The vertical limbs of a $U$ shaped tube are filled with a liquid of density $\rho$ upto a height $h$ on each side. The horizontal portion of the $U$ tube having length $2h$ contains a liquid of density $2\rho$ . The $U$ tube is moved horizontally with an accelerator $g/2$ parallel to the horizontal arm. The difference in heights in liquid levels in the two vertical limbs, at steady state will be
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Given $: \quad a=\frac{g}{2}$
Pressure at $A \quad P_{A}=P_{o}+\rho g h+(2 \rho) g(h-x)=P_{o}+3 \rho g h-2 \rho g x$
Pressure at $\mathrm{B} \quad P_{B}=P_{o}+\rho g x$
Using $\quad P_{A}-P_{B}=[2 \rho(h+x)+\rho(h-x)] a$
$\left(P_{o}+3 \rho g h-2 \rho g x\right)-\left(P_{o}+\rho g x\right)=[3 \rho h+\rho x] \times \frac{g}{2}$
OR $\quad 3\ \rho gh -3 \rho g x=\frac{3}{2} \rho g h+\frac{1}{2} \rho g x$
oR $\quad \frac{3}{2} \rho g h=\frac{7}{2} \rho g x \quad \Longrightarrow x=\frac{3}{7} h$
$\therefore$ Difference in the heights between two columns $\Delta H=(2 h-x)-x=2 h-2 x$
$\Longrightarrow \quad \Delta H=2 h-\frac{6 h}{7}=\frac{8 h}{7}$
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