The breaking stress of a material is `10^(8) Nm^(-2)`. Find the greatest length of a wire that could hang vertically without breaking? Density of material `= 3000 kg m^(-3)`.

Explanation

To find the greatest length of a wire that could hang vertically without breaking, we will use the relationship between stress, density, gravitational acceleration, and the length of the wire.

### Step-by-Step Solution:

1. **Understand the Concept of Stress**:
The breaking stress is defined as the maximum stress that a material can withstand before failure. It is given as \( \sigma_{breaking} = 10^8 \, \text{N/m}^2 \).

2. **Define the Variables**:
- Density of the material, \( \rho = 3000 \, \text{kg/m}^3 \)
- Gravitational acceleration, \( g = 10 \, \text{m/s}^2 \)
- Let \( L \) be the length of the wire.

3. **Calculate the Stress in the Wire**:
The stress at a point in the wire can be expressed as:
\[ \sigma = \frac{Weight \, of \, the \, wire}{Area} \]
The weight of the wire can be expressed in terms of its density and volume:
\[ Weight = \rho \cdot Volume = \rho \cdot (Area \cdot Length) = \rho \cdot A \cdot L \]
Therefore, the stress can be rewritten as:
\[ \sigma = \frac{\rho \cdot A \cdot L \cdot g}{A} = \rho \cdot L \cdot g \]

4. **Set Up the Equation for Breaking Stress**:
For the wire to not break, the stress must be less than or equal to the breaking stress:
\[ \rho \cdot L \cdot g \leq 10^8 \]
For the maximum length before breaking, we set:
\[ \rho \cdot L \cdot g = 10^8 \]

5. **Solve for Length \( L \)**:
Rearranging the equation gives:
\[ L = \frac{10^8}{\rho \cdot g} \]
Substituting the known values:
\[ L = \frac{10^8}{3000 \cdot 10} = \frac{10^8}{30000} = \frac{10^4}{3} \approx 3333.33 \, \text{m} \]

6. **Convert to Kilometers**:
To express the length in kilometers:
\[ L \approx 3.33 \, \text{km} \]

### Final Answer:
The greatest length of the wire that could hang vertically without breaking is approximately **3.33 km**.