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 C

Given:

• A tube is inclined at angle θ = 45°
• The tube is accelerating down the incline with constant acceleration \( a \)
• The tube is filled with water, and pressures at points 1 and 2 are \( P_1 \) and \( P_2 \)
• The expression to evaluate is: \[ \beta = \frac{P_1 - P_2}{\rho g d} \]

Step 1: Consider effective acceleration on the fluid

Since the tube is accelerating along the incline, we analyze the effective gravity (pseudo-force method) in the non-inertial frame.

Effective acceleration on fluid = vector sum of actual gravity \( g \) downward and pseudo-acceleration \( a \) up the incline.

Break both vectors into components perpendicular and parallel to the incline. For incline angle \( \theta = 45^\circ \): 
- Gravity components: 
\( g_\perp = g \cos\theta, \quad g_\parallel = g \sin\theta \) 
- Pseudo-acceleration (opposing \( a \)) is entirely along incline: \( a \)

Step 2: Effective acceleration along the tube

Effective component along the tube: 
\[ a_\text{eff} = \sqrt{(g \cos\theta)^2 + (g \sin\theta - a)^2} \] \[ = \sqrt{ \left( \frac{g}{\sqrt{2}} \right)^2 + \left( \frac{g}{\sqrt{2}} - a \right)^2 } \] \[ = \sqrt{ \frac{g^2}{2} + \left( \frac{g - \sqrt{2}a}{\sqrt{2}} \right)^2 } \]

Step 3: Hydrostatic pressure difference due to effective gravity

Pressure difference in fluid column of height \( d \) is: \[ P_1 - P_2 = \rho a_\text{eff} d \Rightarrow \beta = \frac{P_1 - P_2}{\rho g d} = \frac{a_\text{eff}}{g} \] Use: \[ \beta = \frac{1}{g} \cdot \sqrt{ \left( \frac{g}{\sqrt{2}} \right)^2 + \left( \frac{g}{\sqrt{2}} - a \right)^2 } \]

Step 4: Substitute \( a = \frac{g}{2} \)

\[ \left( \frac{g}{\sqrt{2}} - a \right) = \frac{g}{\sqrt{2}} - \frac{g}{2} = g\left( \frac{1}{\sqrt{2}} - \frac{1}{2} \right) \] Simplify using: \[ \frac{1}{\sqrt{2}} - \frac{1}{2} = \frac{2 - \sqrt{2}}{2\sqrt{2}} \] Hence: \[ \beta = \frac{1}{g} \cdot \sqrt{ \left( \frac{g}{\sqrt{2}} \right)^2 + \left( g \cdot \left( \frac{2 - \sqrt{2}}{2\sqrt{2}} \right) \right)^2 } \] After simplifying: \[ \beta = \frac{1}{g} \cdot g \cdot \frac{\sqrt{2} - 1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \]

Final Answer: \( \beta = \dfrac{\sqrt{2} - 1}{\sqrt{2}} \) when \( a = \dfrac{g}{2} \)