A stress of `10^(6) Nm^(-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 length of the wire made of this material, so that it breakes under its own weight?

Options

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

Explanation

To solve the problem of determining the length of the wire made of a material that breaks under its own weight, we can follow these steps:

### Step 1: Understand the Given Data
We are given:
- Breaking stress (σ) = \(10^6 \, \text{N/m}^2\)
- Density (ρ) = \(3 \times 10^3 \, \text{kg/m}^3\)
- Acceleration due to gravity (g) can be approximated as \(9.8 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\) for simplicity.

### Step 2: Relate Breaking Stress to Weight of the Wire
The breaking stress is defined as:
\[ \sigma = \frac{F}{A} \]
where \(F\) is the breaking force and \(A\) is the cross-sectional area of the wire.

Since the wire breaks under its own weight, the breaking force \(F\) can be expressed as:
\[ F = mg \]
where \(m\) is the mass of the wire.

### Step 3: Express Mass in Terms of Density and Volume
The mass \(m\) of the wire can be expressed using its density and volume:
\[ m = \rho \cdot V \]
The volume \(V\) of the wire can be expressed as:
\[ V = A \cdot L \]
where \(L\) is the length of the wire.

Thus, we can write:
\[ m = \rho \cdot (A \cdot L) \]

### Step 4: Substitute Mass into the Breaking Stress Equation
Substituting \(m\) into the breaking stress equation gives:
\[ \sigma = \frac{\rho \cdot (A \cdot L) \cdot g}{A} \]
This simplifies to:
\[ \sigma = \rho \cdot L \cdot g \]

### Step 5: Solve for Length \(L\)
Rearranging the equation to solve for \(L\):
\[ L = \frac{\sigma}{\rho \cdot g} \]

### Step 6: Substitute the Values
Now, substituting the known values:
- \(\sigma = 10^6 \, \text{N/m}^2\)
- \(\rho = 3 \times 10^3 \, \text{kg/m}^3\)
- \(g = 9.8 \, \text{m/s}^2\) (or \(10 \, \text{m/s}^2\) for simplicity)

Using \(g = 9.8 \, \text{m/s}^2\):
\[ L = \frac{10^6}{(3 \times 10^3) \cdot (9.8)} \approx \frac{10^6}{29400} \approx 34.01 \, \text{m} \]

Using \(g = 10 \, \text{m/s}^2\):
\[ L = \frac{10^6}{(3 \times 10^3) \cdot (10)} = \frac{10^6}{30000} \approx 33.33 \, \text{m} \]

### Conclusion
Thus, the length of the wire that will break under its own weight is approximately:
- \(34.01 \, \text{m}\) (using \(g = 9.8 \, \text{m/s}^2\))
- \(33.33 \, \text{m}\) (using \(g = 10 \, \text{m/s}^2\))

Given the options, \(33.33 \, \text{m}\) is likely the answer.