A block of mass M is suspended from a wire of length L, area of cross-section A and Young's modulus Y. The elastic potential energy stored in the wire is

Options

A. `(1)/(2) (M^(2) g^(2) L)/(AY)`
B. `(1)/(2) (Mg)/(ALY)`
C. `(1)/(2) (M^(2)g^(2)A)/(YL)`
D. `(1)/(2) (MgY)/(AL)`

Answer

To find the elastic potential energy stored in the wire when a block of mass M is suspended from it, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces and Parameters:**
- The block of mass \( M \) exerts a force due to gravity, which is \( F = Mg \).
- The wire has a length \( L \), cross-sectional area \( A \), and Young's modulus \( Y \).

2. **Understand Young's Modulus:**
- Young's modulus \( Y \) is defined as the ratio of stress to strain:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{\frac{F}{A}}{\frac{\Delta L}{L}} \]
- Rearranging gives:
\[ \Delta L = \frac{F L}{A Y} \]

3. **Substitute the Force:**
- Substitute \( F = Mg \) into the equation for extension \( \Delta L \):
\[ \Delta L = \frac{Mg L}{A Y} \]

4. **Determine the Spring Constant \( k \):**
- The spring constant \( k \) for the wire can be derived from the relationship between force, extension, and spring constant:
\[ F = k \Delta L \implies k = \frac{F}{\Delta L} \]
- Substituting \( F = Mg \) and \( \Delta L = \frac{Mg L}{A Y} \):
\[ k = \frac{Mg}{\frac{Mg L}{A Y}} = \frac{A Y}{L} \]

5. **Calculate the Elastic Potential Energy \( U \):**
- The elastic potential energy stored in the wire is given by:
\[ U = \frac{1}{2} k (\Delta L)^2 \]
- Substitute \( k \) and \( \Delta L \):
\[ U = \frac{1}{2} \left(\frac{A Y}{L}\right) \left(\frac{Mg L}{A Y}\right)^2 \]

6. **Simplify the Expression:**
- Simplifying the expression:
\[ U = \frac{1}{2} \cdot \frac{A Y}{L} \cdot \frac{M^2 g^2 L^2}{A^2 Y^2} \]
- Cancel out terms:
\[ U = \frac{1}{2} \cdot \frac{M^2 g^2 L}{A Y} \]

7. **Final Result:**
- The elastic potential energy stored in the wire is:
\[ U = \frac{M^2 g^2 L}{2 A Y} \]