A hollow spherical ball of radius 20 cm floats in still water, with half of its volume submerged. Taking the density of water as 1000 kg/m³, and the acceleration due to gravity as 10 m/s², the natural frequency of small oscillations of the ball, normal to the water surface is _________ radians/s (roundoff to 2 decimal places).

The correct answer is 8.66.

Step-by-Step Derivation and Explanation:

1. Determine the Mass of the Floating Ball
Let $V_s$ be the total volume of the hollow spherical ball of radius $R$. The volume of a sphere is given by:
$$V_s=\frac{4}{3}\pi R^3$$

When floating in still water in equilibrium, the buoyant force ($F_B$) balances the weight of the ball ($mg$):
$$mg=F_B=\rho V_{\text{displaced}}g$$
where $\rho$ is the density of water and $V_{\text{displaced}}$ is the submerged volume.

Given that the ball floats with half of its volume submerged, we have:
$$V_{\text{displaced}}=\frac{V_s}{2}$$

Thus, we can express the mass $m$ of the ball as:
$$m=\rho \frac{V_s}{2}=\rho ·\frac{1}{2}\left(\frac{4}{3}\pi R^3\right)=\frac{2}{3}\pi \rho R^3$$

2. Set Up the Equation of Motion for Small Oscillations
When the ball is displaced vertically downward by a small distance $x$ from its equilibrium position, it experiences an additional upward buoyant force (restoring force) due to the extra displaced volume of water.

Since the displacement $x$ is extremely small, the cross-sectional area of the sphere at the water level remains approximately constant and equal to the cross-sectional area of the sphere at its center:
$$A=\pi R^2$$

The additional volume of water displaced is:
$$\Delta V=Ax=\pi R^{2}x$$

The extra buoyant force acting on the ball is:
$$F_{\text{restoring}}=-\rho g\Delta V=-\rho g\pi R^{2}x$$

Using Newton's second law ($F=m\stackrel{··}{x}$), the equation of motion is:
$$m\stackrel{··}{x}=-\rho g\pi R^{2}x$$
which can be rearranged as:
$$m\stackrel{··}{x}+\pi R^2\rho gx=0$$

3. Calculate the Natural Frequency
Substitute the expression for $m$ obtained in step 1 into the equation of motion:
$$\left(\frac{2}{3}\pi \rho R^3\right)\stackrel{··}{x}+\pi R^2\rho gx=0$$

Dividing by $\pi \rho R^2$ gives:
$$\frac{2}{3}R\stackrel{··}{x}+gx=0$$
which simplifies to standard simple harmonic motion form:
$$\stackrel{··}{x}+\frac{3g}{2R}x=0$$

The natural frequency ${\omega }_n$ is therefore:
$${\omega }_n=\sqrt{\frac{3g}{2R}}$$

Substituting the given values:
- Radius, $R=20\text{cm}=0.20\text{m}$
- Acceleration due to gravity, $g=10{\text{m/s}}^2$
$${\omega }_n=\sqrt{\frac{3\times 10}{2\times 0.20}}=\sqrt{\frac{30}{0.40}}=\sqrt{75}\approx 8.66\text{rad/s}$$