A steel wire of length 1m and diameter 4mm is stretched horizontally between two rigit supports attached to its end. What load would be required to be hung from the mid-point of the wire to produce a depression of 1cm? `(Y= 2 xx 10^(11)Nm^(-2))`

Answer

To solve the problem, we need to determine the load required to produce a depression of 1 cm in a steel wire of length 1 m and diameter 4 mm. We will use the formula for Young's modulus and the relationship between stress, strain, and the applied load.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Length of the wire, \( L = 1 \, \text{m} = 100 \, \text{cm} \)
- Diameter of the wire, \( d = 4 \, \text{mm} = 0.004 \, \text{m} \)
- Depression, \( x = 1 \, \text{cm} = 0.01 \, \text{m} \)
- Young's modulus, \( Y = 2 \times 10^{11} \, \text{N/m}^2 \)

2. **Calculate the Cross-sectional Area of the Wire:**
\[ A = \frac{\pi d^2}{4} = \frac{\pi (0.004)^2}{4} = \frac{\pi \times 0.000016}{4} \approx 1.2566 \times 10^{-5} \, \text{m}^2 \]

3. **Calculate the Strain:**
Strain (ε) is defined as the change in length (depression) divided by the original length.
\[ \text{Strain} \, (\epsilon) = \frac{x}{L} = \frac{0.01}{1} = 0.01 \]

4. **Calculate the Stress:**
Stress (σ) is defined as the force (load) per unit area.
\[ \text{Stress} \, (\sigma) = Y \cdot \epsilon \]
Rearranging gives us:
\[ \sigma = Y \cdot \epsilon = (2 \times 10^{11}) \cdot (0.01) = 2 \times 10^{9} \, \text{N/m}^2 \]

5. **Relate Stress to Load:**
The stress can also be expressed in terms of the load (W):
\[ \sigma = \frac{W}{A} \]
Rearranging gives us:
\[ W = \sigma \cdot A = (2 \times 10^{9}) \cdot (1.2566 \times 10^{-5}) \approx 25120 \, \text{N} \]

6. **Convert Load to Mass:**
To find the mass (m) that corresponds to this load, we use:
\[ W = m \cdot g \]
where \( g \approx 9.81 \, \text{m/s}^2 \). Rearranging gives:
\[ m = \frac{W}{g} = \frac{25120}{9.81} \approx 2564.4 \, \text{kg} \]

### Final Answer:
The load required to produce a depression of 1 cm in the steel wire is approximately **25120 N**, which corresponds to a mass of about **2564.4 kg**.