1. Understanding the Setup and Coordinate System

Let's analyze the physical situation. A cylindrical tube filled with water is accelerating down an inclined plane at an angle $\theta = 45^\circ$ with a constant acceleration $a$. We need to find the pressure difference between two points, 1 and 2, at the base of the tube. These points are at the ends of a diameter that is parallel to the inclined plane, with point 1 being higher up the incline than point 2.

It is convenient to work in a non-inertial frame of reference attached to the tube. Let's define a coordinate system $(x', y')$ where the $x'$-axis points down the incline and the $y'$-axis is perpendicular to the incline, pointing outwards.

2. Effective Gravity in the Accelerating Frame

In this non-inertial frame, the fluid is in equilibrium under an effective gravitational acceleration, $\vec{g}_{eff}$. This is the vector sum of the true gravitational acceleration, $\vec{g}$, and the pseudo-acceleration, $-\vec{a}$.

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

Let's find the components of $\vec{g}$ and $\vec{a}$ in our chosen $(x', y')$ coordinate system:

• The acceleration of the frame is $\vec{a} = (a, 0)$.
• The gravitational acceleration is $\vec{g}$. Its component along the incline is $g \sin\theta$ and perpendicular to it is $-g \cos\theta$. So, $\vec{g} = (g \sin\theta, -g \cos\theta)$.

Now, we can find the components of $\vec{g}_{eff}$: $\vec{g}_{eff} = (g \sin\theta - a, -g \cos\theta)$

Given $\theta = 45^\circ$, we have $\sin\theta = \cos\theta = 1/\sqrt{2}$. $\vec{g}_{eff} = (g/\sqrt{2} - a, -g/\sqrt{2})$

3. Pressure Gradient and Pressure Difference

The pressure gradient in the fluid in this accelerating frame is given by: $\nabla P = \rho \vec{g}_{eff}$

This gives us the partial derivatives of pressure: $\frac{\partial P}{\partial x'} = \rho (g/\sqrt{2} - a)$ $\frac{\partial P}{\partial y'} = \rho (-g/\sqrt{2})$

Points 1 and 2 are located at the base, along the diameter parallel to the incline. We can set their coordinates at the center of the diameter to be $x'=0$. Since point 1 is up the incline and point 2 is down the incline, their coordinates are:

• Point 1: $x'_1 = -d/2$
• Point 2: $x'_2 = +d/2$

Both points are at the same $y'$ coordinate, so we only need to consider the pressure change along the $x'$-axis.

The pressure difference $P_1 - P_2$ can be found by integrating the pressure gradient from point 2 to point 1: $P_1 - P_2 = \int_{x'_2}^{x'_1} \frac{\partial P}{\partial x'} dx'$ $P_1 - P_2 = \int_{d/2}^{-d/2} \rho (g/\sqrt{2} - a) dx'$ Since the integrand is constant: $P_1 - P_2 = \rho (g/\sqrt{2} - a) [x']_{d/2}^{-d/2}$ $P_1 - P_2 = \rho (g/\sqrt{2} - a) (-d/2 - d/2)$ $P_1 - P_2 = \rho (g/\sqrt{2} - a) (-d)$ $P_1 - P_2 = \rho d (a - g/\sqrt{2})$

4. Calculating β

The problem defines $\beta = (P_1 - P_2)/(\rho gd)$. Substituting our expression for $P_1 - P_2$: $\beta = \frac{\rho d (a - g/\sqrt{2})}{\rho g d}$ $\beta = \frac{a - g/\sqrt{2}}{g} = \frac{a}{g} - \frac{1}{\sqrt{2}}$

5. Evaluating the Options

Now we can check each statement using the derived formula for $\beta$.

A: $\beta$ = 0 when a = g/$\sqrt 2$ Substitute $a = g/\sqrt{2}$ into the formula for $\beta$: $\beta = \frac{g/\sqrt{2}}{g} - \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$ This statement is correct.

B: $\beta$ > 0 when a = g/$\sqrt 2$ As calculated for option A, when $a = g/\sqrt{2}$, $\beta = 0$. So, this statement is incorrect.

C: $\beta = {{\sqrt 2 - 1} \over {\sqrt 2 }}$ when a = g/2 Substitute $a = g/2$ into the formula for $\beta$: $\beta = \frac{g/2}{g} - \frac{1}{\sqrt{2}} = \frac{1}{2} - \frac{1}{\sqrt{2}} = \frac{1 - \sqrt{2}}{2}$ The value given in the option is $\frac{\sqrt{2} - 1}{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}}$. Since $\frac{1 - \sqrt{2}}{2} \approx \frac{1 - 1.414}{2} = -0.207$ and $\frac{\sqrt{2} - 1}{\sqrt{2}} \approx \frac{1.414 - 1}{1.414} \approx 0.293$, the values are not equal. Thus, this statement is incorrect.

D: $\beta = {1 \over {\sqrt 2 }}$ when a = g/2 As calculated for option C, when $a = g/2$, $\beta = \frac{1 - \sqrt{2}}{2}$. This is not equal to $\frac{1}{\sqrt{2}}$. So, this statement is incorrect.

Conclusion

Based on the derivation, only option A is correct. The stored answer [A, C] appears to be incorrect. It's worth noting that if the condition in option C were $a = g$ instead of $a = g/2$, the option would be correct: $\beta = \frac{g}{g} - \frac{1}{\sqrt{2}} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$. This suggests a possible typo in the question statement.