Young's modules of material of a wire of length ' "L" ' and cross-sectional area "A" is "Y" .If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be :

Options

A. `Y/4`
B. Y
C. 4Y
D. 2Y

Answer

To solve the problem, we need to understand the definition of Young's modulus and how it relates to the dimensions of the wire.

### Step-by-Step Solution:

1. **Understanding Young's Modulus**:
Young's modulus (Y) is defined as the ratio of stress to strain in a material. It is given by the formula:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
where:
- Stress = \(\frac{F}{A}\) (Force per unit area)
- Strain = \(\frac{\Delta L}{L}\) (Change in length per original length)

2. **Initial Conditions**:
Let the initial length of the wire be \(L\) and the initial cross-sectional area be \(A\). The Young's modulus for this wire is given as \(Y\).

3. **New Conditions**:
- The length of the wire is doubled: New length \(L' = 2L\)
- The cross-sectional area is halved: New area \(A' = \frac{A}{2}\)

4. **Calculating New Stress**:
The stress in the wire can be calculated as:
\[ \text{Stress} = \frac{F}{A'} \]
Substituting the new area:
\[ \text{Stress} = \frac{F}{\frac{A}{2}} = \frac{2F}{A} \]

5. **Calculating New Strain**:
The strain in the wire can be calculated as:
\[ \text{Strain} = \frac{\Delta L}{L'} \]
Here, if we assume that the extension \(\Delta L\) remains the same, we can write:
\[ \text{Strain} = \frac{\Delta L}{2L} \]

6. **Calculating New Young's Modulus**:
Now, substituting the new stress and strain into the Young's modulus formula:
\[ Y' = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{2F}{A}}{\frac{\Delta L}{2L}} = \frac{2F \cdot 2L}{A \cdot \Delta L} = \frac{4FL}{A \Delta L} \]

Since the original Young's modulus \(Y = \frac{F L}{A \Delta L}\), we can express the new Young's modulus as:
\[ Y' = 4Y \]

7. **Conclusion**:
The Young's modulus of the material remains a property of the material itself and does not change with the dimensions of the wire. Therefore, even though we have changed the dimensions, the Young's modulus remains the same:
\[ Y' = Y \]

### Final Answer:
The Young's modulus will remain \(Y\).