Find the increment in the length of a steel wire of length `5 m` and radius `6 mm` under its own weight. Density of steel `= 8000 kg//m^(3)` and young's modulus of steel `= 2xx10^(11) N//m^(2)`. What is the energy stored in the wire ? (Take g `= 9.8 m//s^(2))`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the area of cross-section (A) of the wire
The radius \( R \) of the wire is given as \( 6 \, \text{mm} = 6 \times 10^{-3} \, \text{m} \).

The area of cross-section \( A \) can be calculated using the formula for the area of a circle:
\[ A = \pi R^2 \]
Substituting the value of \( R \):
\[ A = \pi (6 \times 10^{-3})^2 = \pi (36 \times 10^{-6}) \approx 1.131 \times 10^{-5} \, \text{m}^2 \]

### Step 2: Calculate the mass (m) of the wire
The volume \( V \) of the wire can be calculated as:
\[ V = A \times L \]
Where \( L = 5 \, \text{m} \).
Substituting the values:
\[ V = (1.131 \times 10^{-5}) \times 5 = 5.655 \times 10^{-5} \, \text{m}^3 \]

Now, using the density \( \rho = 8000 \, \text{kg/m}^3 \):
\[ m = V \times \rho = (5.655 \times 10^{-5}) \times 8000 \approx 0.4524 \, \text{kg} \]

### Step 3: Calculate the weight (W) of the wire
The weight \( W \) can be calculated using:
\[ W = m \times g \]
Where \( g = 9.8 \, \text{m/s}^2 \):
\[ W = 0.4524 \times 9.8 \approx 4.43 \, \text{N} \]

### Step 4: Calculate the increment in length (\( \Delta L \))
Using the formula:
\[ \Delta L = \frac{W L}{2 A Y} \]
Substituting the values:
\[ \Delta L = \frac{(4.43)(5)}{2 (1.131 \times 10^{-5}) (2 \times 10^{11})} \]
Calculating this gives:
\[ \Delta L = \frac{22.15}{4.524 \times 10^{6}} \approx 4.9 \times 10^{-6} \, \text{m} \, \text{or} \, 4.9 \, \mu m \]

### Step 5: Calculate the energy stored (U) in the wire
The energy stored in the wire can be calculated using the formula:
\[ U = \frac{W^2 L}{6 A Y} \]
Substituting the values:
\[ U = \frac{(4.43)^2 (5)}{6 (1.131 \times 10^{-5}) (2 \times 10^{11})} \]
Calculating this gives:
\[ U = \frac{19.6249 \times 5}{6.792 \times 10^{6}} \approx 7.24 \times 10^{-5} \, \text{J} \]

### Final Answer
- The increment in length of the wire is approximately \( 4.9 \, \mu m \).
- The energy stored in the wire is approximately \( 7.24 \times 10^{-5} \, \text{J} \).