Question

The vertical limbs of a $U$ shaped tube are filled with a liquid of density $\rho$ upto height $h$ on each side. The horizontal portion of $U$ tube having $2h$ contains a liquid of density $2\rho$. the $U$ tube is moved horizontally with an accelerator $g/2$ parallel to the horizontal arm. Find the difference in liquid levels in the two vertical limbs, at steady state.

Answer 1

Given : $a=\frac{g}{2}$

Pressure at A $P_A=P_o+\rho gh+\left(2\rho \right)g(h-x)=P_o+3\rho gh-2\rho gx$

Pressure at B $P_B=P_o+\rho gx$

Using $P_A-P_B=\left[2\rho \right(h+x)+\rho (h-x\left)\right]a$

$\therefore$ $(P_o+3\rho gh-2\rho gx)-(P_o+\rho gx)=[3\rho h+\rho x]\times \frac{g}{2}$

OR $3\rho gh-3\rho gx=\frac{3}{2}\rho gh+\frac{1}{2}\rho gx$

OR $\frac{3}{2}\rho gh=\frac{7}{2}\rho gx$ $⟹x=\frac{3}{7}h$

$\therefore$ Difference in the heights between two columns $\Delta H=(2h-x)-x=2h-2x$

$⟹$ $\Delta H=2h-\frac{6h}{7}=\frac{8h}{7}$