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:

We begin with the cube at an initial temperature of $$0^{\circ}C$$ and at its original volume $$V_0$$. An external pressure $$P$$ is applied uniformly from all directions, so the cube is compressed equally on all sides.

First, we recall the definition of the bulk modulus. The bulk modulus $$K$$ of a material is given by the formula

$$K = -\,\frac{P}{\displaystyle\frac{\Delta V}{V_0}}$$

where $$\Delta V = V - V_0$$ is the change in volume caused by the pressure $$P$$. Because the pressure compresses the cube, the change in volume is negative, so we may write

$$\frac{\Delta V}{V_0} = -\,\frac{P}{K}.$$

Hence the fractional decrease in volume produced by the pressure is

$$\left|\frac{\Delta V}{V_0}\right| = \frac{P}{K}.$$

Next, we wish to restore the cube to its original size by heating it while the same pressure $$P$$ continues to act. For an isotropic solid, the coefficient of linear expansion is $$\alpha$$, so the coefficient of volume expansion is

$$\beta = 3\alpha.$$

If we raise the temperature by an amount $$\Delta T$$, the fractional change in volume produced by thermal expansion is

$$\frac{\Delta V_{\text{thermal}}}{V_0} = \beta\Delta T = 3\alpha\Delta T.$$

To bring the cube back to its original volume, the volume increase produced by heating must exactly cancel the volume decrease caused by the applied pressure. Therefore we set

$$3\alpha\Delta T = \frac{P}{K}.$$

Solving for $$\Delta T$$, we obtain

$$\Delta T = \frac{P}{3\alpha K}.$$

This is the temperature rise required for the cube to regain its original size under the continued external pressure $$P$$.

Hence, the correct answer is Option 2.