A rubber rope of length `8 m` is hung from the ceiling of a room. What is the increase in length of rope due to its own weight? (Given: Young's modulus of elasticity of rubber `= 5 xx 10^(6) N//m` and density of rubber `=1.5xx10^(3)kg//m^(3)`. Take `g=10ms^(-12))`

Options

A. `1.5mm`
B. `6mm`
C. `24mm`
D. `96mm`

Explanation

To find the increase in length of a rubber rope due to its own weight, we can use the formula derived from Young's modulus. Here are the steps to solve the problem:

### Step-by-Step Solution

1. **Identify the Given Values:**
- Length of the rubber rope, \( L = 8 \, \text{m} \)
- Young's modulus of elasticity of rubber, \( Y = 5 \times 10^6 \, \text{N/m}^2 \)
- Density of rubber, \( \rho = 1.5 \times 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Calculate the Weight of the Rope:**
The weight \( W \) of the rope can be expressed as:
\[ W = m \cdot g \]
where \( m \) is the mass of the rope.

3. **Find the Mass of the Rope:**
The mass \( m \) can be calculated using the density and volume of the rope:
\[ m = \rho \cdot V \]
The volume \( V \) of the rope can be expressed as:
\[ V = A \cdot L \]
where \( A \) is the cross-sectional area of the rope. Thus,
\[ m = \rho \cdot A \cdot L \]

4. **Substituting the Mass in the Weight Equation:**
Substitute \( m \) in the weight equation:
\[ W = \rho \cdot A \cdot L \cdot g \]

5. **Determine the Force Acting on the Rope:**
The effective force acting on the rope due to its weight is \( W \) acting at the center of gravity, which is at \( L/2 \):
\[ F = \frac{W}{2} = \frac{\rho \cdot A \cdot L \cdot g}{2} \]

6. **Use the Formula for Increase in Length:**
The increase in length \( \Delta L \) due to the force \( F \) is given by:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
Substituting \( F \) into this equation:
\[ \Delta L = \frac{\left(\frac{\rho \cdot A \cdot L \cdot g}{2}\right) \cdot L}{A \cdot Y} \]
The area \( A \) cancels out:
\[ \Delta L = \frac{\rho \cdot g \cdot L^2}{2Y} \]

7. **Substituting the Known Values:**
Now substitute the known values into the equation:
\[ \Delta L = \frac{(1.5 \times 10^3 \, \text{kg/m}^3) \cdot (10 \, \text{m/s}^2) \cdot (8 \, \text{m})^2}{2 \cdot (5 \times 10^6 \, \text{N/m}^2)} \]

8. **Calculating \( \Delta L \):**
\[ \Delta L = \frac{(1.5 \times 10^3) \cdot (10) \cdot (64)}{2 \cdot (5 \times 10^6)} \]
\[ \Delta L = \frac{96000}{10 \times 10^6} = \frac{96000}{10000000} = 0.0096 \, \text{m} \]
\[ \Delta L = 9.6 \times 10^{-3} \, \text{m} = 96 \, \text{mm} \]

### Final Answer:
The increase in length of the rubber rope due to its own weight is \( 96 \, \text{mm} \).