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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