When at rest, a liquid stands at the same level in the tubes as shown in the figure. But as indicated, a height difference h occurs when the system is given an acceleration a towards the right. Then h is equal to:

\frac{a L}{2 g}

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\frac{g L}{2 a}

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\frac{g L}{a}

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\frac{a L}{g}

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Solution

The correct option is D $\frac{a L}{g}$
Let $P_{1}$ and $P_{2}$ be the pressures at the bottom of the left and right ends of the tube, respectively.
Then $F = (P_{1} - P_{2}) A = ρ g h A$
where A is the cross section of the tube.
The mass of the liquid in the horizontal portion is
$m = ρ L A$
Now, $F = m a$
$\Rightarrow ρ g h A = ρ L A a$
$∴ h = \frac{a L}{g}$

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When at rest, a liquid stands at the same level in the tubes as shown in the figure. But as indicated, a height difference h occurs when the system is given an acceleration a towards the right. Then h is equal to:

The correct option is D aLgLet P1 and P2 be the pressures at the bottom of the left and right ends of the tube, respectively.Then F=(P1−P2)A=ρghAwhere A is the cross section of the tube.The mass of the liquid in the horizontal portion ism=ρLANow, F=ma⇒ρghA=ρLAa∴h=aLg