Question
A U- tube containing a liquid is accelerated horizontally with a constant acceleration a0. If the separation between the vertical limbs is 1, find the difference in the heights of the liquid in the two arms.

Answer

Let the pressure at the base of the back limb at A = P, and the pressure at the base of the front limb at B = Ps. The separation between the vertical limbs AB = (given). When the U- tube is accelerated at an acceleration a, the difference of pressure between horizontal points A and B is given as,

$\text{P}_\text{a}-\text{P}_\beta=\text{l}\rho\text{a}_0$

where p is the density of the liquid. Due to this pressure difference, the liquid in limb A will rise to a heighth more than in limb B. The pressure difference due to this height difference balances the difference of pressure between points A and B.

The pressure due to thhis height $\text{h}=\rho\text{gh},$ Equating we get,

$\rho\text{gh}=\text{p}_\text{a}-\text{p}_\beta=\text{l}\rho\text{a}_0$

$\Rightarrow\text{gh}=\text{la}_0$

$\text{h}=\text{a}_0\text{l}/\text{g}$

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