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Question: A U-tube containing a liquid is accelerated horizontally with a constant acceleration \({a_0}\) . ...

A U-tube containing a liquid is accelerated horizontally with a constant acceleration
a0{a_0} . If the separation between the vertical limbs is ll , find the difference in the heights of the liquid in the two arms.

Explanation

Solution

Here, we will use the formula of the pressure to calculate the height of the liquid in the two arms of the container. Pressure is the force that is applied at the right angle to the surface. Firstly, we will calculate the pressure difference when the fluid is accelerated, then we will equate this equation to the common formula of the pressure difference.

FORMULA USED:
The formula used for the pressure in the liquid accelerated with an acceleration aa is given by
P2P1=ρla{P_2} - {P_1} = \rho la
Here, P2P1{P_2} - {P_1} is the difference in pressures, ρ\rho is the density of the liquid, ll is the distance between the limbs of the container and aa is the acceleration of the liquid.
Also, the common formula of the pressure is given by

P=P2P1=hρgP = {P_2} - {P_1} = h\rho g

Here, ρ\rho is the density of the liquid, hh is the height of the liquid in the two arms of the container and gg is the acceleration due to gravity.

COMPLETE STEP BY STEP ANSWER:
When any liquid will be accelerated with an acceleration aa , then the difference in the pressures acting on the liquid is given by
P2P1=ρla{P_2} - {P_1} = \rho la
Here, ρ\rho is the density of the liquid, ll is the distance between the limbs of the container and
aa is the acceleration of the liquid.

Now, as given in the question, the liquid is accelerated with an acceleration a0{a_0} , then the difference in pressures is given by
P2P1=ρla0{P_2} - {P_1} = \rho l{a_0}

Now, the common formula used for the pressure difference is given by
P2P1=hρg{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
ρla0=hρg\rho l{a_0} = h\rho g
hg=la0\Rightarrow \,hg = l{a_0}
h=la0g\Rightarrow \,h = \dfrac{{l{a_0}}}{g}
Therefore, the difference in the heights of the liquid in the two arms is la0g\dfrac{{l{a_0}}}{g}

NOTE: Now, you might think that from where the formula P2P1=ρla{P_2} - {P_1} = \rho la has come. This is nothing but the common formula of pressure in which height hh is replaced with length ll and the acceleration of gravity gg is replaced with the acceleration aa . Therefore, you might not get confused about the formula.