Step 1:
To determine which material results in the lowest weight of the rod, we will use the strength-based failure criterion. The weight of the rod is determined by the following relationship:
\[
{Weight} = \frac{P}{\sigma_y} \cdot g \cdot L
\]
Where:
\( \sigma_y \) is the yield strength of the material,
\( P \) is the tensile load,
\( g \) is the gravitational constant (which does not affect the comparison),
\( L \) is the length of the rod (1 m in this case).
To minimize the weight of the rod, we want to maximize the ratio \( \frac{P}{\sigma_y} \). This corresponds to choosing the material with the highest yield strength relative to its density.
Step 2:
To calculate this, we use the following equation for the ratio of yield strength to density:
\[
\frac{\sigma_y}{\rho}
\]
Where:
\( \sigma_y \) is the yield strength,
\( \rho \) is the density of the material.
Now, let’s calculate \( \frac{\sigma_y}{\rho} \) for each material:
- For Material \( \alpha \):
\[
\frac{\sigma_y}{\rho} = \frac{270}{2700} = 0.1 \, {MPa/(kg/m}^3)
\]
- For Material \( \beta \):
\[
\frac{\sigma_y}{\rho} = \frac{900}{4500} = 0.2 \, {MPa/(kg/m}^3)
\]
- For Material \( \gamma \):
\[
\frac{\sigma_y}{\rho} = \frac{520}{7800} = 0.067 \, {MPa/(kg/m}^3)
\]
- For Material \( \delta \):
\[
\frac{\sigma_y}{\rho} = \frac{540}{9000} = 0.06 \, {MPa/(kg/m}^3)
\]
Step 3:
Comparing the Ratios:
From the calculations above, we see that Material \( \beta \) has the highest yield strength to density ratio (0.2 MPa/(kg/m³)).
Step 4:
Conclusion:
Thus, Material \( \beta \) will result in the lowest weight for the rod, as it provides the highest yield strength per unit density.
The correct answer is (B) Material \( \beta \).