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 dimensions of the wire.

### Step-by-Step Solution:

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

2. **Initial Conditions**:
- Original length of the wire = \( L \)
- Original radius of the wire = \( r \)
- Original cross-sectional area \( A = \pi r^2 \)
- Original Young's modulus = \( Y \)

3. **New Conditions After Changes**:
- New length of the wire = \( \frac{L}{2} \)
- New radius of the wire = \( \frac{r}{2} \)
- New cross-sectional area \( A' = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} = \frac{A}{4} \)

4. **Calculating New Young's Modulus**:
Since Young's modulus is a property of the material and does not depend on the dimensions of the wire, the Young's modulus remains the same regardless of the changes in length and radius. Therefore, the new Young's modulus \( Y' \) is still equal to the original Young's modulus \( Y \).

5. **Conclusion**:
Thus, after reducing the length and radius of the wire, the Young's modulus remains:
\[ Y' = Y \]

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

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