Question

A tube consisting of two vertical arms and a horizontal arm completely filled with water is held at rest. Inner diameter of the tube is much smaller than the length of each segment of the tube. The tube is made to move horizontally in its plane with a constant acceleration so that 1/16 of water flows out of the tube. Neglecting capillary action, deduce suitable expression for pressure at the closed end. Atmospheric pressure is $p_0$ and acceleration due to gravity is g.

Answer 1

The pressure at the closed end is $p_0 + \frac{3}{16}\rho gl$.

Answer 2

In an accelerating frame with acceleration $a$ horizontally, the effective gravity is $\vec{g}_{eff} = -a\hat{i} - g\hat{j}$. The free surface of the liquid is perpendicular to $\vec{g}_{eff}$, with equation $ax + gy = C$.

Assuming the horizontal arm connects $(0,0)$ to $(l,0)$ and vertical arms go up to $(0,l)$ and $(l,l)$, and water spills from the left arm, the free surface passes through $(0,l)$. Thus, $C = gl$. The water level in the right arm is $h_R = (C - al)/g = l - al/g$.

The volume of water remaining is $V_{rem} = 2Al + A(l - al/g) = 3Al - A(al/g)$. Initially, $V_{initial} = 3Al$. Since $1/16$ flows out, $V_{rem} = \frac{15}{16}V_{initial} = \frac{45Al}{16}$. Equating the volumes: $\frac{45Al}{16} = 3Al - \frac{Aal}{g} \implies \frac{al}{g} = 3l - \frac{45l}{16} = \frac{3l}{16} \implies \frac{a}{g} = \frac{3}{16}$.

The pressure at the top of the right arm $(l,l)$ is calculated from the free surface at $(l, h_R)$ where pressure is $p_0$: $P(l,l) = p_0 + \rho g (l - h_R) = p_0 + \rho g (l - (l - al/g)) = p_0 + \rho al$. Substituting $a = \frac{3g}{16}$: $P(l,l) = p_0 + \rho \left(\frac{3g}{16}\right) l = p_0 + \frac{3}{16}\rho gl$.