A uniform pressure P is exerted on all sides of a solid cube at temperature t `""^(@)C`. By what amount should the temperature of the cube be raised in order to bring its volume back to the value it had before the pressure was applied? The coefficient of volume expansion of cube is `alpha` and the bulk modulus is K.

Options

A. `P/(3alphaK)`
B. `(Palpha)/(K)`
C. `(PK)/(alpha)`
D. `(2K)/(P)`

Answer

To solve the problem, we need to determine the change in temperature required to return the volume of a solid cube to its original value after a uniform pressure \( P \) has been applied. We will use the concepts of volume expansion and bulk modulus in our calculations.

### Step-by-Step Solution:

1. **Understanding Volume Change Due to Pressure:**
When a pressure \( P \) is applied to the cube, it causes a change in volume \( \Delta V \). The relationship between the change in volume and the applied pressure can be expressed using the bulk modulus \( K \):
\[ \Delta V = -\frac{V P}{K} \]
Here, \( V \) is the original volume of the cube.

2. **Volume Expansion Due to Temperature Change:**
The volume of a solid changes with temperature according to the coefficient of volume expansion \( \gamma \) (which is related to the coefficient of linear expansion \( \alpha \) by \( \gamma = 3\alpha \)):
\[ \Delta V = V \gamma \Delta T \]
where \( \Delta T \) is the change in temperature.

3. **Setting the Equations Equal:**
To bring the volume back to its original value, the change in volume due to temperature increase must equal the volume change due to pressure:
\[ V \gamma \Delta T = -\Delta V \]
Substituting the expression for \( \Delta V \) from step 1:
\[ V \gamma \Delta T = \frac{V P}{K} \]

4. **Cancelling Volume \( V \):**
Since \( V \) is common on both sides, we can cancel it out (assuming \( V \neq 0 \)):
\[ \gamma \Delta T = \frac{P}{K} \]

5. **Solving for \( \Delta T \):**
Rearranging the equation to solve for \( \Delta T \):
\[ \Delta T = \frac{P}{\gamma K} \]

6. **Substituting for \( \gamma \):**
Since \( \gamma = 3\alpha \), we substitute this into the equation:
\[ \Delta T = \frac{P}{3\alpha K} \]

### Final Result:
The amount by which the temperature of the cube should be raised to bring its volume back to the original value is:
\[ \Delta T = \frac{P}{3\alpha K} \]