In a U-tube,the liquid level stands at the sa | vedclass

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(D) When the $U$-tube is accelerated horizontally with acceleration $a$,a pseudo force acts on the liquid in the opposite direction of acceleration.Th

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$\frac{aL}{g}$

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(D) When the $U$-tube is accelerated horizontally with acceleration $a$,a pseudo force acts on the liquid in the opposite direction of acceleration.The effective acceleration $g_{	ext{eff}}$ is the vector sum of the acceleration due to gravity $g$ (acting downwards) and the pseudo acceleration $a$ (acting towards the left).The free surface of the liquid becomes perpendicular to the effective gravity vector $g_{	ext{eff}}$.The angle $	heta$ that the free surface makes with the horizontal is given by $	an 	heta = \frac{a}{g}$.From the geometry of the $U$-tube,the slope of the liquid surface is also $	an 	heta = \frac{h}{L}$,where $h$ is the height difference and $L$ is the horizontal distance between the two arms.Equating the two expressions for $	an 	heta$,we get $\frac{a}{g} = \frac{h}{L}$.Therefore,the height difference is $h = \frac{aL}{g}$.

In a $U$-tube,the liquid level stands at the same level when it is at rest. When the $U$-tube is accelerated towards the right with acceleration $a$,as shown in the figure,the difference $h$ between the levels of the two arms is:

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