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

Options

A. `(palpha)/(beta)`
B. `(3palpha)/(beta)`
C. `(p)/(alphabeta)`
D. `(p)/(3alphabeta)`

Explanation

To find the amount by which the temperature of the cube should be raised in order to bring its original volume back to the value it had before the pressure was applied, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Volume Change Due to Pressure:**
When a uniform pressure \( P \) is applied to the cube, it causes a change in volume \( \Delta V \). The relationship between pressure, bulk modulus \( K \), and volume change is given by:
\[ K = -\frac{P}{\frac{\Delta V}{V_0}} \]
where \( V_0 \) is the original volume of the cube.

2. **Expressing Volume Change:**
Rearranging the above equation gives:
\[ \frac{\Delta V}{V_0} = -\frac{P}{K} \]
This indicates that the volume decreases when pressure is applied.

3. **Thermal Expansion Relation:**
The change in volume due to a change in temperature \( \Delta \theta \) can be expressed using the coefficient of volume expansion \( \gamma \):
\[ \Delta V = V_0 \gamma \Delta \theta \]
For solids, the coefficient of volume expansion \( \gamma \) is related to the coefficient of linear expansion \( \alpha \) by:
\[ \gamma = 3\alpha \]
Therefore, we can write:
\[ \Delta V = V_0 (3\alpha) \Delta \theta \]

4. **Setting Up the Equation:**
Since we want to bring the volume back to its original value, we set the volume change due to thermal expansion equal to the volume change due to pressure:
\[ -\frac{P}{K} = 3\alpha \Delta \theta \]

5. **Solving for Temperature Change:**
Rearranging the equation to solve for \( \Delta \theta \):
\[ \Delta \theta = -\frac{P}{3\alpha K} \]

6. **Final Result:**
The amount by which the temperature should be raised to restore the original volume is:
\[ \Delta \theta = -\frac{P}{3\alpha K} \]