For steel, the breaking stress is `8 xx 10^(6) N//m^(2)`. What is the maximum length of a steel wire, which can be suspended without breaking under its own weight?
[`g = 10 m//s^(2)`, density of steel `= 8 xx 10^(3) kg//m^(3)`]

Options

A. 50m
B. 75m
C. 100m
D. 125m

Explanation

To solve the problem of finding the maximum length of a steel wire that can be suspended without breaking under its own weight, we will follow these steps:

### Step 1: Understand the Breaking Stress
The breaking stress (σ_max) is given as:
\[ \sigma_{\text{max}} = 8 \times 10^6 \, \text{N/m}^2 \]

### Step 2: Define the Weight of the Wire
The weight (W) of the wire can be expressed as:
\[ W = m \cdot g \]
where \( m \) is the mass of the wire and \( g \) is the acceleration due to gravity.

### Step 3: Calculate the Mass of the Wire
The mass \( m \) of the wire can be calculated using its density (ρ) and volume (V):
\[ m = \rho \cdot V \]
The volume \( V \) of the wire can be expressed in terms of its length \( L \) and cross-sectional area \( A \):
\[ V = A \cdot L \]
Thus, the mass becomes:
\[ m = \rho \cdot A \cdot L \]

### Step 4: Substitute Mass into the Weight Equation
Substituting the expression for mass into the weight equation gives:
\[ W = \rho \cdot A \cdot L \cdot g \]

### Step 5: Relate Breaking Stress to Weight and Area
The breaking stress is defined as the weight per unit area:
\[ \sigma_{\text{max}} = \frac{W}{A} \]
Substituting for \( W \):
\[ \sigma_{\text{max}} = \frac{\rho \cdot A \cdot L \cdot g}{A} \]
This simplifies to:
\[ \sigma_{\text{max}} = \rho \cdot L \cdot g \]

### Step 6: Solve for Maximum Length (L)
Rearranging the equation to solve for \( L \):
\[ L = \frac{\sigma_{\text{max}}}{\rho \cdot g} \]

### Step 7: Substitute Known Values
Now, substituting the known values:
- \( \sigma_{\text{max}} = 8 \times 10^6 \, \text{N/m}^2 \)
- \( \rho = 8 \times 10^3 \, \text{kg/m}^3 \)
- \( g = 10 \, \text{m/s}^2 \)

Calculating \( L \):
\[ L = \frac{8 \times 10^6}{8 \times 10^3 \cdot 10} \]
\[ L = \frac{8 \times 10^6}{8 \times 10^4} = 100 \, \text{m} \]

### Final Answer
The maximum length of the steel wire that can be suspended without breaking under its own weight is:
\[ \boxed{100 \, \text{m}} \]