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 can use the formula for elastic potential energy stored in the material due to deformation. The formula for the energy (U) stored in the rod is given by:

\[ U = \frac{1}{2} F \Delta L \]

Where:
- \( F \) is the force applied,
- \( \Delta L \) is the change in length of the rod.

### Step 1: Calculate the Force (F)
The force applied on the rod can be calculated using the weight of the object compressing it. The weight (force due to gravity) is given by:

\[ F = mg \]

Where:
- \( m = 5 \, \text{kg} \) (mass of the weight),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).

Calculating the force:

\[ F = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N} \]

### Step 2: Calculate the Change in Length (ΔL)
The change in length can be calculated using Young's modulus (Y), which is defined as:

\[ Y = \frac{F/A}{\Delta L/L} \]

Rearranging this gives us:

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

Where:
- \( L = 0.2 \, \text{m} \) (original length of the rod),
- \( A = 1 \, \text{cm}^2 = 1 \times 10^{-4} \, \text{m}^2 \) (cross-sectional area),
- \( Y = 1 \times 10^{11} \, \text{N/m}^2 \) (Young's modulus of brass).

Substituting the values into the formula:

\[ \Delta L = \frac{50 \, \text{N} \times 0.2 \, \text{m}}{1 \times 10^{-4} \, \text{m}^2 \times 1 \times 10^{11} \, \text{N/m}^2} \]

Calculating \( \Delta L \):

\[ \Delta L = \frac{10}{1 \times 10^{7}} = 1 \times 10^{-6} \, \text{m} \]

### Step 3: Calculate the Increase in Energy (U)
Now, we can substitute \( F \) and \( \Delta L \) back into the energy formula:

\[ U = \frac{1}{2} F \Delta L \]

Substituting the values:

\[ U = \frac{1}{2} \times 50 \, \text{N} \times 1 \times 10^{-6} \, \text{m} \]

Calculating \( U \):

\[ U = 25 \times 10^{-6} \, \text{J} = 2.5 \times 10^{-5} \, \text{J} \]

Thus, the increase in energy of the rod is:

\[ \boxed{2.5 \times 10^{-5} \, \text{J}} \]