A wire of length `L_0` is supplied heat to raise its temperature by T. if `gamma` is the coefficient of volume expansion of the wire and Y is Young's modulus of the wire then the energy density stored in the wire is

Options

A. `(1)/(2)gamma^2T^2Y`
B. `(1)/(3)gamma^2T^2Y^3`
C. `(1)/(18)(gamma^2T^2)/(T)`
D. `(1)/(18)gamma^2T^2Y`

Explanation

To solve the problem of finding the energy density stored in a wire when it is heated, we can follow these steps:

### Step 1: Understand the Problem
We are given:
- Length of the wire: \( L_0 \)
- Temperature increase: \( T \)
- Coefficient of volume expansion: \( \gamma \)
- Young's modulus: \( Y \)

We need to find the energy density stored in the wire due to the temperature increase.

### Step 2: Relate Temperature Change to Length Change
When the wire is heated, it expands. The longitudinal strain (\( \epsilon \)) produced in the wire due to the temperature change can be expressed as:
\[ \epsilon = \frac{\Delta L}{L_0} = \alpha \Delta T \]
where \( \alpha \) is the coefficient of linear expansion and \( \Delta T = T \).

### Step 3: Substitute for Longitudinal Strain
Substituting \( \Delta T \) into the equation gives:
\[ \epsilon = \alpha T \]

### Step 4: Relate Stress to Strain
The stress (\( \sigma \)) in the wire is related to Young's modulus and strain:
\[ \sigma = Y \cdot \epsilon \]
Substituting the expression for strain, we have:
\[ \sigma = Y \cdot (\alpha T) \]

### Step 5: Calculate the Energy Density
The energy density (\( E \)) stored in the wire can be calculated using the formula for elastic potential energy per unit volume:
\[ E = \frac{1}{2} \sigma \epsilon \]
Substituting the expressions for stress and strain:
\[ E = \frac{1}{2} (Y \cdot (\alpha T)) \cdot (\alpha T) \]
This simplifies to:
\[ E = \frac{1}{2} Y \alpha^2 T^2 \]

### Step 6: Relate Linear Expansion to Volume Expansion
We know that the coefficient of linear expansion (\( \alpha \)) is related to the coefficient of volume expansion (\( \gamma \)) by:
\[ \alpha = \frac{\gamma}{3} \]
Substituting this into the energy density equation gives:
\[ E = \frac{1}{2} Y \left(\frac{\gamma}{3}\right)^2 T^2 \]
This simplifies to:
\[ E = \frac{1}{2} Y \cdot \frac{\gamma^2}{9} T^2 \]

### Step 7: Final Expression for Energy Density
Thus, the energy density stored in the wire is:
\[ E = \frac{1}{18} \gamma^2 Y T^2 \]

### Conclusion
The final expression for the energy density stored in the wire when heated is:
\[ \boxed{\frac{1}{18} \gamma^2 Y T^2} \]