In which case there is maximum extension in the wire, if same force is applied on each wire

Options

A. L = 500 cm, d = 0.05 mm
B. L = 200 cm,d = 0.02 mm
C. L = 300 cm,d = 0.03 mm
D. L = 400 cm,d = 0.01 mm

Answer

To determine in which case there is maximum extension in the wire when the same force is applied to each wire, we can use the formula derived from Young's modulus of elasticity:

1. **Understanding Young's Modulus**:
Young's modulus (Y) is defined as:
\[ Y = \frac{F \cdot L}{A \cdot \Delta L} \]
where:
- \(F\) = applied force
- \(L\) = original length of the wire
- \(A\) = cross-sectional area of the wire
- \(\Delta L\) = extension in the wire

2. **Expressing Area**:
The cross-sectional area \(A\) of a wire can be expressed in terms of its diameter \(D\):
\[ A = \frac{\pi D^2}{4} \]

3. **Rearranging the Formula**:
Rearranging the Young's modulus formula to find the extension \(\Delta L\):
\[ \Delta L = \frac{F \cdot L}{Y \cdot A} \]
Substituting the expression for area:
\[ \Delta L = \frac{F \cdot L}{Y \cdot \frac{\pi D^2}{4}} = \frac{4F \cdot L}{Y \cdot \pi D^2} \]

4. **Analyzing the Extension**:
Since the force \(F\), Young's modulus \(Y\), and \(\pi\) are constants for all cases, we can simplify the relationship to focus on the ratio:
\[ \Delta L \propto \frac{L}{D^2} \]
This means that the extension \(\Delta L\) is directly proportional to the length \(L\) of the wire and inversely proportional to the square of the diameter \(D\).

5. **Calculating the Ratios**:
We will calculate the ratio \(\frac{L}{D^2}\) for each case given in the problem:

- **Case 1**: \(L = 500 \, \text{cm}, D = 5 \, \text{cm}\)
\[ \frac{L}{D^2} = \frac{500}{5^2} = \frac{500}{25} = 20 \]

- **Case 2**: \(L = 200 \, \text{cm}, D = 2 \, \text{cm}\)
\[ \frac{L}{D^2} = \frac{200}{2^2} = \frac{200}{4} = 50 \]

- **Case 3**: \(L = 300 \, \text{cm}, D = 3 \, \text{cm}\)
\[ \frac{L}{D^2} = \frac{300}{3^2} = \frac{300}{9} \approx 33.33 \]

- **Case 4**: \(L = 400 \, \text{cm}, D = 1 \, \text{cm}\)
\[ \frac{L}{D^2} = \frac{400}{1^2} = \frac{400}{1} = 400 \]

6. **Comparing the Results**:
Now we compare the calculated ratios:
- Case 1: 20
- Case 2: 50
- Case 3: 33.33
- Case 4: 400

The maximum ratio is found in **Case 4**, which gives the maximum extension.

7. **Conclusion**:
Therefore, the case with maximum extension in the wire when the same force is applied is **Case 4**.