A uniform steel rod of cross- sectional area `A` and `L` is suspended so that it hangs vertically. The stress at the middle point of the rod is

Options

A. `1/2rhogL`
B. `1/4rhogL`
C. `rhogL`
D. None of these

Explanation

To find the stress at the middle point of a uniform steel rod of cross-sectional area \( A \) and length \( L \) that is suspended vertically, we can follow these steps:

### Step 1: Understand the Concept of Stress
Stress (\( \sigma \)) is defined as the force (\( F \)) applied per unit area (\( A \)). Mathematically, it is expressed as:
\[ \sigma = \frac{F}{A} \]

### Step 2: Identify the Force at the Middle Point
The force acting on the middle point of the rod is due to the weight of the portion of the rod that is below the middle point. Since the rod is uniform, we can consider the mass of the lower half of the rod.

### Step 3: Calculate the Mass of the Lower Half of the Rod
The mass (\( m \)) of the lower half of the rod can be calculated using the formula:
\[ m = \text{Volume} \times \text{Density} \]
The volume of the lower half of the rod is given by:
\[ \text{Volume} = A \times \frac{L}{2} \]
Thus, the mass can be expressed as:
\[ m = A \times \frac{L}{2} \times \rho \]
where \( \rho \) is the density of the material (steel in this case).

### Step 4: Calculate the Force Due to Gravity
The force due to gravity acting on this mass is:
\[ F = m \times g = \left(A \times \frac{L}{2} \times \rho\right) \times g \]

### Step 5: Substitute the Force into the Stress Formula
Now we substitute the expression for force into the stress formula:
\[ \sigma = \frac{F}{A} = \frac{\left(A \times \frac{L}{2} \times \rho \times g\right)}{A} \]

### Step 6: Simplify the Expression
Since the area \( A \) cancels out, we have:
\[ \sigma = \frac{L}{2} \times \rho \times g \]

### Step 7: Final Result
Thus, the stress at the middle point of the rod is:
\[ \sigma = \frac{1}{2} \rho g L \]

### Conclusion
The correct answer is \( \frac{1}{2} \rho g L \), which corresponds to option 1.

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