The Young's modulus of a rubber string 8 cm long and density `1.5kg//m^(3)` is `5xx10^(8)N//m^(2)` is suspended on the ceiling in a room. The increase in length due to its own weight will be-

Options

A. `9.6xx10^(-5)m`
B. `9.6xx10^(-11)m`
C. `9.6xx10^(-3)m`
D. `9.6m`

Answer

To solve the problem of finding the increase in length of a rubber string due to its own weight, we can follow these steps:

### Step 1: Identify the given values
- Length of the rubber string (L) = 8 cm = 0.08 m
- Density of the rubber string (ρ) = 1.5 kg/m³
- Young's modulus (Y) = 5 × 10⁸ N/m²
- Acceleration due to gravity (g) = 10 m/s² (approximately)

### Step 2: Use the formula for the increase in length due to its own weight
The formula to calculate the increase in length (ΔL) of a material under its own weight is given by:

\[ \Delta L = \frac{L^2 \cdot \rho \cdot g}{2Y} \]

### Step 3: Substitute the values into the formula
Substituting the values into the formula:

\[ \Delta L = \frac{(0.08)^2 \cdot (1.5) \cdot (10)}{2 \cdot (5 \times 10^8)} \]

### Step 4: Calculate the numerator
First, calculate \( (0.08)^2 \):

\[ (0.08)^2 = 0.0064 \]

Now, calculate the numerator:

\[ 0.0064 \cdot 1.5 \cdot 10 = 0.096 \]

### Step 5: Calculate the denominator
Now calculate the denominator:

\[ 2 \cdot (5 \times 10^8) = 10 \times 10^8 = 1 \times 10^9 \]

### Step 6: Combine the results
Now, substitute the results back into the equation for ΔL:

\[ \Delta L = \frac{0.096}{1 \times 10^9} \]

### Step 7: Final calculation
Calculating ΔL gives:

\[ \Delta L = 9.6 \times 10^{-11} \text{ m} \]

### Conclusion
The increase in length of the rubber string due to its own weight is \( 9.6 \times 10^{-11} \text{ m} \).