Rotating U-tube pressure and water lost

Reading of the figure: axis through limb AB (A sealed at the top, on the axis); base BC has length $L=2$ cm; limb CD is vertical with the open end D at the same level as A, $H=4$ cm above the base. The tube is full of water. The clause "1600 rpm" is not needed; the question is for $N=300$ rpm, so $\omega=\dfrac{2\pi\times300}{60}=31.42$ rad/s.

 A (sealed)          D (open)
 |                    |
 | H = 4 cm           |
 B+------- L = 2 cm ---+C

Pressure at A

Pressure at D is zero. The pressure varies with radius and height as $p=\tfrac12\rho\omega^{2}r^{2}-\rho g z+\text{const}$. Between D ($r=L$, $z=H$) and A ($r=0$, $z=H$), the heights are equal, so only the rotational term acts:

$$p_A=p_D-\tfrac{1}{2}\rho\omega^{2}L^{2}$$ $$p_A=-\tfrac12(1000)(31.42)^{2}(0.02)^{2}=-197.4\ \text{Pa}$$

Pressure at A $\approx-197$ Pa (gauge), i.e. a small suction.

Water lost if A leaks

When A is opened to the atmosphere ($p_A=0$), the water surface at A (on the axis) and the surface at D (at $r=L$) must differ in level by

$$\Delta z=\frac{\omega^{2}L^{2}}{2g}=\frac{(31.42)^{2}(0.02)^{2}}{2(9.81)}=0.0201\ \text{m}=20.1\ \text{mm}$$

The surface at D cannot rise above the rim, so the water surface in AB falls by 20.1 mm and the same volume spills out at D.

Answer: $p_A\approx-197$ Pa; water lost at D equals a length of 20.1 mm of tube, i.e. volume $=\dfrac{\pi}{4}d^{2}\times0.0201\ \text{m}^3$ for tube bore $d$.