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

Options

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

Answer

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 stress to strain in a material. It is a measure of the stiffness of a material. Mathematically, it is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L_0} \]
where \( F \) is the force applied, \( A \) is the cross-sectional area, \( \Delta L \) is the change in length, and \( L_0 \) is the original length.

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

3. **Changing Dimensions**:
The problem states that the length of the wire is doubled, so the new length \( L' = 2L \). The radius is reduced to half, so the new radius \( r' = \frac{r}{2} \).

4. **Calculating Cross-Sectional Area**:
The cross-sectional area \( A \) of the wire is given by:
\[ A = \pi r^2 \]
The new cross-sectional area \( A' \) after changing the radius will be:
\[ A' = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} = \frac{A}{4} \]

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

6. **Conclusion**:
Since the Young's modulus is a property of the material, it will remain \( Y \) regardless of the changes in length and radius. Thus, the Young's modulus after the changes is still \( Y \).

### Final Answer:
The Young's modulus of the wire after doubling its length and halving its radius remains \( Y \) newton/m².

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