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 follow these steps:

### Step 1: Understand the relationship between tension, weight, and breaking stress
The tension at the top of the wire (T_max) is equal to the weight of the wire hanging below it. The breaking stress is defined as the force per unit area that the material can withstand before failing.

### Step 2: Write the expression for T_max
The maximum tension (T_max) at the top of the wire can be expressed as:
\[ T_{\text{max}} = \text{Weight of the wire} = \text{Density} \times \text{Volume} \times g \]

### Step 3: Express the volume of the wire
The volume (V) of the wire can be expressed in terms of its length (L) and cross-sectional area (A):
\[ V = A \times L \]

### Step 4: Substitute the volume into the weight equation
Substituting the volume into the weight equation gives:
\[ T_{\text{max}} = \text{Density} \times (A \times L) \times g \]

### Step 5: Relate T_max to breaking stress
The breaking stress (σ) is defined as:
\[ \sigma = \frac{T_{\text{max}}}{A} \]
Setting T_max equal to the breaking stress, we have:
\[ T_{\text{max}} \leq \sigma \]

### Step 6: Combine the equations
Substituting the expression for T_max into the breaking stress equation gives:
\[ \text{Density} \times (A \times L) \times g \leq \sigma \]
This simplifies to:
\[ \text{Density} \times L \times g \leq \sigma \]

### Step 7: Solve for L
Rearranging the equation to solve for L:
\[ L \leq \frac{\sigma}{\text{Density} \times g} \]

### Step 8: Substitute the known values
Now we can substitute the given values into the equation:
- Breaking stress (σ) = \( 8.0 \times 10^8 \, \text{N/m}^2 \)
- Density of steel = \( 8.0 \times 10^3 \, \text{kg/m}^3 \)
- g = \( 10 \, \text{m/s}^2 \)

Substituting these values:
\[ L \leq \frac{8.0 \times 10^8}{8.0 \times 10^3 \times 10} \]

### Step 9: Calculate L
Calculating the right-hand side:
\[ L \leq \frac{8.0 \times 10^8}{8.0 \times 10^4} = 10^4 \, \text{m} \]

### Step 10: Conclusion
Thus, the greatest length of the steel wire that can hang vertically without breaking is:
\[ L = 10,000 \, \text{m} \]

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