If the strain in a wire is not more than `1//1000` and Y `=2xx10^(11)N//m^(2)` Diameter of wire is 1 mm. The maximum weight hung from the wire is

Options

A. 110 N
B. 125 N
C. 157 N
D. 168 N

Answer

To find the maximum weight that can be hung from a wire given the strain, Young's modulus, and diameter, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Strain (ε) = \( \frac{1}{1000} \)
- Young's Modulus (Y) = \( 2 \times 10^{11} \, \text{N/m}^2 \)
- Diameter of the wire (d) = 1 mm = \( 1 \times 10^{-3} \, \text{m} \)

2. **Calculate the Radius of the Wire:**
\[ r = \frac{d}{2} = \frac{1 \times 10^{-3}}{2} = 0.5 \times 10^{-3} \, \text{m} = 5 \times 10^{-4} \, \text{m} \]

3. **Calculate the Cross-Sectional Area (A) of the Wire:**
\[ A = \pi r^2 = \pi (5 \times 10^{-4})^2 = \pi (25 \times 10^{-8}) = 25\pi \times 10^{-8} \, \text{m}^2 \]

4. **Calculate the Maximum Stress (σ) Using Young's Modulus:**
\[ \sigma = Y \cdot \epsilon = (2 \times 10^{11}) \cdot \left(\frac{1}{1000}\right) = 2 \times 10^{11} \times 10^{-3} = 2 \times 10^{8} \, \text{N/m}^2 \]

5. **Relate Stress to Weight (mg):**
\[ \sigma = \frac{mg}{A} \]
Rearranging gives:
\[ mg = \sigma \cdot A \]

6. **Substituting the Values:**
\[ mg = (2 \times 10^{8}) \cdot (25\pi \times 10^{-8}) \]
\[ mg = 50\pi \, \text{N} \]

7. **Calculate the Numerical Value:**
Using \( \pi \approx 3.14 \):
\[ mg \approx 50 \cdot 3.14 = 157 \, \text{N} \]

### Final Answer:
The maximum weight that can be hung from the wire is approximately **157 N**.

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