Two wires A and B are of the same length. Their diameters are in the ratio `1:2` and the Young's modulii are in ratio `2:1` . If they are pulled by the same force, then their elongtions will be in ratio

Options

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

Explanation

To solve the problem, we will use the relationship between stress, strain, and Young's modulus. The elongation (ΔL) of a wire can be expressed using the formula:

\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]

Where:
- \( F \) = Force applied
- \( L \) = Original length of the wire
- \( A \) = Cross-sectional area of the wire
- \( Y \) = Young's modulus of the material

### Step-by-Step Solution:

1. **Identify the Given Ratios:**
- The diameters of wires A and B are in the ratio \( D_A : D_B = 1 : 2 \).
- The Young's moduli of wires A and B are in the ratio \( Y_A : Y_B = 2 : 1 \).

2. **Express the Cross-Sectional Area:**
- The cross-sectional area \( A \) of a wire is given by:
\[ A = \frac{\pi D^2}{4} \]
- For wire A:
\[ A_A = \frac{\pi D_A^2}{4} \]
- For wire B:
\[ A_B = \frac{\pi D_B^2}{4} \]

3. **Substituting the Diameter Ratios:**
- Let \( D_A = d \) and \( D_B = 2d \) (since \( D_A : D_B = 1 : 2 \)).
- Therefore:
\[ A_A = \frac{\pi d^2}{4} \]
\[ A_B = \frac{\pi (2d)^2}{4} = \frac{\pi \cdot 4d^2}{4} = \pi d^2 \]

4. **Substituting Young's Modulus Ratios:**
- Let \( Y_A = 2Y \) and \( Y_B = Y \) (since \( Y_A : Y_B = 2 : 1 \)).

5. **Calculating Elongations:**
- For wire A:
\[ \Delta L_A = \frac{F \cdot L}{A_A \cdot Y_A} = \frac{F \cdot L}{\left(\frac{\pi d^2}{4}\right) \cdot (2Y)} = \frac{4F \cdot L}{2\pi d^2 Y} = \frac{2F \cdot L}{\pi d^2 Y} \]
- For wire B:
\[ \Delta L_B = \frac{F \cdot L}{A_B \cdot Y_B} = \frac{F \cdot L}{(\pi d^2) \cdot Y} = \frac{F \cdot L}{\pi d^2 Y} \]

6. **Finding the Ratio of Elongations:**
- Now, we can find the ratio of elongations:
\[ \frac{\Delta L_A}{\Delta L_B} = \frac{\frac{2F \cdot L}{\pi d^2 Y}}{\frac{F \cdot L}{\pi d^2 Y}} = \frac{2F \cdot L}{F \cdot L} = 2 \]

### Final Answer:
The ratio of the elongations of wires A and B is:
\[ \Delta L_A : \Delta L_B = 2 : 1 \]