An external pressure P is applied on a cube at `0^(@)C` so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and `alpha` is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by

Options

A. `(3alpha)/(PK)`
B. `3 PK alpha`
C. `(P)/(3alpha K)`
D. `(P)/(alpha K)`

Explanation

To solve the problem, we need to determine the temperature change required to bring the cube back to its original size after it has been compressed by an external pressure \( P \).

### Step-by-Step Solution:

1. **Understand the Bulk Modulus**:
The bulk modulus \( K \) is defined as:
\[ K = -\frac{\Delta P}{\frac{\Delta V}{V_0}} \]
where \( \Delta P \) is the change in pressure, \( \Delta V \) is the change in volume, and \( V_0 \) is the original volume.

2. **Relate Volume Change to Pressure**:
Rearranging the formula gives us:
\[ \frac{\Delta V}{V_0} = -\frac{\Delta P}{K} \]

3. **Thermal Expansion**:
When the cube is heated, it expands. The volume change due to thermal expansion can be expressed as:
\[ \frac{\Delta V}{V_0} = \gamma \Delta T \]
where \( \gamma \) is the volume expansion coefficient. For a cube, \( \gamma \) is related to the coefficient of linear expansion \( \alpha \) by:
\[ \gamma = 3\alpha \]
Therefore, we can write:
\[ \frac{\Delta V}{V_0} = 3\alpha \Delta T \]

4. **Set the Two Expressions for Volume Change Equal**:
From steps 2 and 3, we have:
\[ -\frac{\Delta P}{K} = 3\alpha \Delta T \]

5. **Solve for the Temperature Change \( \Delta T \)**:
Rearranging the equation gives:
\[ \Delta T = -\frac{\Delta P}{3\alpha K} \]
Since \( \Delta P \) is the pressure applied, we can substitute \( \Delta P = P \):
\[ \Delta T = -\frac{P}{3\alpha K} \]

6. **Final Result**:
The temperature should be raised by:
\[ \Delta T = \frac{P}{3\alpha K} \]