if `rho` is the density of the meterial of a wire and `sigma` is the breaking stress. The greatest length of the wire that can hang freely without breaking is

Options

A. `(2 sigma)/(rho g)`
B. `(rho)/(sigma g)`
C. `(rho g)/(2 sigma)`
D. `(sigma)/(rho g)`

Answer

To find the greatest length of the wire that can hang freely without breaking, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Wire**:
- When the wire hangs freely, the only force acting on it is its weight, which acts downward.

2. **Express the Weight of the Wire**:
- The weight \( F \) of the wire can be expressed as:
\[ F = mg \]
- Here, \( m \) is the mass of the wire and \( g \) is the acceleration due to gravity.

3. **Relate Mass to Density and Volume**:
- The mass \( m \) of the wire can be expressed in terms of its density \( \rho \), cross-sectional area \( A \), and length \( L \):
\[ m = \rho \cdot V = \rho \cdot (A \cdot L) \]
- Therefore, the weight can be rewritten as:
\[ F = \rho A L g \]

4. **Define Stress**:
- Stress \( \sigma \) is defined as the force per unit area:
\[ \sigma = \frac{F}{A} \]

5. **Substitute the Expression for Weight into the Stress Formula**:
- Substituting the expression for \( F \) into the stress equation gives:
\[ \sigma = \frac{\rho A L g}{A} = \rho L g \]

6. **Relate Breaking Stress to Length**:
- Since the breaking stress is given as \( \sigma \), we can set up the equation:
\[ \sigma = \rho L g \]
- Rearranging this equation to solve for \( L \) gives:
\[ L = \frac{\sigma}{\rho g} \]

7. **Conclusion**:
- The greatest length of the wire that can hang freely without breaking is:
\[ L = \frac{\sigma}{\rho g} \]