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${a_0}$ . If the separation between the vertical limbs is $l$ , find the difference in the heights of the liquid in the two arms.

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The formula used for the pressure in the liquid accelerated with an acceleration $a$ is given by

${P_2} - {P_1} = \rho la$

Here, ${P_2} - {P_1}$ is the difference in pressures, $\rho $ is the density of the liquid, $l$ is the distance between the limbs of the container and $a$ is the acceleration of the liquid.

Also, the common formula of the pressure is given by

$P = {P_2} - {P_1} = h\rho g$

Here, $\rho $ is the density of the liquid, $h$ is the height of the liquid in the two arms of the container and $g$ is the acceleration due to gravity.

When any liquid will be accelerated with an acceleration $a$ , then the difference in the pressures acting on the liquid is given by

${P_2} - {P_1} = \rho la$

Here, $\rho $ is the density of the liquid, $l$ is the distance between the limbs of the container and

$a$ is the acceleration of the liquid.

Now, as given in the question, the liquid is accelerated with an acceleration ${a_0}$ , then the difference in pressures is given by

${P_2} - {P_1} = \rho l{a_0}$

Now, the common formula used for the pressure difference is given by

${P_2} - {P_1} = h\rho g$

Now, equating the above equation of the pressure difference in the liquids to find the height of the liquid, we get

$\rho l{a_0} = h\rho g$

$ \Rightarrow \,hg = l{a_0}$

$ \Rightarrow \,h = \dfrac{{l{a_0}}}{g}$