A horizontal rod is supported at both ends and loaded at the middle. If L and Y are length and Young's modulus repectively, then depression at the middle is directly proportional to

Options

A. L
B. `L^(2)`
C. Y
D. `(1)/(Y)`

Answer

To solve the problem, we need to understand how the depression (deflection) of a horizontal rod, supported at both ends and loaded at the middle, relates to its length (L) and Young's modulus (Y).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A horizontal rod is supported at both ends and has a load applied at its center. This setup is known as a simply supported beam.

2. **Identifying the Variables**:
- Let \( L \) be the length of the rod.
- Let \( Y \) be the Young's modulus of the material of the rod.
- The depression (deflection) at the middle of the rod is denoted as \( \delta \).

3. **Using the Formula for Deflection**:
- The deflection \( \delta \) at the center of a simply supported beam loaded at the center can be expressed using the formula:
\[ \delta = \frac{W L^3}{48 Y I} \]
- Here, \( W \) is the load applied at the center, \( I \) is the moment of inertia of the beam's cross-section.

4. **Analyzing the Proportionality**:
- From the formula, we can see that:
- The deflection \( \delta \) is directly proportional to \( L^3 \) (the length cubed).
- The deflection \( \delta \) is inversely proportional to \( Y \) (Young's modulus).
- Therefore, we can express this relationship as:
\[ \delta \propto \frac{L^3}{Y} \]

5. **Conclusion**:
- Since the question asks for what the depression at the middle is directly proportional to, we can conclude that:
- The depression \( \delta \) is directly proportional to \( L^3 \) and inversely proportional to \( Y \).
- Therefore, the correct answer is that the depression is directly proportional to \( L^3 \) and \( \frac{1}{Y} \).

### Final Answer:
The depression at the middle is directly proportional to \( L^3 \) and \( \frac{1}{Y} \).