Explanation
The pressure just above the meniscus is atmospheric pressure and is greater than the pressure just below the meniscus since the pressure on the concave side is greater than that on the convex side.$$P_{concave} - P_{convex} = S(\dfrac{1}{R_1} + \dfrac{1}{R_2})$$For a spherical surface $$R_1=R_2=r$$Hence, $$P_{concave} - P_{convex} =\dfrac{2S}{r}$$In the given scenario, $$P_{concave}= P_0$$ where $$P_0$$ is the atmospheric pressure.$$P_{convex} = P_0 -\dfrac{2S}{r}$$
As the tube rotates, the centrifugal force (as seen in the rotating frame of the tube) creates a pressure head causing the water to be jetted out of the orifice at the end of tube
Consider, an infinitesimally thin slice of water of width $$dx$$ at a distance $$x$$ from the one end of the tube. Let the area of cross-section of the tube be $$s$$. The mass of water contained in this infinitesimally thin slice of water is given by $$dm=\rho sdx$$, here $$\rho$$ is the density of water. the centrifugal force by this slice of water on the next is given by,
$$df=(dm)w^2x=\rho s \omega^2xdx$$...............(1)
the net force acting at the end of the tube just before the orifice can be obtained by integrating over all such thin slice as,
$$\int^f_0 df = \int^l _ {l-h} \rho s \omega^2xdx$$ or,
$$f=\dfrac{1}{2}\rho s \omega^2h(2l-h)$$ or,
$$P=f/s = \dfrac{1}{2}\rho \omega^2h(2l-h)$$.................(2)
In (2), $$P$$ is the pressure head generated by the centrifugal force acting on the water. the net pressure at the end of the tube just before the orifice is given by $$P + P_o$$, when $$P_o$$ is the atmosphereic pressure. We assume that at this point the water is still i.e its velocity is zero. Right outside the tube, the water jets into an area of pressure $$P_i$$ (atmospheric pressure), say at a velocity $$v$$. Using Bernoulli's equation at point just before and just outside the orifice we have,
$$P+P_o+\dfrac{1}{2}\rho (0)^2=P_0+\dfrac{1}{2}\rho v^2$$ or,
$$P=\dfrac{1}{2}\rho v^2$$
or, $$v=\omega \sqrt{h(2l-h)}$$
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