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 $\beta =\frac{(P_1–P_2)}{\left(\rho gd\right)}$, 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. $\beta =\frac{1}{\sqrt{2}}$ if $a=\frac{g}{2}$
• B. $\beta =0$ if $a=\frac{g}{\sqrt{2}}$
• C. $\beta >0$ if $a=\frac{g}{\sqrt{2}}$
• D. $\beta =\frac{1}{\sqrt{2}}-1$ if $a=\frac{g}{2}$

Answer 1

Answer: (B), (D)

$\beta =\frac{1}{\sqrt{2}}-1$ if $a=\frac{g}{2}$


If we resolve the components of acceleration, we have
$a_x=a_y=\frac{a}{\sqrt{2}}$
So, we have pressure difference between $1$ and $3$ as
$P_1-P_3=\rho (g-\frac{a}{\sqrt{2}})d$
and $P_2-P_3=\frac{\rho ad}{\sqrt{2}}$
$\therefore P_1-P_2=\rho d[g-\frac{2a}{\sqrt{2}}]$
$\therefore \frac{P_1-P_2}{\rho gd}=[1-\sqrt{2}\frac{a}{g}]=\beta$
So, if $\beta =0,a=\frac{g}{\sqrt{2}}$
Hence, (A) is correct.
and if $\beta =(\frac{1}{\sqrt{2}}-1),a=\frac{g}{2}$
Hence, (D) is correct.