Question

A cylindrical tank containing water is rotating about the z-axis at a constant angular velocity of 10 rad s$^{-1}$. The schematic of the isobaric surface is shown below, where ‘A’ and ‘B’ are two points on the isobaric surface at heights $z_1$ and $z_2$, respectively. Assuming negligible atmospheric pressure and no transient flow, estimate the elevation difference in m between $z_1$ and $z_2$. (Given: $r_1 = 0.5 \, m$, $r_2 = 1.0 \, m$, $g = 9.8 \, m/s^2$, $\omega = 10 \, rad/s$).

Hint:

In rotating fluids, free surface elevation varies quadratically with radius: $z \propto r^2$.

Answer 1

Correct Answer: 3.82

Step 1: Equation for free surface of rotating liquid.
The surface shape is given by: \[ z(r) = \frac{\omega^2 r^2}{2g} + C \] where $\omega$ = angular velocity, $r$ = radial distance.
Step 2: Elevation difference between $r_1$ and $r_2$.
\[ \Delta z = z_2 - z_1 = \frac{\omega^2}{2g} (r_2^2 - r_1^2) \]
Step 3: Substitute values.
\[ \Delta z = \frac{10^2}{2 \times 9.8} \left(1.0^2 - 0.5^2\right) \] \[ = \frac{100}{19.6} \times (1 - 0.25) = 5.102 \times 0.75 = 3.83 \, m \] Final Answer: \[ \boxed{3.83 \, m} \]