An external pressure P is applied on a cube at `0^(@)C` so that is it equally compressed from all sides. K is the bulk modulus o 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:-

Answer

To solve the problem, we need to determine the temperature increase required to restore the original size of a cube that has been compressed under an external pressure \( P \). The bulk modulus \( K \) and the coefficient of linear expansion \( \alpha \) of the material are given.

### Step-by-step Solution:

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

2. **Relating Pressure to Volume Change**:
Rearranging the formula gives us:
\[ \frac{\Delta V}{V} = -\frac{P}{K} \]
Since the cube is compressed, \( \Delta V \) will be negative, hence the negative sign in the equation.

3. **Volume Change Due to Temperature Increase**:
When the temperature is increased, the volume change \( \Delta V \) can also be expressed in terms of the coefficient of volume expansion. For a cube, the volume change can be approximated as:
\[ \Delta V = V_0 \cdot 3\alpha \Delta \theta \]
where \( V_0 \) is the original volume, \( \alpha \) is the coefficient of linear expansion, and \( \Delta \theta \) is the change in temperature.

4. **Equating Volume Changes**:
Since the volume change due to pressure and temperature must be equal in magnitude (but opposite in sign), we can set the two expressions for \( \Delta V \) equal to each other:
\[ -\frac{P}{K} = 3\alpha \Delta \theta \]

5. **Solving for Temperature Change**:
Rearranging the equation to solve for \( \Delta \theta \):
\[ \Delta \theta = -\frac{P}{3K\alpha} \]
Since we are interested in the increase in temperature (which is a positive quantity), we can express it as:
\[ \Delta \theta = \frac{P}{3K\alpha} \]

### Final Answer:
The temperature should be raised by:
\[ \Delta \theta = \frac{P}{3K\alpha} \]