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Water in a rotating container

A thin cylindrical container of radius $R$, wall thickness $2d\ll R$, and height $h=d$, filled with water, rotates at angular velocity $\Omega$. Six radial holes of $h \times h$ square cross-section are made through the wall at $60^\circ$ intervals (at clock positions 1, 3, 5, 7, 9, 11). Initially the holes are plugged so that water fills the region of radial distance $r < R + d$ as shown in the figure. The surface tension $\sigma \gg \rho d^2 R \Omega^2$, where $\rho$ denotes the water density, and the contact angle is $90^\circ$. Assume $R \gg d$ and neglect gravity. Outside air pressure is atmospheric.

All plugs are removed simultaneously; immediately after removal, the water is still at rest in the co-rotating frame, but now water can flow through the holes. Let $t=0$ be when water in one of the holes has displaced by $x$ with $x \ll d$; it appears that by $t=T_1$​, it has displaced by $2x$.

A radial thin partition is now installed along the line connecting the clock positions of 12 and 6, creating two semicircular compartments with three holes in each. The experiment is repeated.

Find $T_2^2−T_1^2$​, where $T_2$ is the time from displacement $x$ to displacement $2x$ in the hole at 3 o’clock of the partitioned case.