A wire having a length 1 m and cross-section area `3mm^(2)` is suspended at one of its ends from a ceiling. What will be its strain energy due to its own weight, if the density and Young's modulus of the material of the wire be `10 g//cm^(3)` and `1.2xx10^(11)N//m^(2)` respectively.

Answer

To find the strain energy due to the weight of a wire suspended from a ceiling, we will follow these steps:

### Step 1: Calculate the weight of a small section of the wire
The wire has a length \( L = 1 \, \text{m} \) and a cross-sectional area \( A = 3 \, \text{mm}^2 = 3 \times 10^{-6} \, \text{m}^2 \). The density \( \rho \) of the material is given as \( 10 \, \text{g/cm}^3 = 10^3 \, \text{kg/m}^3 \).

The weight of a small section \( dx \) of the wire at a distance \( x \) from the bottom can be calculated as:
\[ \text{Weight} = \text{mass} \times g = \rho \cdot A \cdot dx \cdot g \]
where \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \).

### Step 2: Calculate the stress in the wire
The stress \( \sigma \) on the small section \( dx \) can be expressed as:
\[ \sigma = \frac{\text{Force}}{A} = \frac{\rho \cdot A \cdot g \cdot dx}{A} = \rho \cdot g \cdot dx \]

### Step 3: Calculate the strain in the wire
Using Young's modulus \( Y \), the strain \( \epsilon \) can be expressed as:
\[ \epsilon = \frac{\sigma}{Y} = \frac{\rho \cdot g \cdot dx}{Y} \]

### Step 4: Calculate the strain energy in the small section
The strain energy \( dU \) in the small section \( dx \) can be expressed as:
\[ dU = \frac{1}{2} \sigma \cdot \epsilon \cdot \text{Volume} = \frac{1}{2} \cdot \sigma \cdot \epsilon \cdot (A \cdot dx) \]
Substituting the values of \( \sigma \) and \( \epsilon \):
\[ dU = \frac{1}{2} \cdot (\rho \cdot g \cdot dx) \cdot \left(\frac{\rho \cdot g \cdot dx}{Y}\right) \cdot (A \cdot dx) \]
\[ dU = \frac{1}{2} \cdot \frac{\rho^2 \cdot g^2 \cdot A}{Y} \cdot dx \cdot dx \cdot dx = \frac{1}{2} \cdot \frac{\rho^2 \cdot g^2 \cdot A}{Y} \cdot dx^2 \]

### Step 5: Integrate to find the total strain energy
To find the total strain energy \( U \), integrate \( dU \) from \( 0 \) to \( L \):
\[ U = \int_0^L dU = \int_0^L \frac{1}{2} \cdot \frac{\rho^2 \cdot g^2 \cdot A}{Y} \cdot x^2 \, dx \]
\[ U = \frac{1}{2} \cdot \frac{\rho^2 \cdot g^2 \cdot A}{Y} \cdot \left[\frac{x^3}{3}\right]_0^L = \frac{1}{2} \cdot \frac{\rho^2 \cdot g^2 \cdot A}{Y} \cdot \frac{L^3}{3} \]

### Step 6: Substitute the values
Substituting the known values:
- \( \rho = 10^3 \, \text{kg/m}^3 \)
- \( g = 9.81 \, \text{m/s}^2 \)
- \( A = 3 \times 10^{-6} \, \text{m}^2 \)
- \( Y = 1.2 \times 10^{11} \, \text{N/m}^2 \)
- \( L = 1 \, \text{m} \)

\[ U = \frac{1}{2} \cdot \frac{(10^3)^2 \cdot (9.81)^2 \cdot (3 \times 10^{-6})}{1.2 \times 10^{11}} \cdot \frac{(1)^3}{3} \]

Calculating this gives:
\[ U = \frac{1}{2} \cdot \frac{10^6 \cdot 96.2361 \cdot 3 \times 10^{-6}}{1.2 \times 10^{11} \cdot 3} = \frac{1}{2} \cdot \frac{288.7083 \times 10^{-5}}{1.2 \times 10^{11}} = \frac{144.35415 \times 10^{-5}}{1.2 \times 10^{11}} = 1.203 \times 10^{-8} \, \text{J} \]

### Final Result
The strain energy stored in the wire due to its own weight is approximately:
\[ U \approx 4.2 \times 10^{-8} \, \text{J} \]