If the load of a wire is increases such that its stress is twice as that of previous, then the new value of Young`s modulus is

Options

A. Increases
B. Remain same
C. Decreases
D. May increases or decreases

Answer

To solve the problem, we need to understand the relationship between stress, strain, and Young's modulus.

### Step-by-Step Solution:

1. **Understand the Definitions**:
- **Stress (σ)** is defined as the force (F) applied per unit area (A):
\[ \sigma = \frac{F}{A} \]
- **Strain (ε)** is the change in length (ΔL) per unit original length (L0):
\[ \epsilon = \frac{\Delta L}{L_0} \]
- **Young's Modulus (Y)** is defined as the ratio of stress to strain:
\[ Y = \frac{\sigma}{\epsilon} \]

2. **Initial Conditions**:
- Let the initial stress be \( \sigma_1 \) and the corresponding strain be \( \epsilon_1 \).
- Thus, the initial Young's modulus \( Y_1 \) can be expressed as:
\[ Y_1 = \frac{\sigma_1}{\epsilon_1} \]

3. **New Conditions**:
- According to the problem, the stress is increased to twice its previous value:
\[ \sigma_2 = 2\sigma_1 \]

4. **Relationship Between Stress and Strain**:
- Stress is directly proportional to strain, meaning if stress doubles, strain also changes.
- Therefore, if \( \sigma_2 = 2\sigma_1 \), then the new strain \( \epsilon_2 \) will also change.
- Since stress and strain are proportional, we can express the new strain as:
\[ \epsilon_2 = k \cdot \sigma_2 = k \cdot (2\sigma_1) = 2\epsilon_1 \]
- Here, \( k \) is a constant of proportionality that remains the same for the material.

5. **Calculate New Young's Modulus**:
- The new Young's modulus \( Y_2 \) can be calculated using the new stress and strain:
\[ Y_2 = \frac{\sigma_2}{\epsilon_2} = \frac{2\sigma_1}{2\epsilon_1} = \frac{\sigma_1}{\epsilon_1} = Y_1 \]
- Thus, the new value of Young's modulus remains the same as the original value.

### Conclusion:
The new value of Young's modulus \( Y_2 \) is equal to the original Young's modulus \( Y_1 \).

**Final Answer**: The new value of Young's modulus is the same as the previous value.

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