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 find the increase in length of a rubber string due to its own weight, we will use the formula derived from the definition of Young's modulus. Here’s a step-by-step solution:

### Step 1: Understand the Given Data
- Length of the rubber string, \( L = 8 \, \text{cm} = 0.08 \, \text{m} \)
- Density of the rubber string, \( \rho = 1.5 \, \text{kg/m}^3 \)
- Young's modulus, \( Y = 5 \times 10^8 \, \text{N/m}^2 \)

### Step 2: Calculate the Mass of the Rubber String
The mass \( m \) of the rubber string can be calculated using the formula:
\[ m = \text{Volume} \times \text{Density} \]
The volume \( V \) of the string can be expressed as:
\[ V = A \times L \]
where \( A \) is the cross-sectional area. Therefore,
\[ m = A \times L \times \rho \]

### Step 3: Calculate the Force Due to Weight
The force \( F \) acting on the string due to its own weight is given by:
\[ F = m \times g = A \times L \times \rho \times g \]
where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).

### Step 4: Use the Young's Modulus Formula
The increase in length \( \Delta L \) due to the weight of the string can be calculated using the formula:
\[ \Delta L = \frac{F \times L}{2 \times A \times Y} \]
Substituting \( F \) from the previous step:
\[ \Delta L = \frac{(A \times L \times \rho \times g) \times L}{2 \times A \times Y} \]

### Step 5: Simplify the Equation
The cross-sectional area \( A \) cancels out:
\[ \Delta L = \frac{L^2 \times \rho \times g}{2 \times Y} \]

### Step 6: Substitute the Values
Now, substituting the known values into the equation:
\[ \Delta L = \frac{(0.08)^2 \times 1.5 \times 9.81}{2 \times 5 \times 10^8} \]

Calculating \( (0.08)^2 \):
\[ (0.08)^2 = 0.0064 \]

Now substituting this value:
\[ \Delta L = \frac{0.0064 \times 1.5 \times 9.81}{10 \times 10^8} \]

Calculating \( 0.0064 \times 1.5 \times 9.81 \):
\[ 0.0064 \times 1.5 = 0.0096 \]
\[ 0.0096 \times 9.81 \approx 0.0943 \]

Now substituting back:
\[ \Delta L = \frac{0.0943}{10 \times 10^8} = 0.0943 \times 10^{-8} \, \text{m} = 9.43 \times 10^{-10} \, \text{m} \]

### Step 7: Final Result
Thus, the increase in length due to its own weight is:
\[ \Delta L \approx 9.43 \times 10^{-10} \, \text{m} \]