Young's modulus for a wire of length L and area of cross-section A is Y. What will be Young's Modulus for wire of same material, but half its original length and double its area?

Options

A. `Y//2`
B. 2Y
C. y
D. `4Y`

Answer

To solve the problem of finding the Young's modulus for a wire of the same material but with half its original length and double its area, we can follow these steps:

### Step 1: Understand the definition of Young's Modulus
Young's modulus (Y) is defined as the ratio of stress to strain in a material. Mathematically, it is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]

### Step 2: Define stress and strain
- **Stress (σ)** is defined as the force (F) applied per unit area (A):
\[ \sigma = \frac{F}{A} \]
- **Strain (ε)** is defined as the change in length (ΔL) divided by the original length (L):
\[ \epsilon = \frac{\Delta L}{L} \]

### Step 3: Calculate the new dimensions of the wire
- The original length of the wire is \( L \).
- The new length of the wire is \( \frac{L}{2} \).
- The original area of cross-section is \( A \).
- The new area of cross-section is \( 2A \).

### Step 4: Calculate the new stress
Using the definition of stress, the new stress (σ') in the wire can be calculated as:
\[ \sigma' = \frac{F}{2A} \]

### Step 5: Calculate the new strain
The new strain (ε') in the wire can be calculated as:
\[ \epsilon' = \frac{\Delta L}{\frac{L}{2}} = \frac{2\Delta L}{L} \]

### Step 6: Substitute into the Young's Modulus formula
Now we can substitute the new stress and strain into the Young's modulus formula:
\[ Y' = \frac{\sigma'}{\epsilon'} \]
Substituting the values we have:
\[ Y' = \frac{\frac{F}{2A}}{\frac{2\Delta L}{L}} \]

### Step 7: Simplify the expression
This simplifies to:
\[ Y' = \frac{F \cdot L}{2A \cdot 2\Delta L} = \frac{F \cdot L}{4A \cdot \Delta L} \]

### Step 8: Relate it to the original Young's modulus
Since the original Young's modulus \( Y \) is given by:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
We can express the new Young's modulus in terms of the original:
\[ Y' = \frac{Y}{4} \]

### Conclusion
Thus, the Young's modulus for the wire of the same material, but half its original length and double its area, is:
\[ Y' = \frac{Y}{4} \]