \( \frac{A_1}{A_2} = \frac{2}{3} \):
\[ \frac{\Delta l_1}{\Delta l_2} = \frac{2 \cdot A_2}{3 \cdot \left(\frac{3}{2} A_2\right)} = \frac{2}{3} \cdot \frac{2}{3} = \frac{4}{9} \]

7. **Conclusion:**
- The ratio of elongations of the two wires is:
\[ \frac{\Delta l_1}{\Delta l_2} = \frac{4}{9} \]

### Final Answer:
The elongations of the two wires are in the ratio \( \Delta l_1 : \Delta l_2 = 4 : 9 \).