A thick rope of density `rho` and length `L` is hung from a rigid support. The increase in length of the rope due to its own weight is (`Y` is the Young's modulus)

Options

A. `0.1/(4Y)rhoL^(2)g`
B. `1/(2Y)rhoL^(2)g`
C. `(rhoL^(2)g)/Y`
D. `(rhoLg)/Y`

Explanation

To solve the problem of finding the increase in length (ΔL) of a thick rope of density ρ and length L hung from a rigid support, we will follow these steps:

### Step-by-Step Solution

1. **Understanding the Setup**:
- We have a thick rope of length L and density ρ hanging from a rigid support. The weight of the rope will cause it to stretch.

2. **Consider a Small Element of the Rope**:
- Let's take a small segment of the rope of length dy at a distance y from the top. The weight of this small segment will contribute to the stress experienced by the rope.

3. **Calculate the Weight of the Small Segment**:
- The mass (m) of the small segment can be expressed as:
\[ m = \text{Volume} \times \text{Density} = A \cdot dy \cdot \rho \]
where A is the cross-sectional area of the rope.
- The weight (W) of this small segment is given by:
\[ W = m \cdot g = A \cdot dy \cdot \rho \cdot g \]

4. **Calculate the Stress on the Small Segment**:
- Stress (σ) is defined as force (F) per unit area (A):
\[ \sigma = \frac{F}{A} = \frac{W}{A} = \frac{A \cdot dy \cdot \rho \cdot g}{A} = \rho \cdot g \cdot dy \]

5. **Calculate the Strain on the Small Segment**:
- Strain (ε) is defined as the change in length (ΔL) divided by the original length (L):
\[ \epsilon = \frac{\Delta L}{y} \]

6. **Using Young's Modulus**:
- Young's modulus (Y) relates stress and strain:
\[ Y = \frac{\sigma}{\epsilon} \]
- Substituting the expressions for stress and strain:
\[ Y = \frac{\rho \cdot g \cdot dy}{\frac{\Delta L}{y}} \]
- Rearranging gives:
\[ \Delta L = \frac{\rho \cdot g \cdot y \cdot dy}{Y} \]

7. **Total Elongation of the Rope**:
- To find the total elongation (ΔL_total), we need to integrate from 0 to L:
\[ \Delta L_{\text{total}} = \int_0^L \frac{\rho \cdot g \cdot y}{Y} \, dy \]
- This integral evaluates to:
\[ \Delta L_{\text{total}} = \frac{\rho \cdot g}{Y} \cdot \left[\frac{y^2}{2}\right]_0^L = \frac{\rho \cdot g}{Y} \cdot \frac{L^2}{2} \]

8. **Final Expression for Elongation**:
- Thus, the increase in length of the rope is:
\[ \Delta L = \frac{\rho \cdot g \cdot L^2}{2Y} \]

### Conclusion
The increase in length of the rope due to its own weight is given by:
\[ \Delta L = \frac{\rho \cdot g \cdot L^2}{2Y} \]