The breaking stress for a substance is `10^(6)N//m^(2)`. What length of the wire of this substance should be suspended verticaly so that the wire breaks under its own weight? (Given: density of material of the wire `=4xx10^(3)kg//m^(3)` and `g=10 ms^(-12))`

Options

A. `10m`
B. `15m`
C. `25m`
D. `34m`

Explanation

To solve the problem, we need to determine the length of the wire that will break under its own weight given the breaking stress, density of the material, and acceleration due to gravity.

### Step-by-Step Solution:

1. **Understand Breaking Stress**:
The breaking stress (σ) is defined as the maximum stress that a material can withstand before failure. It is given by the formula:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force applied (in this case, the weight of the wire) and \( A \) is the cross-sectional area of the wire.

2. **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.

3. **Mass of the Wire**:
The mass \( m \) can be calculated using the density \( \rho \) and the volume \( V \):
\[ m = \rho \cdot V \]
The volume \( V \) of the wire can be expressed as:
\[ V = A \cdot L \]
where \( L \) is the length of the wire. Thus, we have:
\[ m = \rho \cdot A \cdot L \]

4. **Substituting Mass into Weight**:
Substituting the expression for mass into the weight formula gives:
\[ W = \rho \cdot A \cdot L \cdot g \]

5. **Substituting Weight into Breaking Stress**:
Now substituting \( W \) into the breaking stress formula:
\[ \sigma = \frac{\rho \cdot A \cdot L \cdot g}{A} \]
The area \( A \) cancels out:
\[ \sigma = \rho \cdot L \cdot g \]

6. **Rearranging for Length \( L \)**:
We can rearrange this equation to solve for the length \( L \):
\[ L = \frac{\sigma}{\rho \cdot g} \]

7. **Substituting Given Values**:
Now we substitute the given values into the equation:
- Breaking stress \( \sigma = 10^6 \, \text{N/m}^2 \)
- Density \( \rho = 4 \times 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)

Thus, we have:
\[ L = \frac{10^6}{4 \times 10^3 \cdot 10} \]

8. **Calculating Length**:
Now we calculate:
\[ L = \frac{10^6}{4 \times 10^4} = \frac{10^6}{4 \times 10^4} = \frac{10^6}{4 \times 10^4} = \frac{10^6}{4 \times 10^4} = 25 \, \text{m} \]

### Final Answer:
The length of the wire that should be suspended vertically so that it breaks under its own weight is **25 meters**.

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