1. Given data

• Volume of water in bottle: $V = 310\,\text{ml} = 310\times 10^{-6}\,\text{m}^3 = 3.10\times 10^{-4}\,\text{m}^3$
• Radius of bottle: $r = 2.5\,\text{cm} = 2.5\times 10^{-2}\,\text{m}$
• Density of water: $\rho = 10^3\,\text{kg m}^{-3}$

Since the bottle has negligible mass, the oscillating mass is only the water inside it.

Thus, $m = \rho V = 10^3 \times 3.10\times 10^{-4} = 0.31\,\text{kg}$

2. Restoring force when bottle is pushed down by $x$

If the bottle is pushed downward by a small distance $x$, the extra volume submerged is $\Delta V = A x$ where $A$ is the cross-sectional area of the cylindrical bottle.

So the increase in buoyant force is $\Delta F = \rho g \Delta V = \rho g A x$

This acts upward, so the restoring force is $F = -\rho g A x$

This is of the form $F=-kx$, hence $k = \rho g A$

3. Area of cross-section

$A = \pi r^2 = \pi (2.5\times 10^{-2})^2$ $A = \pi \times 6.25\times 10^{-4} \approx 1.963\times 10^{-3}\,\text{m}^2$

4. Angular frequency of SHM

For SHM, $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{\rho g A}{m}}$

Substitute values: $\omega = \sqrt{\frac{(10^3)(9.8)(1.963\times 10^{-3})}{0.31}}$

First calculate numerator: $10^3 \times 9.8 \times 1.963\times 10^{-3} \approx 19.24$

Then, $\frac{19.24}{0.31} \approx 62.06$

Hence, $\omega = \sqrt{62.06} \approx 7.88\,\text{rad s}^{-1}$

So, $\omega \approx 7.9\,\text{rad s}^{-1}$

5. Checking options

• A: $2.50\,\text{rad s}^{-1}$
• B: $3.75\,\text{rad s}^{-1}$
• C: $5.00\,\text{rad s}^{-1}$
• D: $7.90\,\text{rad s}^{-1}$

Therefore, the correct option is D.