The young's modulus of a wire of length (L) and radius (r ) is Y. If the length is reduced to` L/2` and radius `r/2` , then its young's modulus will be

Options

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

Explanation

To solve the problem, we need to understand the concept of Young's modulus and how it relates to the properties of the material.

### Step-by-Step Solution:

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

2. **Initial Conditions**:
Let the initial length of the wire be \( L \) and the initial radius be \( r \). The initial Young's modulus is given as \( Y \).

3. **New Conditions**:
The length of the wire is reduced to \( L/2 \) and the radius is reduced to \( r/2 \).

4. **Cross-Sectional Area Calculation**:
The cross-sectional area \( A \) of a wire is given by:
\[ A = \pi r^2 \]
- Initial area \( A_1 = \pi r^2 \)
- New area \( A_2 = \pi (r/2)^2 = \pi \frac{r^2}{4} = \frac{1}{4} A_1 \)

5. **Effect on Young's Modulus**:
Young's modulus is a property of the material and does not depend on the dimensions of the wire. Therefore, even after changing the length and radius, the Young's modulus remains the same.

6. **Conclusion**:
The Young's modulus after reducing the length to \( L/2 \) and the radius to \( r/2 \) will still be:
\[ Y' = Y \]

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

---