Question

The narrow U-tube shown in figure containing two immiscible liquids of specific gravities $\rho_1$ & $\rho_2$ is given an acceleration 'a' in the horizontal direction and final heights of the liquid columns are shown in figure. Find $\frac{\rho_1}{\rho_2}$.

Answer 1

$\frac{h_2}{h_1}$

Answer 2

In a U-tube accelerated horizontally with acceleration $a$, the effective gravity in the horizontal direction is $a$ and in the vertical direction is $g$. The pressure gradient in the fluid is given by $\nabla P = \rho (\vec{g}_{eff})$, where $\vec{g}_{eff} = \vec{g} - \vec{a}$.

The slope of the free surface is given by $\frac{dy}{dx} = \frac{a}{g}$. For liquid 1, the free surface passes through $(x_1, h_1)$. Assuming the interface is at $x=0$, the equation of the free surface is $y = \frac{a}{g} x$. This implies $h_1 = \frac{a}{g} x_1$, so $ax_1 = gh_1$. For liquid 2, the free surface passes through $(x_2, h_2)$. Assuming the interface is at $x=0$, the equation of the free surface is $y = \frac{a}{g} x$. This implies $h_2 = \frac{a}{g} x_2$, so $ax_2 = gh_2$.

Equating the pressure at the interface of the two liquids: $\rho_1 a x_1 + \rho_1 g h_1 = \rho_2 a x_2 + \rho_2 g h_2$ Substituting $ax_1 = gh_1$ and $ax_2 = gh_2$: $\rho_1 (gh_1) + \rho_1 g h_1 = \rho_2 (gh_2) + \rho_2 g h_2$ $2 \rho_1 g h_1 = 2 \rho_2 g h_2$ $\rho_1 h_1 = \rho_2 h_2$ $\frac{\rho_1}{\rho_2} = \frac{h_2}{h_1}$.