A heavy rope is suspended from the ceiling of a room. If `phi` is the density of the rope, L be its original length and Y be its. Young's modulus, then increase `DeltaL` in the length of the rope due to its own weight is

Options

A. a. `DeltaL=(phi gL^(2))/(Y)`
B. b. `DeltaL=(2phi g L^(2))/(Y)`
C. c. `DeltaL=(phi g L^(2))/(2Y)`
D. d. `DeltaL=(phigL)/(Y)`

Answer

To find the increase in length \( \Delta L \) of a heavy rope suspended from the ceiling due to its own weight, we can follow these steps:

### Step 1: Understand the Problem
We have a rope of original length \( L \), density \( \phi \), and Young's modulus \( Y \). The rope is subjected to its own weight, which causes it to stretch. We need to find the increase in length \( \Delta L \).

### Step 2: Consider an Element of the Rope
Consider a small element of the rope at a distance \( x \) from the bottom with a small length \( dx \). The weight of this element will contribute to the tension in the rope.

### Step 3: Calculate the Weight of the Element
The mass \( m \) of the small element can be expressed as:
\[ m = \text{Volume} \times \text{Density} = A \cdot dx \cdot \phi \]
where \( A \) is the cross-sectional area of the rope.

The weight \( W \) of this small element is given by:
\[ W = m \cdot g = A \cdot dx \cdot \phi \cdot g \]
where \( g \) is the acceleration due to gravity.

### Step 4: Determine the Stress on the Element
The stress \( \sigma \) on the element is defined as the force (weight) per unit area:
\[ \sigma = \frac{W}{A} = \frac{A \cdot dx \cdot \phi \cdot g}{A} = \phi \cdot g \cdot dx \]

### Step 5: Calculate the Strain in the Element
Strain \( \epsilon \) is defined as the change in length per unit length:
\[ \epsilon = \frac{\Delta l}{dx} \]
where \( \Delta l \) is the change in length of the small element.

### Step 6: Relate Stress and Strain using Young's Modulus
According to Hooke's Law, Young's modulus \( Y \) relates stress and strain:
\[ Y = \frac{\sigma}{\epsilon} \]
Substituting the expressions for stress and strain:
\[ Y = \frac{\phi \cdot g \cdot dx}{\Delta l / dx} \]
Rearranging gives:
\[ \Delta l = \frac{\phi \cdot g \cdot dx}{Y} \]

### Step 7: Integrate to Find Total Change in Length
To find the total increase in length \( \Delta L \), we need to integrate \( \Delta l \) from \( x = 0 \) to \( x = L \):
\[ \Delta L = \int_0^L \frac{\phi \cdot g \cdot x}{Y} \, dx \]
This simplifies to:
\[ \Delta L = \frac{\phi \cdot g}{Y} \int_0^L x \, dx \]
Calculating the integral:
\[ \int_0^L x \, dx = \frac{L^2}{2} \]
Thus,
\[ \Delta L = \frac{\phi \cdot g}{Y} \cdot \frac{L^2}{2} \]

### Step 8: Final Expression
The final expression for the increase in length \( \Delta L \) is:
\[ \Delta L = \frac{\phi \cdot g \cdot L^2}{2Y} \]

### Conclusion
The increase in length \( \Delta L \) in the rope due to its own weight is:
\[ \Delta L = \frac{\phi g L^2}{2Y} \]