A metallic rod of length l and cross-sectional area A is made of a material of Young's modulus Y. If the rod is elongated by an amount y, then the work done is proportional to

Options

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

Explanation

To solve the problem of determining how the work done on a metallic rod is proportional to the elongation, we can follow these steps:

### Step 1: Understand the Definitions
- **Young's Modulus (Y)**: It is defined as the ratio of stress to strain.
- **Stress**: It is defined as force (F) per unit area (A), i.e., \( \text{Stress} = \frac{F}{A} \).
- **Strain**: It is defined as the change in length (elongation, y) divided by the original length (l), i.e., \( \text{Strain} = \frac{y}{l} \).

### Step 2: Relate Stress and Strain
Using the definition of Young's modulus:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{y/l} \]
Rearranging gives us:
\[ F = Y \cdot \frac{A \cdot y}{l} \]

### Step 3: Calculate the Work Done
The work done (W) on the rod when it is elongated by an amount y can be calculated using the formula:
\[ W = \text{Force} \times \text{displacement} = F \cdot y \]
Substituting the expression for force from Step 2:
\[ W = \left(Y \cdot \frac{A \cdot y}{l}\right) \cdot y = \frac{Y \cdot A \cdot y^2}{l} \]

### Step 4: Determine the Proportionality
From the expression derived for work done:
\[ W = \frac{Y \cdot A}{l} \cdot y^2 \]
We can see that the work done (W) is proportional to the square of the elongation (y):
\[ W \propto y^2 \]

### Conclusion
Thus, the work done is proportional to \( y^2 \).

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