Question
A $U-$ tube containing a liquid is accelerated horizontally with a constant acceleration $a_0.$ 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,
$\rho\text{gh}=\text{p}_\text{a}-\text{p}_\beta=\text{l}\rho\text{a}_0$
$\Rightarrow\text{gh}=\text{la}_0$
$\text{h}=\text{a}_0\text{l}/\text{g}$
$\text{P}_\text{a}-\text{P}_\beta=\text{l}\rho\text{a}_0$
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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