Question

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°. P1 and P2 are pressures at points 1 and 2, respectively, located at the base of the tube. Let B= (P1-P2)/(pgd), where p is the 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?

• A. B = 0 when a = g/√2
• B. B > 0 when a = 9//2
• C. B = √2-1 when a = 9/2
• D. B = 1/√2 when a = g/2

Answer 1

Answer: (A), (B)

A and B

Answer 2

The pressure gradient in the fluid is given by $\nabla P = \rho (\vec{g} - \vec{a}_{fluid})$. Assuming the fluid is in equilibrium in the accelerating frame, $\vec{a}_{fluid} = \vec{a}_{tube} = a \hat{i}$. The gravitational acceleration is $\vec{g} = g \sin \theta \hat{i} - g \cos \theta \hat{j}$. The acceleration of the tube is $\vec{a} = a \hat{i}$.

The pressure gradient components are: $\frac{\partial P}{\partial x} = \rho (g \sin \theta - a)$ $\frac{\partial P}{\partial y} = -\rho g \cos \theta$

Points 1 and 2 are at the base of the tube, and the line connecting them is along the inclined plane, with point 1 further down the incline than point 2. The distance between them along the incline is $d$. Thus, the pressure difference $P_1 - P_2$ is given by: $P_1 - P_2 = \int_{x_2}^{x_1} \frac{\partial P}{\partial x} dx = \int_{x_2}^{x_1} \rho (g \sin \theta - a) dx$ Since the distance along the incline is $d$, and point 1 is at $x_1 = x_2 + d$: $P_1 - P_2 = \rho (g \sin \theta - a) d$

We are given $\theta = 45^\circ$, so $\sin \theta = \frac{1}{\sqrt{2}}$. $P_1 - P_2 = \rho \left( \frac{g}{\sqrt{2}} - a \right) d$

The quantity $B$ is defined as $B = \frac{P_1 - P_2}{\rho g d}$. Substituting the expression for $P_1 - P_2$: $B = \frac{\rho \left( \frac{g}{\sqrt{2}} - a \right) d}{\rho g d} = \frac{\frac{g}{\sqrt{2}} - a}{g} = \frac{1}{\sqrt{2}} - \frac{a}{g}$

Now we evaluate the given options:

A. $B = 0$ when $a = g/\sqrt{2}$. If $a = g/\sqrt{2}$, then $B = \frac{1}{\sqrt{2}} - \frac{g/\sqrt{2}}{g} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$. This statement is correct.

B. $B > 0$ when $a = 9/\sqrt{2}$. Assuming $g > 9$ (which is true, as $g \approx 9.8 \, m/s^2$), then $a = 9/\sqrt{2} < g/\sqrt{2}$. In this case, $\frac{a}{g} < \frac{1}{\sqrt{2}}$, so $B = \frac{1}{\sqrt{2}} - \frac{a}{g} > 0$. This statement is correct.

C. $B = \sqrt{2}-1$ when $a = 9/2$. If $a = 9/2$, then $B = \frac{1}{\sqrt{2}} - \frac{9/2}{g} = \frac{1}{\sqrt{2}} - \frac{9}{2g}$. For $B = \sqrt{2}-1$, we would need $\frac{1}{\sqrt{2}} - \frac{9}{2g} = \sqrt{2}-1$. $\frac{9}{2g} = \frac{1}{\sqrt{2}} - (\sqrt{2}-1) = 1 - \frac{1}{\sqrt{2}} = 1 - \frac{\sqrt{2}}{2}$. $g = \frac{9}{2(1 - \sqrt{2}/2)} = \frac{9}{2-\sqrt{2}} = \frac{9(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})} = \frac{9(2+\sqrt{2})}{4-2} = \frac{9(2+\sqrt{2})}{2} \approx 15.36$. This value of $g$ is not standard. Thus, this statement is incorrect for the usual value of $g$.

D. $B = 1/\sqrt{2}$ when $a = g/2$. If $a = g/2$, then $B = \frac{1}{\sqrt{2}} - \frac{g/2}{g} = \frac{1}{\sqrt{2}} - \frac{1}{2}$. This is not equal to $1/\sqrt{2}$. This statement is incorrect.

Therefore, statements A and B are correct.