A wire of length `L` and radius `r` is fixed at one end. When a stretching force `F` is applied at free end, the elongation in the wire is `l`. When another wire of same material but of length `2L` and radius `2r`, also fixed at one end is stretched by a force `2F` applied at free end, then elongation in the second wire will be

Options

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

Explanation

To solve the problem, we need to understand the relationship between the elongation of a wire, the force applied, its length, and its cross-sectional area. We will use the formula for elongation derived from Young's modulus.

### Step-by-Step Solution:

1. **Understanding the First Wire:**
- For the first wire of length \( L \) and radius \( r \), when a force \( F \) is applied, the elongation \( l \) can be expressed using Young's modulus \( Y \):
\[ l = \frac{F \cdot L}{A \cdot Y} \]
where \( A \) is the cross-sectional area of the wire, given by \( A = \pi r^2 \).

2. **Substituting for Area:**
- The area \( A \) can be substituted into the elongation formula:
\[ l = \frac{F \cdot L}{\pi r^2 \cdot Y} \]

3. **Understanding the Second Wire:**
- For the second wire, which has a length of \( 2L \) and a radius of \( 2r \), and is subjected to a force of \( 2F \), we need to find the elongation \( l_2 \):
\[ l_2 = \frac{2F \cdot (2L)}{A_2 \cdot Y} \]
where \( A_2 = \pi (2r)^2 = 4\pi r^2 \).

4. **Substituting for Area in the Second Wire:**
- Now substituting the area for the second wire:
\[ l_2 = \frac{2F \cdot (2L)}{4\pi r^2 \cdot Y} \]

5. **Simplifying the Expression:**
- Simplifying the expression for \( l_2 \):
\[ l_2 = \frac{4F \cdot L}{4\pi r^2 \cdot Y} = \frac{F \cdot L}{\pi r^2 \cdot Y} \]

6. **Relating \( l_2 \) to \( l \):**
- Notice that the expression for \( l_2 \) is the same as the expression for \( l \):
\[ l_2 = l \]

### Conclusion:
The elongation in the second wire \( l_2 \) is equal to the elongation in the first wire \( l \):
\[ l_2 = l \]