The figure shows a three-arm tube in which a | vedclass

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(C) Let the cross-sectional area of each arm be $S$. When the tube rotates,the liquid level in arm $B$ drops to zero,and the liquid rises in arms $A$

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$\sqrt{\frac{3g}{l}}$

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(C) Let the cross-sectional area of each arm be $S$. When the tube rotates,the liquid level in arm $B$ drops to zero,and the liquid rises in arms $A$ and $C$. Since the total volume of liquid is conserved,the liquid rises by $l/2$ in arms $A$ and $C$,reaching a height of $3l/2$.The pressure at the base of arm $A$ (or $C$) is $P = P_0 + \rho g (3l/2)$.The pressure difference between the base of arm $A$ and the axis of rotation (arm $B$) provides the centripetal force for the liquid in the horizontal section of length $l$. The center of mass of this liquid column is at a distance $l/2$ from the axis.Equating the pressure difference to the centripetal force per unit area:$(P - P_0) S = (\rho S l) \omega^2 (l/2)$Substituting $P - P_0 = \rho g (3l/2)$:$\rho g (3l/2) S = \rho S l^2 \omega^2 / 2$Canceling $\rho, S,$ and $l$ from both sides:$3g/2 = l \omega^2 / 2$$\omega^2 = 3g/l$$\omega = \sqrt{\frac{3g}{l}}$

The figure shows a three-arm tube in which a liquid is filled up to a height $l$. It is now rotated at an angular frequency $\omega$ about an axis passing through arm $B$. Find the angular frequency $\omega$ at which the level of liquid in arm $B$ becomes zero.

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