A steel wire of original length `1 m` and cross-sectional area `4.00 mm^2` is clamped at the two ends so that it lies horizontally and without tension. If a load of `2.16 kg` is suspended from the middle point of the wire, what would be its vertical depression ?
Y of the steel `=2.0xx10^11Nm^-2`. Take `g=10 ms^-2`

Explanation

To find the vertical depression of the steel wire when a load is suspended from its middle point, we can follow these steps:

### Step 1: Identify the given values
- Original length of the wire, \( L = 1 \, \text{m} \)
- Cross-sectional area, \( A = 4.00 \, \text{mm}^2 = 4.00 \times 10^{-6} \, \text{m}^2 \)
- Mass of the load, \( m = 2.16 \, \text{kg} \)
- Young's modulus of steel, \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the force exerted by the load
The force \( F \) exerted by the load can be calculated using:
\[ F = m \cdot g \]
Substituting the values:
\[ F = 2.16 \, \text{kg} \times 10 \, \text{m/s}^2 = 21.6 \, \text{N} \]

### Step 3: Set up the equation for depression
When the load is applied, it creates a tension in the wire. The relationship between stress, strain, and Young's modulus is given by:
\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L} \]
Where:
- \( \Delta L \) is the change in length (depression),
- \( L \) is the original length.

Rearranging gives:
\[ \Delta L = \frac{F \cdot L}{Y \cdot A} \]

### Step 4: Substitute the values into the equation
Now, substituting the known values:
\[ \Delta L = \frac{21.6 \, \text{N} \cdot 1 \, \text{m}}{(2.0 \times 10^{11} \, \text{N/m}^2) \cdot (4.00 \times 10^{-6} \, \text{m}^2)} \]

### Step 5: Calculate the denominator
Calculating the denominator:
\[ Y \cdot A = 2.0 \times 10^{11} \, \text{N/m}^2 \cdot 4.00 \times 10^{-6} \, \text{m}^2 = 800 \, \text{N} \]

### Step 6: Calculate the depression
Now substituting back into the equation for \( \Delta L \):
\[ \Delta L = \frac{21.6 \, \text{N}}{800 \, \text{N}} = 0.027 \, \text{m} = 2.7 \, \text{cm} \]

### Step 7: Conclusion
Thus, the vertical depression of the steel wire is approximately:
\[ \Delta L \approx 2.7 \, \text{cm} \]