A test tube containing mercury oscillates vertically in water. Calculate the frequency of oscillation if the test tube has a mass of 0.006 kg and the mercury has a mass of 0.01 kg. The external diamet

The total mass $M$ is the sum of the mass of the test tube and the mass of the mercury.

$$M = m_{tube} + m_{mercury}$$ $$M = 0.006 \text{ kg} + 0.01 \text{ kg}$$ $$M = 0.016 \text{ kg}$$

The external diameter $D$ is $0.02 \text{ m}$, so the radius $r$ is $D/2$.

$$r = \frac{0.02 \text{ m}}{2} = 0.01 \text{ m}$$

The cross-sectional area $A$ is given by the formula for the area of a circle.

$$A = \pi r^2$$ $$A = \pi (0.01 \text{ m})^2$$ $$A = 0.0001\pi \text{ m}^2$$

When the test tube is depressed by a small amount $x$, an additional volume $V_{additional} = A x$ of water is displaced.
The additional buoyant force $F_{buoyant, additional}$ acts as a restoring force.

$$F_{buoyant, additional} = \rho_{water} V_{additional} g$$ $$F_{buoyant, additional} = \rho_{water} (A x) g$$

This restoring force is proportional to the displacement $x$, which is characteristic of Simple Harmonic Motion (SHM). Comparing this to $F = -kx$, the effective spring constant $k$ is:

$$k = \rho_{water} A g$$

Using the density of water $\rho_{water} = 1000 \text{ kg/m}^3$ and acceleration due to gravity $g = 9.81 \text{ m/s}^2$:

$$k = (1000 \text{ kg/m}^3) (0.0001\pi \text{ m}^2) (9.81 \text{ m/s}^2)$$ $$k = 0.1\pi \times 9.81 \text{ N/m}$$ $$k = 0.981\pi \text{ N/m}$$

For a system undergoing SHM, the period $T$ is given by:

$$T = 2\pi \sqrt{\frac{M}{k}}$$

Substitute the values of $M$ and $k$:

$$T = 2\pi \sqrt{\frac{0.016 \text{ kg}}{0.981\pi \text{ N/m}}}$$ $$T = 2\pi \sqrt{\frac{0.016}{0.981\pi}}$$ $$T = 2\pi \sqrt{\frac{0.016}{3.0827}}$$ $$T = 2\pi \sqrt{0.005189}$$ $$T = 2\pi \times 0.07203$$ $$T = 0.4525 \text{ s}$$

The period of its oscillation is $\text{0.453 s}$.