A brass rod of cross-sectional area `1cm^(2)` and length 0.2 m is compressed lengthwise by a weight of 5 kg. if young's modulus of elasticity of brass is `1xx10^(11)N//m^(2)` and `g=10m//sec^(2)` then increase in the energy of the rod will be

Options

A. `10^(-5)`joule
B. `2.5xx10^(-5)` joule
C. `5xx10^(-5)` joule
D. `2.5xx10^(-4)` joule

Answer

To find the increase in energy of the brass rod when it is compressed, we will use the formula for elastic potential energy stored in the material. The steps to solve the problem are as follows:

### Step 1: Calculate the Stress
The stress (σ) on the rod can be calculated using the formula:
\[ \sigma = \frac{F}{A} \]
where:
- \( F \) is the force applied (weight of the object),
- \( A \) is the cross-sectional area of the rod.

Given:
- Weight \( F = 5 \, \text{kg} \times g = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N} \)
- Cross-sectional area \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \)

Now, substituting the values:
\[ \sigma = \frac{50 \, \text{N}}{1 \times 10^{-4} \, \text{m}^2} = 5 \times 10^{5} \, \text{N/m}^2 \]

### Step 2: Calculate the Strain
The strain (ε) can be calculated using Young's modulus (E):
\[ \epsilon = \frac{\sigma}{E} \]
Given Young's modulus for brass:
- \( E = 1 \times 10^{11} \, \text{N/m}^2 \)

Now, substituting the values:
\[ \epsilon = \frac{5 \times 10^{5} \, \text{N/m}^2}{1 \times 10^{11} \, \text{N/m}^2} = 5 \times 10^{-6} \]

### Step 3: Calculate the Volume of the Rod
The volume (V) of the rod can be calculated using the formula:
\[ V = A \times L \]
where:
- \( L = 0.2 \, \text{m} \)

Substituting the values:
\[ V = 1 \times 10^{-4} \, \text{m}^2 \times 0.2 \, \text{m} = 2 \times 10^{-5} \, \text{m}^3 \]

### Step 4: Calculate the Elastic Potential Energy
The elastic potential energy (U) stored in the rod can be calculated using the formula:
\[ U = \frac{1}{2} \times \sigma \times \epsilon \times V \]

Substituting the values:
\[ U = \frac{1}{2} \times (5 \times 10^{5} \, \text{N/m}^2) \times (5 \times 10^{-6}) \times (2 \times 10^{-5} \, \text{m}^3) \]

Calculating:
\[ U = \frac{1}{2} \times 5 \times 10^{5} \times 5 \times 10^{-6} \times 2 \times 10^{-5} \]
\[ U = \frac{1}{2} \times 5 \times 5 \times 2 \times 10^{-6} \times 10^{5} \times 10^{-5} \]
\[ U = \frac{1}{2} \times 50 \times 10^{-6} = 25 \times 10^{-6} \, \text{J} \]
\[ U = 2.5 \times 10^{-5} \, \text{J} \]

### Final Answer
The increase in the energy of the rod is:
\[ \text{Increase in Energy} = 2.5 \times 10^{-5} \, \text{J} \]