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, we need to find the relationship between the work done on a metallic rod and the elongation it experiences when subjected to a tensile force. Let's break it down step by step.

### Step 1: Understand the Definitions
- **Young's Modulus (Y)**: It is defined as the ratio of stress to strain in a material. Mathematically, it is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
- **Stress**: It is defined as the force (F) applied per unit area (A):
\[ \text{Stress} = \frac{F}{A} \]
- **Strain**: It is defined as the change in length (ΔL or y) divided by the original length (L):
\[ \text{Strain} = \frac{y}{L} \]

### Step 2: Relate Stress and Strain
From the definition of Young's modulus, we can express stress in terms of Young's modulus and strain:
\[ \text{Stress} = Y \times \text{Strain} = Y \times \frac{y}{L} \]

### Step 3: Calculate the Work Done
The work done (W) on the rod when it is elongated can be expressed using the formula:
\[ W = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume} \]
The volume (V) of the rod is given by:
\[ V = A \times L \]
Substituting the expressions for stress and strain into the work done formula:
\[ W = \frac{1}{2} \times \left(Y \times \frac{y}{L}\right) \times \left(\frac{y}{L}\right) \times (A \times L) \]

### Step 4: Simplify the Expression
Now, substituting the volume into the work done expression:
\[ W = \frac{1}{2} \times Y \times \frac{y}{L} \times \frac{y}{L} \times (A \times L) \]
This simplifies to:
\[ W = \frac{1}{2} \times Y \times A \times \frac{y^2}{L} \]

### Step 5: Identify Proportionality
From the simplified expression, 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 the square of the elongation of the rod.

### Final Answer
The work done is proportional to \( y^2 \).

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