A uniform metal rod fixed at its ends of `2 mm^(2)` cross-section is cooled from `40@^C` to `20^@C`. The coefficient of the linear expansion of the rod is `12xx10^(-6) ` per degree celsius and its young's modulus of elasticity is `10^11 N//m^(2).` The energy stored per unit volume of the rod is

Options

A. `2880 J//m^(3)`
B. `1500 J//m^(3)`
C. `5760 J//m^(3)`
D. `1440 J//m^(3)`

Explanation

To find the energy stored per unit volume of the metal rod when it is cooled, we can follow these steps:

### Step 1: Understand the relationship between stress, strain, and Young's modulus
The energy density (energy stored per unit volume) in a material can be expressed using the formula:
\[ \text{Energy Density} = \frac{1}{2} \times \text{Stress} \times \text{Strain} \]
We also know that Young's modulus (Y) is defined as:
\[ Y = \frac{\text{Stress}}{\text{Strain}} \]
From this, we can express Stress in terms of Young's modulus and Strain:
\[ \text{Stress} = Y \times \text{Strain} \]

### Step 2: Substitute Stress into the energy density formula
Substituting the expression for Stress into the energy density formula gives:
\[ \text{Energy Density} = \frac{1}{2} \times (Y \times \text{Strain}) \times \text{Strain} = \frac{1}{2} Y \times \text{Strain}^2 \]

### Step 3: Calculate the strain
The strain (ε) due to a change in temperature can be calculated using the coefficient of linear expansion (α):
\[ \text{Strain} = \alpha \times \Delta T \]
Where:
- \(\alpha = 12 \times 10^{-6} \, \text{per degree Celsius}\)
- \(\Delta T = T_{\text{final}} - T_{\text{initial}} = 20°C - 40°C = -20°C\)

Now, substituting the values:
\[ \text{Strain} = 12 \times 10^{-6} \times (-20) = -240 \times 10^{-6} \]

### Step 4: Calculate the energy density
Now we can substitute the values into the energy density formula:
\[ \text{Energy Density} = \frac{1}{2} \times Y \times \text{Strain}^2 \]
Where:
- \(Y = 10^{11} \, \text{N/m}^2\)
- \(\text{Strain} = -240 \times 10^{-6}\)

Calculating \(\text{Strain}^2\):
\[ \text{Strain}^2 = (-240 \times 10^{-6})^2 = 57600 \times 10^{-12} = 5.76 \times 10^{-8} \]

Now substituting back into the energy density formula:
\[ \text{Energy Density} = \frac{1}{2} \times 10^{11} \times 5.76 \times 10^{-8} \]
\[ = \frac{1}{2} \times 5.76 \times 10^3 = 2880 \, \text{J/m}^3 \]

### Final Answer
The energy stored per unit volume of the rod is:
\[ \text{Energy Density} = 2880 \, \text{J/m}^3 \]

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