Two wires A and B of same dimensions are stretched by same amount of force .Young's modulus of A is twice that of B. Which wire will get more elongation?

Explanation

To solve the problem, we need to analyze the relationship between the elongation of the two wires A and B when they are subjected to the same force. We will use the formula for elongation based on Young's modulus.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two wires, A and B, with the same dimensions (same length and cross-sectional area).
- The Young's modulus of wire A (E1) is twice that of wire B (E2). Therefore, we can express this as:
\[ E1 = 2E2 \]

2. **Formula for Elongation**:
- The elongation (ΔL) of a wire when a force (F) is applied can be calculated using the formula:
\[ \Delta L = \frac{F \cdot L}{A \cdot E} \]
- Where:
- \( F \) = applied force
- \( L \) = original length of the wire
- \( A \) = cross-sectional area
- \( E \) = Young's modulus of the material

3. **Calculating Elongation for Wire A**:
- For wire A, the elongation (ΔL1) can be expressed as:
\[ \Delta L1 = \frac{F \cdot L}{A \cdot E1} \]

4. **Calculating Elongation for Wire B**:
- For wire B, the elongation (ΔL2) can be expressed as:
\[ \Delta L2 = \frac{F \cdot L}{A \cdot E2} \]

5. **Finding the Ratio of Elongations**:
- To compare the elongations, we can take the ratio of ΔL1 to ΔL2:
\[ \frac{\Delta L1}{\Delta L2} = \frac{\frac{F \cdot L}{A \cdot E1}}{\frac{F \cdot L}{A \cdot E2}} = \frac{E2}{E1} \]
- Since \( E1 = 2E2 \), we can substitute this into the equation:
\[ \frac{\Delta L1}{\Delta L2} = \frac{E2}{2E2} = \frac{1}{2} \]

6. **Conclusion**:
- From the ratio, we find that:
\[ \Delta L2 = 2 \Delta L1 \]
- This means that the elongation of wire B (ΔL2) is twice that of wire A (ΔL1). Therefore, wire B will experience more elongation than wire A.

### Final Answer:
Wire B will get more elongated.