Figure shows a three arm tube in which a liqu | vedclass

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Question

$P=P_{0}+\rho g\left(\frac{3 l}{2}\right)=P_{0}+\frac{3 \rho g l}{2}$

$\left(\mathrm{P}-\mathrm{P}_{0}\right) \mathrm{S}=(\rho \mathrm{S} l) \frac{

Vedclass · Accepted Answer

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

Vedclass · Answer

$P=P_{0}+\rho g\left(\frac{3 l}{2}\right)=P_{0}+\frac{3 \rho g l}{2}$

$\left(\mathrm{P}-\mathrm{P}_{0}\right) \mathrm{S}=(\rho \mathrm{S} l) \frac{l}{2} \omega^{2}$

$\frac{3 \rho g l S}{2}=\frac{\rho S l^{2}}{2} \omega^{2}$

$\frac{3}{2} g=\frac{l \omega^{2}}{2}$

$\omega^{2}=\frac{3 g}{l}$

$\omega=\sqrt{\frac{3 g}{l}}$

Figure shows a three arm tube in which a liquid is filled upto levels of height $l$. It is now rotated at an angular frequency $\omega$ about an axis passing through arm $B$. The angular frequency $\omega$ at which level of liquid in arm $B$ becomes zero.

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