To solve the problem, we need to determine the length of the wire that will break under its own weight when suspended vertically. We will use the concepts of density, breaking stress, and the relationship between stress, force, and area.
### Step-by-Step Solution:
1. **Understand the Given Data:**
- Density of the material, \( \rho = 10 \, \text{g/cm}^3 \)
- Breaking stress, \( \sigma = 5 \times 10^9 \, \text{N/m}^2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
2. **Convert Density to SI Units:**
- Since \( 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \), we convert the density:
\[
\rho = 10 \, \text{g/cm}^3 = 10 \times 1000 \, \text{kg/m}^3 = 10000 \, \text{kg/m}^3
\]
3. **Define the Stress Equation:**
- The stress in the wire due to its own weight can be expressed as:
\[
\sigma = \frac{F}{A}
\]
where \( F \) is the force acting on the wire and \( A \) is the cross-sectional area.
4. **Calculate the Force (Weight) of the Wire:**
- The weight of the wire can be expressed as:
\[
F = mg = \rho V g
\]
where \( V \) is the volume of the wire. For a wire of length \( L \) and cross-sectional area \( A \):
\[
V = A \cdot L
\]
Thus, the weight becomes:
\[
F = \rho (A \cdot L) g
\]
5. **Substituting into the Stress Equation:**
- The stress can now be rewritten as:
\[
\sigma = \frac{\rho (A \cdot L) g}{A} = \rho L g
\]
6. **Set the Stress Equal to the Breaking Stress:**
- The wire will break when the stress reaches the breaking stress:
\[
\sigma = \rho L g = 5 \times 10^9 \, \text{N/m}^2
\]
7. **Substituting Known Values:**
- Plug in the values of \( \rho \) and \( g \):
\[
10000 \, \text{kg/m}^3 \cdot L \cdot 10 \, \text{m/s}^2 = 5 \times 10^9 \, \text{N/m}^2
\]
8. **Solving for Length \( L \):**
- Rearranging the equation gives:
\[
L = \frac{5 \times 10^9}{10000 \cdot 10}
\]
\[
L = \frac{5 \times 10^9}{10^4 \cdot 10} = \frac{5 \times 10^9}{10^5} = 5 \times 10^4 \, \text{m}
\]
9. **Final Answer:**
- The length of the wire that will break under its own weight is:
\[
L = 5 \times 10^4 \, \text{m}
\]
