A cylindrical tube, with its base as shown in the figure, is filled with water. It is moving down with a constant acceleration $a$ along a fixed inclined plane with angle $\theta = 45^\circ$. $P_1$ and $P_2$ are pressures at points 1 and 2, respectively, located at the base of the tube. Let $\beta = (P_1 - P_2)/(\rho g d)$, where $\rho$ is density of water, $d$ is the inner diameter of the tube and $g$ is the acceleration due to gravity. Which of the following statement(s) is(are) correct?

Step 1: Define the effective acceleration in the tube's frame.

In the frame of the tube moving down the incline at $\theta = 45^\circ$, the effective acceleration due to gravity is the vector sum of gravity and the pseudo-acceleration.

$$\vec{g}_{\text{eff}} = \vec{g} - \vec{a}$$

Using a coordinate system where $\hat{i}$ is horizontal and $\hat{j}$ is vertical:

$$\vec{g} = -g\hat{j}$$ $$\vec{a} = \frac{a}{\sqrt{2}}\hat{i} - \frac{a}{\sqrt{2}}\hat{j}$$ $$\vec{g}_{\text{eff}} = -\frac{a}{\sqrt{2}}\hat{i} - \left(g - \frac{a}{\sqrt{2}}\right)\hat{j}$$

Step 2: Determine the separation vector between points 1 and 2.

Point 2 is uphill and point 1 is downhill along the base of the tube, separated by a distance $d$ along the incline.

$$\vec{r}_1 - \vec{r}_2 = d\hat{i} - d\hat{j}$$

Step 3: Calculate the pressure difference.

The pressure difference is determined by the dot product of the effective gravity and the separation vector.

$$P_1 - P_2 = \rho \vec{g}_{\text{eff}} \cdot (\vec{r}_1 - \vec{r}_2)$$ $$P_1 - P_2 = \rho \left[ -\frac{a}{\sqrt{2}}\hat{i} - \left(g - \frac{a}{\sqrt{2}}\right)\hat{j} \right] \cdot (d\hat{i} - d\hat{j})$$ $$P_1 - P_2 = \rho \left[ -\frac{a}{\sqrt{2}}d + \left(g - \frac{a}{\sqrt{2}}\right)d \right]$$ $$P_1 - P_2 = \rho d (g - \sqrt{2}a)$$

Step 4: Express the parameter $\beta$.

Substitute the pressure difference into the definition of $\beta$.

$$\beta = \frac{P_1 - P_2}{\rho g d}$$ $$\beta = \frac{\rho d (g - \sqrt{2}a)}{\rho g d}$$ $$\beta = 1 - \sqrt{2}\frac{a}{g}$$

Step 5: Evaluate the given cases.

Substitute the values of acceleration $a$ to find the corresponding values of $\beta$.

For $a = \frac{g}{\sqrt{2}}$:

$$\beta = 1 - \sqrt{2}\left(\frac{g/\sqrt{2}}{g}\right)$$ $$\beta = 0$$

For $a = \frac{g}{2}$:

$$\beta = 1 - \sqrt{2}\left(\frac{g/2}{g}\right)$$ $$\beta = \frac{\sqrt{2} - 1}{\sqrt{2}}$$

Answer: The correct statements are $\beta = 0$ for $a = \frac{g}{\sqrt{2}}$ and $\beta = \frac{\sqrt{2}-1}{\sqrt{2}}$ for $a = \frac{g}{2}$.