Fig. shows a U-tube of uniform cross-sectional area $A$ accelerated with acceleration $a$ as shown. If $d$ is the separation between the limbs, then the difference in the levels of the liquid in the U-tube is

\frac{a d}{g}

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\frac{g}{a d}

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a d g

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a d + g

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Solution

The correct option is D $\frac{a d}{g}$
Treating this analogues to acceleration due to gravity in Y direction and a in X- direction,
$p_{a t} h_{1} = f a d + p_{o}$
$p_{a t} h_{2} = p_{o}$
So,
Difference in level of liquid be h,
$f g h = f a d$
$\Rightarrow h = \frac{a d}{g}$

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Fig. shows a U-tube of uniform cross-sectional area A accelerated with acceleration a as shown. If d is the separation between the limbs, then the difference in the levels of the liquid in the U-tube is

The correct option is D adgTreating this analogues to acceleration due to gravity in Y direction and a in X- direction,path1=fad+popath2=poSo,Difference in level of liquid be h,fgh=fad⇒h=adg

Fig. shows a U-tube of uniform cross-sectional area $A$ accelerated with acceleration $a$ as shown. If $d$ is the separation between the limbs, then the difference in the levels of the liquid in the U-tube is

The correct option is D $\frac{a d}{g}$
Treating this analogues to acceleration due to gravity in Y direction and a in X- direction,
$p_{a t} h_{1} = f a d + p_{o}$
$p_{a t} h_{2} = p_{o}$
So,
Difference in level of liquid be h,
$f g h = f a d$
$\Rightarrow h = \frac{a d}{g}$