A force of `10^6(N)/(m^2)` is required for breaking material, If the density is `3xx10^3kgm^-3` then what should be the maximum length which can be hanged so that it is the point of breaking by its own weight?

Options

A. `34m`
B. `340m`
C. `3.4m`
D. `0.34m`

Explanation

To solve the problem step by step, we need to determine the maximum length of a material that can be suspended without breaking due to its own weight.

### Step 1: Understand the Given Data
- **Breaking Stress (σ)**: \( 10^6 \, \text{N/m}^2 \)
- **Density (ρ)**: \( 3 \times 10^3 \, \text{kg/m}^3 \)
- **Acceleration due to gravity (g)**: \( 10 \, \text{m/s}^2 \) (approximate value)

### Step 2: Relate Stress to Weight
The stress (σ) experienced by the material is given by the formula:
\[ \sigma = \frac{\text{Weight}}{\text{Area}} \]
The weight of the hanging material can be expressed as:
\[ \text{Weight} = \text{Volume} \times \text{Density} \times g \]
The volume of the material can be expressed as:
\[ \text{Volume} = \text{Area} \times \text{Length} \]
Thus, the weight can be rewritten as:
\[ \text{Weight} = \text{Area} \times \text{Length} \times \rho \times g \]

### Step 3: Substitute Weight into the Stress Formula
Substituting the expression for weight into the stress formula gives:
\[ \sigma = \frac{\text{Area} \times \text{Length} \times \rho \times g}{\text{Area}} \]
This simplifies to:
\[ \sigma = \text{Length} \times \rho \times g \]

### Step 4: Solve for Length
Now, we can set the stress equal to the breaking stress:
\[ 10^6 = \text{Length} \times (3 \times 10^3) \times 10 \]
This simplifies to:
\[ 10^6 = \text{Length} \times 3 \times 10^4 \]
Now, solve for Length:
\[ \text{Length} = \frac{10^6}{3 \times 10^4} \]
\[ \text{Length} = \frac{10^6}{3 \times 10^4} = \frac{10^2}{3} = \frac{100}{3} \approx 33.33 \, \text{m} \]

### Step 5: Conclusion
The maximum length of the material that can be hanged without breaking due to its own weight is approximately:
\[ \text{Length} \approx 33.33 \, \text{m} \]