The length of a rod is `20 cm` and area of cross-section `2 cm^(2)`. The Young's modulus of the material of wire is `1.4 xx 10^(11) N//m^(2)`. If the rod is compressed by `5 kg-wt` along its length, then increase in the energy of the rod in joules will be

Explanation

To solve the problem step by step, we will follow the principles of elasticity and work done in compressing the rod.

### Given Data:
- Length of the rod, \( L = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Area of cross-section, \( A = 2 \, \text{cm}^2 = 2 \times 10^{-4} \, \text{m}^2 \)
- Young's modulus, \( Y = 1.4 \times 10^{11} \, \text{N/m}^2 \)
- Force applied, \( F = 5 \, \text{kg-wt} = 5 \times 9.81 \, \text{N} = 49.05 \, \text{N} \)

### Step 1: Calculate the change in length (\( \Delta L \))
Using the formula for change in length due to stress:
\[ \Delta L = \frac{F \cdot L}{A \cdot Y} \]
Substituting the values:
\[ \Delta L = \frac{49.05 \, \text{N} \cdot 0.2 \, \text{m}}{2 \times 10^{-4} \, \text{m}^2 \cdot 1.4 \times 10^{11} \, \text{N/m}^2} \]

### Step 2: Calculate the denominator
Calculating the denominator:
\[ 2 \times 10^{-4} \, \text{m}^2 \cdot 1.4 \times 10^{11} \, \text{N/m}^2 = 2.8 \times 10^{7} \, \text{N} \]

### Step 3: Calculate \( \Delta L \)
Now substituting back into the equation:
\[ \Delta L = \frac{49.05 \, \text{N} \cdot 0.2 \, \text{m}}{2.8 \times 10^{7} \, \text{N}} = \frac{9.81 \times 10^{-2} \, \text{N m}}{2.8 \times 10^{7} \, \text{N}} \approx 3.5 \times 10^{-9} \, \text{m} \]

### Step 4: Calculate the work done (energy increase)
The work done (or energy increase) in compressing the rod can be calculated using the formula:
\[ E = \frac{1}{2} F \Delta L \]
Substituting the values:
\[ E = \frac{1}{2} \cdot 49.05 \, \text{N} \cdot 3.5 \times 10^{-9} \, \text{m} \]

### Step 5: Calculate \( E \)
Calculating the energy:
\[ E = \frac{1}{2} \cdot 49.05 \cdot 3.5 \times 10^{-9} \approx 8.57 \times 10^{-8} \, \text{J} \]

### Final Result
The increase in the energy of the rod is approximately:
\[ E \approx 8.57 \times 10^{-8} \, \text{J} \text{ or } 8.57 \, \mu J \]