Find the greatest length of steel wire that can hang vertically without breaking.Breaking stress of steel `=8.0xx10^(8) N//m^(2)`. Density of steel `=8.0xx10^(3) kg//m^(3)`. Take `g =10 m//s^(2)`.

Explanation

To find the greatest length of a steel wire that can hang vertically without breaking, we will use the given properties of steel: breaking stress, density, and the acceleration due to gravity.

### Step-by-Step Solution:

1. **Understand the Problem**:
We need to determine the maximum length \( L \) of a steel wire that can hang vertically without breaking due to its own weight.

2. **Identify the Given Data**:
- Breaking stress of steel, \( \sigma_M = 8.0 \times 10^8 \, \text{N/m}^2 \)
- Density of steel, \( \rho = 8.0 \times 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

3. **Relate Stress, Force, and Area**:
The stress experienced by the wire due to its own weight can be expressed as:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force (weight of the wire) and \( A \) is the cross-sectional area of the wire.

4. **Calculate the Weight of the Wire**:
The weight \( F \) of the wire can be expressed in terms of its mass \( M \):
\[ F = M \cdot g \]
The mass \( M \) of the wire can be expressed as:
\[ M = \rho \cdot V \]
where \( V \) is the volume of the wire. The volume can be expressed in terms of the cross-sectional area \( A \) and the length \( L \):
\[ V = A \cdot L \]
Thus, we have:
\[ M = \rho \cdot A \cdot L \]
Therefore, the weight of the wire becomes:
\[ F = \rho \cdot A \cdot L \cdot g \]

5. **Substitute into the Stress Equation**:
Now substituting \( F \) into the stress equation:
\[ \sigma = \frac{\rho \cdot A \cdot L \cdot g}{A} = \rho \cdot L \cdot g \]

6. **Set the Stress Equal to Breaking Stress**:
To avoid breaking, the stress must not exceed the breaking stress:
\[ \sigma_M = \rho \cdot L \cdot g \]
Therefore, we can set up the equation:
\[ 8.0 \times 10^8 = 8.0 \times 10^3 \cdot L \cdot 10 \]

7. **Solve for Length \( L \)**:
Rearranging the equation gives:
\[ L = \frac{8.0 \times 10^8}{8.0 \times 10^3 \cdot 10} \]
Simplifying this:
\[ L = \frac{8.0 \times 10^8}{8.0 \times 10^4} = 10^4 \, \text{m} \]

8. **Final Result**:
The greatest length of the steel wire that can hang vertically without breaking is:
\[ L = 10,000 \, \text{m} \text{ or } 10 \, \text{km} \]