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

Answer 1

Answer: (A), (C)

$\left(P_1-P_2\right)ds=\rho dsd\sqrt{2}(g\sin 45-a)$
$\left(P_1-P_2\right)=\rho d(g-a\sqrt{2})$
$\beta =\frac{\left(P_1-P_2\right)}{\rho gd}=\left(1-\frac{a\sqrt{2}}{g}\right)$
When $a=g/\sqrt{2},\beta =0$
When $a=g/2,\beta =\left(\frac{\sqrt{2}-1}{\sqrt{2}}\right)$