A force of `6 xx 10^(6) Nm^(-2)` required for breaking a material. The denisty `rho` of the material is `3 xx 10^(3) kg m^(-3)`. If the wire is to break under its own weight, then the length of the wire made of that material should be (Given, g = 10 `ms^(-2)`)

Options

A. 20 m
B. 200 m
C. 100 m
D. 2000 m

Explanation

To solve the problem, we need to determine the length of a wire made from a material that will break under its own weight. Given the breaking stress, density of the material, and the acceleration due to gravity, we can derive the required length step by step.

### Step-by-Step Solution:

1. **Understand the Given Values**:
- Breaking stress, \( S = 6 \times 10^6 \, \text{N/m}^2 \)
- Density, \( \rho = 3 \times 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

2. **Relate Stress to Force and Area**:
The breaking stress is defined as:
\[ S = \frac{F}{A} \]
where \( F \) is the force acting on the wire and \( A \) is the cross-sectional area of the wire.

3. **Determine the Force Acting on the Wire**:
Since the wire is to break under its own weight, the force \( F \) can be expressed as:
\[ F = m \cdot g \]
where \( m \) is the mass of the wire. The mass can be calculated using the volume and density:
\[ m = \rho \cdot V \]
The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
where \( L \) is the length of the wire.

4. **Substituting for Mass**:
Substituting the expression for volume into the mass equation:
\[ m = \rho \cdot (A \cdot L) = \rho A L \]
Therefore, the force becomes:
\[ F = \rho A L \cdot g \]

5. **Substituting into the Stress Equation**:
Now substituting \( F \) back into the stress equation:
\[ S = \frac{\rho A L \cdot g}{A} \]
The area \( A \) cancels out:
\[ S = \rho L g \]

6. **Rearranging for Length \( L \)**:
To find the length \( L \):
\[ L = \frac{S}{\rho g} \]

7. **Substituting the Known Values**:
Now substitute the values of \( S \), \( \rho \), and \( g \):
\[ L = \frac{6 \times 10^6}{3 \times 10^3 \cdot 10} \]

8. **Calculating**:
\[ L = \frac{6 \times 10^6}{3 \times 10^4} = \frac{6}{3} \times 10^{6-4} = 2 \times 10^2 = 200 \, \text{m} \]

### Final Answer:
The length of the wire should be \( L = 200 \, \text{m} \).

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