A wire of mass m and length l is suspended from a ceiling. Due to its own wieght it elongates, consider cross-section are of wire as A ans Young's modulus of material of wire as Y. Th eelongation in the wire is

Options

A. `(2mg)/(3YA)`
B. `(mgl)/(YA)`
C. `(mgl)/(2YA)`
D. cannot be calcuated

Explanation

To find the elongation in a wire of mass \( m \) and length \( l \) suspended from a ceiling due to its own weight, we can follow these steps:

### Step 1: Understand the Forces Acting on the Wire
The wire is subjected to its own weight, which creates a tension that varies along its length. The weight of the wire acts downward, and the tension at any point in the wire depends on the weight of the portion of the wire below that point.

### Step 2: Consider an Element of the Wire
Let’s consider a small element of the wire of length \( dx \) at a distance \( x \) from the bottom of the wire. The weight of the wire below this point contributes to the tension at this point.

### Step 3: Calculate the Weight of the Wire Below the Element
The mass of the wire below the element can be expressed as:
\[ m_{\text{below}} = \frac{m}{l} (l - x) \]
Thus, the weight of this portion is:
\[ F = m_{\text{below}} \cdot g = \frac{m}{l} (l - x) \cdot g \]

### Step 4: Determine the Tension in the Wire
The tension \( T \) at the point \( x \) is equal to the weight of the wire below it:
\[ T = \frac{m}{l} (l - x) \cdot g \]

### Step 5: Relate the Tension to the Elongation
Using the definition of Young's modulus \( Y \):
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{T/A}{\Delta L / L} \]
Rearranging gives:
\[ \Delta L = \frac{T \cdot L}{Y \cdot A} \]

### Step 6: Substitute for Tension
Substituting the expression for tension \( T \):
\[ \Delta L = \frac{\left(\frac{m}{l} (l - x) \cdot g\right) \cdot L}{Y \cdot A} \]

### Step 7: Integrate to Find Total Elongation
To find the total elongation, we need to integrate from \( x = 0 \) to \( x = l \):
\[ \Delta L = \int_0^l \frac{\left(\frac{m}{l} (l - x) \cdot g\right) \cdot L}{Y \cdot A} \, dx \]
This simplifies to:
\[ \Delta L = \frac{mgL}{Y \cdot A} \int_0^l \frac{(l - x)}{l} \, dx \]
Calculating the integral:
\[ \int_0^l (l - x) \, dx = \left[ lx - \frac{x^2}{2} \right]_0^l = l^2 - \frac{l^2}{2} = \frac{l^2}{2} \]
Thus,
\[ \Delta L = \frac{mgL}{Y \cdot A} \cdot \frac{l^2}{2l} = \frac{mgl}{2YA} \]

### Final Result
The elongation in the wire is given by:
\[ \Delta L = \frac{mgl}{2YA} \]