A uniform rod of length `L`, has a mass per unit length `lambda` and area of cross section `A`. The elongation in the rod is `l` due to its own weight, if it suspended form the celing of a room. The Young's modulus of the rod is

Options

A. `(3 lambda g L^(2))/(Al)`
B. `(lambda g L^(2))/(2Al)`
C. `(2lambda g L)/(Al)`
D. `(lambda g L^(2))/(Al)`

Explanation

To find the Young's modulus of a uniform rod of length \( L \), mass per unit length \( \lambda \), cross-sectional area \( A \), and elongation \( l \) due to its own weight when suspended from the ceiling, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
The rod is uniform, meaning its mass is distributed evenly along its length. The weight of the rod causes it to elongate when suspended.

2. **Define the Variables**:
- Length of the rod: \( L \)
- Mass per unit length: \( \lambda \)
- Cross-sectional area: \( A \)
- Elongation due to its weight: \( l \)
- Young's modulus: \( Y \)

3. **Calculate the Force on a Small Element**:
Consider a small element of the rod at a distance \( x \) from the top, with thickness \( dx \). The weight of the portion of the rod below this element is:
\[ F = \lambda (L - x) g \]
where \( g \) is the acceleration due to gravity.

4. **Determine the Stress and Strain**:
The stress \( \sigma \) on the small element is given by:
\[ \sigma = \frac{F}{A} = \frac{\lambda (L - x) g}{A} \]
The strain \( \epsilon \) is defined as:
\[ \epsilon = \frac{dl}{L} \]

5. **Relate Stress and Strain**:
According to Hooke's Law:
\[ \sigma = Y \epsilon \]
Substituting the expressions for stress and strain, we have:
\[ \frac{\lambda (L - x) g}{A} = Y \frac{dl}{L} \]

6. **Integrate to Find Total Elongation**:
The total elongation \( l \) of the rod can be found by integrating the above equation from \( x = 0 \) to \( x = L \):
\[ l = \int_0^L \frac{\lambda (L - x) g}{AY} \, dx \]

7. **Perform the Integration**:
The integral simplifies to:
\[ l = \frac{\lambda g}{AY} \int_0^L (L - x) \, dx \]
The integral \( \int_0^L (L - x) \, dx = \left[ Lx - \frac{x^2}{2} \right]_0^L = L^2 - \frac{L^2}{2} = \frac{L^2}{2} \).

8. **Substituting Back**:
Thus, we have:
\[ l = \frac{\lambda g}{AY} \cdot \frac{L^2}{2} \]

9. **Rearranging for Young's Modulus**:
Rearranging the equation to solve for \( Y \):
\[ Y = \frac{\lambda g L^2}{2Al} \]

### Final Answer:
The Young's modulus \( Y \) of the rod is given by:
\[ Y = \frac{\lambda g L^2}{2Al} \]