$\omega = 90(\pi/30) = 3\pi ~ \text{rad/sec}$
$V_{tank} = \frac{1}{4}\pi D^2 H = \frac{1}{4}\pi (1.3^2)(2.1)$
$V_{tank} = 2.7874 ~ \text{m}^2$
$V_{w1} = \frac{1}{4}\pi D^2 d = \frac{1}{4}\pi (1.3^2)(1.4)$
$V_{w1} = 1.8582 ~ \text{m}^3$
$h = \dfrac{\omega^2 r^2}{2g} = \dfrac{(3\pi)^2 (0.65^2)}{2(9.81)}$
$h = 1.9128 ~ \text{m}$ ← [ B ] answer for part 1
$h/2 = 0.9564 ~ \text{m}$
Since h/2 > 0.7 m, water are spilled from the tank
$V_{air} = \frac{1}{2}\pi r^2 h = \frac{1}{2}\pi (0.65^2)(1.9128)$
$V_{air} = 1.2694 ~ \text{m}^3$
$V_{w2} = V_{tank} - V_{air} = 2.7874 - 1.2694$
$V_{w2} = 1.518 ~ \text{m}^3$
$V_{spilled} = V_{w1} - V_{w2} = 1.8582 - 1.518$
$V_{spilled} = 0.3402 ~ \text{m}^3 = 340.2 ~ \text{Liters}$ ← [ C ] answer for part 2
So that no water is spilled out
$H = 1.4 + h/2 = 1.4 + 0.9564$
$H = 2.3564 ~ \text{m}$ ← [ D ] answer for part 3
