A rubber cord `10 m` long is suspended vertically. How much does it stretch under its own weight (Density of rubber is `1500 kg//m, Y=5xx10 N//m,g = 10 m//s`)

Options

A. `15xx10^(-4) m`
B. `7.5 xx 10^(4) m`
C. `12xx10^(-4)m`
D. `25xx10^(-4)` m

Explanation

To solve the problem of how much the rubber cord stretches under its own weight, we can use the formula for the elongation of a material due to its own weight. The relevant formula is:

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

Where:
- \(\Delta L\) = change in length (stretch)
- \(L\) = original length of the cord
- \(\rho\) = density of the material
- \(g\) = acceleration due to gravity
- \(Y\) = Young's modulus of the material

### Step 1: Identify the given values
- Original length of the rubber cord, \(L = 10 \, \text{m}\)
- Density of rubber, \(\rho = 1500 \, \text{kg/m}^3\)
- Young's modulus, \(Y = 5 \times 10^8 \, \text{N/m}^2\)
- Acceleration due to gravity, \(g = 10 \, \text{m/s}^2\)

### Step 2: Substitute the values into the formula
Now we substitute the values into the formula:

\[ \Delta L = \frac{(10 \, \text{m})^2 \times (1500 \, \text{kg/m}^3) \times (10 \, \text{m/s}^2)}{2 \times (5 \times 10^8 \, \text{N/m}^2)} \]

### Step 3: Calculate the numerator
Calculating the numerator:

\[ (10 \, \text{m})^2 = 100 \, \text{m}^2 \]
\[ 100 \, \text{m}^2 \times 1500 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 = 1500000 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 1500000 \, \text{N} \]

### Step 4: Calculate the denominator
Calculating the denominator:

\[ 2 \times (5 \times 10^8 \, \text{N/m}^2) = 10 \times 10^8 \, \text{N/m}^2 = 10^9 \, \text{N/m}^2 \]

### Step 5: Calculate the change in length
Now substituting the values back into the formula:

\[ \Delta L = \frac{1500000 \, \text{N}}{10^9 \, \text{N/m}^2} \]

Calculating this gives:

\[ \Delta L = 0.0015 \, \text{m} = 1.5 \, \text{mm} \]

### Step 6: Final result
Thus, the rubber cord stretches under its own weight by:

\[ \Delta L = 1.5 \, \text{mm} \]