A glass tube of uniform cross-section is fill | vedclass

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(A) When the tube rotates with angular velocity $\omega$,the water in the horizontal arms experiences a centrifugal force directed outwards.For a smal

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Water levels in both sections $A$ and $B$ go up.

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(A) When the tube rotates with angular velocity $\omega$,the water in the horizontal arms experiences a centrifugal force directed outwards.For a small element of water of mass $dm$ at a distance $r$ from the axis of rotation,the required centripetal force is provided by the pressure gradient: $dP = \rho \omega^2 r dr$.Integrating this from the axis $(r=0)$ to a distance $r$,we get the pressure at distance $r$ as $P(r) = P_0 + \frac{1}{2} \rho \omega^2 r^2$,where $P_0$ is the pressure at the axis.Since the pressure at the surface of the water in the vertical arms must be equal to the atmospheric pressure,the height $h$ of the water column in an arm at distance $r$ from the axis is given by $\rho gh = P(r) - P_{atm}$.As $r$ increases,the pressure $P(r)$ increases,which means the height $h$ of the water column must increase to balance this pressure.Since both arms $A$ and $B$ are at distances $L$ and $2L$ respectively from the axis,the water levels in both arms will rise compared to the level at the central axis to provide the necessary centripetal force for the rotation of the water mass.

Column-$I$	Column-$II$
$(a)$ Rain drops move downwards with constant velocity.	$(i)$ Viscous liquids
$(b)$ Floating clouds at a height in air.	$(ii)$ Viscosity
	$(iii)$ Less density

$A$ glass tube of uniform cross-section is filled with water and mounted on a rotatable shaft as shown in the figure. If the tube is rotated with a constant angular velocity $\omega$,then:

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Column-$I$ Column-$II$

$(a)$ Rain drops move downwards with constant velocity. $(i)$ Viscous liquids

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$(iii)$ Less density

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