A stress of `10^(6) N//m^(2)` is required for breaking a material. If the density of the material is `3 xx 10^(3) Kg//m^(3)`, then what should be the minimum length of the wire made of the same material so that it breaks by its own weight `(g = 10m//s^(2))`

Options

A. 10 m
B. 33.3 m
C. 5 m
D. 66.6 m

Explanation

To find the minimum length of the wire made of the same material that will break under its own weight, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Stress**:
Stress (σ) is defined as the force (F) applied per unit area (A). The formula for stress is given by:
\[ \sigma = \frac{F}{A} \]

2. **Identify the Force due to Weight**:
The force acting on the wire due to its own weight can be expressed as:
\[ F = mg \]
where \( m \) is the mass of the wire and \( g \) is the acceleration due to gravity.

3. **Express Mass in Terms of Density and Volume**:
The mass \( m \) of the wire can be expressed in terms of its density (ρ) and volume (V):
\[ m = \rho V \]
The volume of the wire can be expressed as:
\[ V = A \cdot L \]
where \( A \) is the cross-sectional area and \( L \) is the length of the wire. Thus, we have:
\[ m = \rho A L \]

4. **Substitute Mass into the Force Equation**:
Substituting the expression for mass into the force equation gives:
\[ F = \rho A L g \]

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

6. **Rearranging for Length**:
To find the length \( L \) that will cause the wire to break, we can rearrange the equation:
\[ L = \frac{\sigma}{\rho g} \]

7. **Substituting Given Values**:
We know from the problem:
- Breaking stress \( \sigma = 10^6 \, \text{N/m}^2 \)
- Density \( \rho = 3 \times 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Substituting these values into the equation for \( L \):
\[ L = \frac{10^6}{3 \times 10^3 \times 10} \]

8. **Calculating the Length**:
Performing the calculation:
\[ L = \frac{10^6}{3 \times 10^4} = \frac{10^6}{3 \times 10^4} = \frac{10^2}{3} \approx 33.33 \, \text{m} \]

### Final Answer:
The minimum length of the wire that will break under its own weight is approximately **33.33 meters**.