An aluminium rod with young's modulus `Y= 7.0times 10^(10)N/m^(2)` undergoes elastic strain of `0.04%`.The energy per unit volume stored in the rod in SI unit is:

Options

A. `5600`
B. `8400`
C. `11200`
D. `2800`

Answer

To find the energy per unit volume stored in the aluminium rod, we can use the formula for elastic potential energy per unit volume, which is given by:

\[ U = \frac{1}{2} \sigma \epsilon \]

where \( U \) is the energy per unit volume, \( \sigma \) is the stress, and \( \epsilon \) is the strain.

### Step 1: Convert the strain percentage to a decimal
The strain is given as \( 0.04\% \). To convert this percentage to a decimal, we divide by 100:

\[ \epsilon = \frac{0.04}{100} = 0.0004 \]

### Step 2: Calculate the stress using Young's modulus
The stress \( \sigma \) can be calculated using Young's modulus \( Y \) and the strain \( \epsilon \):

\[ \sigma = Y \cdot \epsilon \]

Substituting the values:

\[ Y = 7.0 \times 10^{10} \, \text{N/m}^2 \]
\[ \sigma = 7.0 \times 10^{10} \times 0.0004 \]

Calculating \( \sigma \):

\[ \sigma = 7.0 \times 10^{10} \times 4 \times 10^{-4} = 28.0 \times 10^{6} \, \text{N/m}^2 = 2.8 \times 10^{7} \, \text{N/m}^2 \]

### Step 3: Calculate the energy per unit volume
Now we can substitute the values of stress and strain into the energy per unit volume formula:

\[ U = \frac{1}{2} \sigma \epsilon \]

Substituting the values we have:

\[ U = \frac{1}{2} \times (2.8 \times 10^{7}) \times (0.0004) \]

Calculating \( U \):

\[ U = \frac{1}{2} \times 2.8 \times 10^{7} \times 4 \times 10^{-4} \]
\[ U = \frac{1}{2} \times 11.2 \times 10^{3} = 5.6 \times 10^{3} \, \text{J/m}^3 \]

### Final Answer
Thus, the energy per unit volume stored in the rod is:

\[ U = 5600 \, \text{J/m}^3 \]