A metal wire of length L, area of cross-section A and young's modulus `Y` is stretched by a variable force `F` such that `F` is always slightly greater than the elastic forces of resistance in the wire. When the elongation of the wire is `l`

Options

A. (a)the work done by `F` is `(YAl^(2))/(2L)`
B. (b)the work done by `F` is `(YAl^(2))/(L)`
C. (c)the elastic potential energy stored in the wire is `(Yal^(2))/(2L)`
D. (d)the elastic potential energy stored in the wire is ` (Yal^(2))/(4L)`

Explanation

To solve the problem, we need to find the work done on a metal wire when it is stretched by a force \( F \) that is slightly greater than the elastic resistance of the wire, resulting in an elongation \( l \). We will use the relationship between stress, strain, and Young's modulus.

### Step-by-Step Solution:

1. **Understand the relationship between stress, strain, and Young's modulus:**
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Where:
- \( Y \) = Young's modulus
- \( F \) = applied force
- \( A \) = area of cross-section
- \( \Delta L \) = elongation (which is \( l \))
- \( L \) = original length of the wire

2. **Rearranging the formula to find the force \( F \):**
From the formula, we can express \( F \):
\[ F = Y \cdot \frac{A \cdot \Delta L}{L} \]
Substituting \( \Delta L \) with \( l \):
\[ F = Y \cdot \frac{A \cdot l}{L} \]

3. **Using the analogy with spring force:**
The force \( F \) can be compared to Hooke's law for springs:
\[ F = k \cdot x \]
where \( k \) is the spring constant and \( x \) is the elongation. Here, we can identify:
\[ k = \frac{Y \cdot A}{L} \]
and \( x \) corresponds to \( l \).

4. **Calculating the work done (or potential energy stored):**
The work done on the wire (or potential energy stored in the wire) when it is stretched can be calculated using the formula for the work done on a spring:
\[ U = \frac{1}{2} k x^2 \]
Substituting \( k \) and \( x \):
\[ U = \frac{1}{2} \left(\frac{Y \cdot A}{L}\right) l^2 \]

5. **Final expression for work done:**
Therefore, the work done \( U \) can be expressed as:
\[ U = \frac{Y \cdot A \cdot l^2}{2L} \]

### Final Answer:
The work done by the force \( F \) is:
\[ U = \frac{Y \cdot A \cdot l^2}{2L} \]