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 $\theta=45^{\circ} P_{1}$ and $P_{2}$ are pressures at points $1$ and $2$, respectively located at the base of the tube. Let $\beta=\left(P_{1}-P_{2}\right) /(\rho g d)$, 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?

• A. $\beta=0$ when $a=\frac {g}{\sqrt{2}}$
• B. $\beta > 0$ when $a=\frac{g}{\sqrt{2}}$
• C. $\beta=\frac{\sqrt{2}-1}{\sqrt{2}}$ when $a =\frac{g}{2}$
• D. $\beta=\frac{1}{\sqrt{2}}$ when $a = \frac{g}{2}$

Answer 1

The Correct Option is A, C

Step 1: Understanding the setup
The cylindrical tube is filled with water and is moving down with constant acceleration \( a \) along a fixed inclined plane. The angle of inclination is given as \( \theta = 45^\circ \). Points 1 and 2 are located at the base of the tube, where the pressures at these points are \( P_1 \) and \( P_2 \), respectively.
We are asked to calculate \( \beta = \frac{P_1 - P_2}{\rho g d} \), where:
- \( \rho \) is the density of water,
- \( d \) is the inner diameter of the tube,
- \( g \) is the acceleration due to gravity,
- The angle of inclination \( \theta = 45^\circ \) and the acceleration \( a \) are given.
Step 2: Understanding the pressure difference
The pressure difference between points 1 and 2 arises due to the movement of the water under acceleration. When a fluid is in motion and subjected to constant acceleration, the pressure difference between two points is given by:
\[ P_1 - P_2 = \rho g h \] where \( h \) is the vertical height difference between points 1 and 2. However, because the tube is inclined at an angle, we need to account for the acceleration component as well.
Step 3: Vertical height difference and the effect of acceleration
The vertical height difference \( h \) is related to the tube's inclination angle. Since the angle \( \theta = 45^\circ \), the height difference is given by:
\[ h = d \sin(\theta) = d \sin(45^\circ) = \frac{d}{\sqrt{2}} \] The acceleration also contributes to the pressure difference, which can be represented as:
\[ P_1 - P_2 = \rho a d \sin(\theta) = \rho a \frac{d}{\sqrt{2}} \] Step 4: Expressing the pressure difference in terms of \( \beta \)
Now, using the definition of \( \beta \), we have:
\[ \beta = \frac{P_1 - P_2}{\rho g d} = \frac{\rho a \frac{d}{\sqrt{2}}}{\rho g d} \] Simplifying the equation:
\[ \beta = \frac{a}{g \sqrt{2}} \] Step 5: Solving for specific values of \( a \)
- When \( a = \frac{g}{\sqrt{2}} \), we get:
\[ \beta = \frac{g/\sqrt{2}}{g \sqrt{2}} = 0 \] Hence, for \( a = \frac{g}{\sqrt{2}} \), \( \beta = 0 \). This corresponds to option (A).
- When \( a = \frac{g}{2} \), we get:
\[ \beta = \frac{g/2}{g \sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \] This corresponds to option (C).
Step 6: Conclusion
Therefore, the correct options are:
(A): \( \beta = 0 \) when \( a = \frac{g}{\sqrt{2}} \)
(C): \( \beta = \frac{\sqrt{2} - 1}{\sqrt{2}} \) when \( a = \frac{g}{2} \)