A $U-$ tube containing a liquid moves with a horizontal acceleration a along a direction joining the two vertical limbs. The separation between these limbs is $d$ . The difference in their liquid levels is
Diffcult
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Let $A$ be the area of cross$-$section of the tube and $\rho$ be the density of the liquid. Consider the section $AB$ of the tube.
Mass of the liquid in $\mathrm{AB}=\mathrm{dA} \rho$
Pressures at $\mathrm{A}$ and $\mathrm{B}=\mathrm{h}_{2} \rho \mathrm{g}$ and $\mathrm{h}_{1} \rho \mathrm{g}$
Net force to the right on $\mathrm{AB}=\left(\mathrm{h}_{2} \rho \mathrm{g}-\mathrm{h}_{1} \rho \mathrm{g}\right) \mathrm{A}$
$\therefore\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right) \rho \mathrm{g} \mathrm{A}=(\mathrm{d} \mathrm{A} \rho) \mathrm{a}$
Or $\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right) \mathrm{g}=\mathrm{d} \mathrm{a}$
Or $\quad \mathrm{h}_{2}-\mathrm{h}_{1}=\left(\frac{\mathrm{da}}{\mathrm{g}}\right)$
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