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. `(2 lambda g L^(2))/(Al)`
B. `(lambda g L^(2))/(2Al)`
C. `(2 lambda gL)/(Al)`
D. `(lambda gl^(2))/(AL)`

Answer

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. **Identify the Forces Acting on the Rod**:
The rod is suspended vertically, and the force acting on a small section of the rod at a distance \( x \) from the free end is due to the weight of the portion of the rod below that point. The weight of the section below \( x \) is given by:
\[ F = \lambda (L - x) g \]
where \( g \) is the acceleration due to gravity.

2. **Calculate the Stress at Section \( x \)**:
Stress (\( \sigma \)) at the section \( x \) is defined as the force acting on the section divided by the area of cross-section \( A \):
\[ \sigma = \frac{F}{A} = \frac{\lambda (L - x) g}{A} \]

3. **Determine the Strain in the Section**:
The elongation \( dL \) of the small section \( dx \) can be expressed in terms of stress and Young's modulus (\( Y \)):
\[ dL = \frac{\sigma \cdot dx}{Y} = \frac{\lambda (L - x) g \cdot dx}{A Y} \]

4. **Integrate to Find Total Elongation**:
To find the total elongation \( l \) of the rod, integrate \( dL \) from \( x = 0 \) to \( x = L \):
\[ l = \int_0^L \frac{\lambda (L - x) g \cdot dx}{A Y} \]

5. **Evaluate the Integral**:
The integral can be simplified:
\[ l = \frac{\lambda g}{A Y} \int_0^L (L - x) \, dx \]
The integral \( \int_0^L (L - x) \, dx = \left[ Lx - \frac{x^2}{2} \right]_0^L = L^2 - \frac{L^2}{2} = \frac{L^2}{2} \).
Thus,
\[ l = \frac{\lambda g}{A Y} \cdot \frac{L^2}{2} \]

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

### Final Expression for Young's Modulus:
The Young's modulus of the rod is given by:
\[ Y = \frac{\lambda g L^2}{2 A l} \]