A cylindrical object with an outer diameter of 20 cm and a mass of...

When the cylindrical object is depressed by a small distance $x$, the additional buoyant force acts as the restoring force.

$F = -(\text{Area}) \times x \times \rho \times g$

$F = -(\pi r^2 \rho g) x$

Comparing this with the standard equation for simple harmonic motion $F = -kx$, the effective force constant is:

$k = \pi r^2 \rho g$

The time period of oscillation is given by:

$T = 2\pi \sqrt{\dfrac{m}{k}} = 2\pi \sqrt{\dfrac{m}{\pi r^2 \rho g}}$

Given values:
Mass $m = 2 \text{ kg}$
Radius $r = \dfrac{D}{2} = 10 \text{ cm} = 0.1 \text{ m}$
Density of water $\rho = 1000 \text{ kg/m}^3$
Acceleration due to gravity $g = 9.8 \text{ m/s}^2$

Substituting the values into the formula:

$T = 2\pi \sqrt{\dfrac{2}{\pi (0.1)^2 \times 1000 \times 9.8}}$

$T = 2\pi \sqrt{\dfrac{2}{98 \pi}} = 2\pi \sqrt{\dfrac{1}{49 \pi}}$

$T = \dfrac{2\pi}{7\sqrt{\pi}} = \dfrac{2\sqrt{\pi}}{7}$

Using $\sqrt{\pi} \approx 1.772$:

$T \approx \dfrac{2 \times 1.772}{7} \approx 0.506 \text{ s} \approx 0.5 \text{ s}$

Answer: $0.5 \text{ s}$