A hollow cylinder with closed ends is \(300 \mathrm{~mm}\) diameter and \(450 \mathrm{~mm}\) high, has a mass of \(27 \mathrm{~kg}\) and has a small hole in the base. It is lowered into water so that its axis remains vertical. Calculate the depth to which it will sink, the height to which the water will rise inside it and the air pressure inside it. Disregard the effect of the thickness of the walls but assume that it is uniform and that the compression of the air is isothermal. (Atmospheric pressure \(=101.3 \mathrm{kPa}\) )

Buoyancy is a force that acts upwards on an object submerged in a fluid, counteracting the gravitational force pulling the object downwards.
It's the reason why objects float or sink in water. The buoyant force on an object is equal to the weight of the fluid displaced by the object.
For our hollow cylinder, this buoyant force is given by the displaced volume of water multiplied by the density of water.
\[ F_b = \rho_{water} \times V_{displaced} \times g \] where:

• \( F_b \) is the buoyant force
• \( \rho_{water} \) is the density of water (approximately 1000 kg/m³)
• \( V_{displaced} \) is the volume of water displaced
• \( g \) is the acceleration due to gravity (9.81 m/s²)

Understanding this concept helps us determine how deep the cylinder will sink in the water.

Isothermal compression occurs when the temperature of the gas remains constant while it is being compressed.
In our problem, the air inside the cylinder gets compressed as water enters. The pressure and volume relationship in isothermal processes is given by Boyle's law:
\[ P_1 V_1 = P_2 V_2 \] where:

• \( P_1 \) is the initial pressure (atmospheric pressure)
• \( V_1 \) is the initial volume of air
• \( P_2 \) is the final pressure
• \( V_2 \) is the final volume of air

Since the temperature remains constant, we use this relationship to determine the new pressure inside the cylinder after water enters it.

To find the depth to which the cylinder sinks, we look at the equilibrium of forces.
At equilibrium, the gravitational force (weight of the cylinder) is balanced by the buoyant force from the displaced water.
Mathematically, this balance can be written as:
\[ m \times g = \rho_{water} \times V_{displaced} \times g \] where:

• \( m \) is the mass of the cylinder (27 kg)
• \( g \) is the acceleration due to gravity (9.81 m/s²)
• \( \rho_{water} \) is the density of water
• \( V_{displaced} \) is the volume of water displaced.

By finding this balance, we can solve for the depth \( h \) to which the cylinder sinks.

To calculate the volume of the hollow cylinder, we start with the geometry of a cylinder.
The volume \( V \) of a cylinder is given by:
\[ V = \text{base area} \times \text{height} \]
For a cylinder with diameter \( d \) and height \( h \):
\[ \text{base area} = \pi \times \left( \frac{d}{2} \right)^2 \]
Plugging in the given diameter of 300 mm (or 0.3 m):
\[ \text{base area} = \pi \times \left( \frac{0.3}{2} \right)^2 \]
\[ = \pi \times 0.0225 \text{ m}^2 \]
From there, you can determine the volume of water displaced by accounting for the height the cylinder sinks.
For our provided problem, the volume of the water displaced \( V_{displaced} \) can be found by the height to which the cylinder sinks multiplied by the base area.

The pressure inside the cylinder changes as water rises in the cylinder due to the hole at the bottom.
Remember the initial pressure of the air inside the cylinder equals atmospheric pressure (101.3 kPa).
The formula for pressure during isothermal compression is:
\[ P_1 V_1 = P_2 V_2 \]
Initially, the volume \( V_1 \) is the initial height of the air column times the base area.
After water rises to a height \( x \), the new air volume \( V_2 \) is the compressed height times base area.
So, the new pressure \( P_2 \) can be found by rearranging the formula:
\[ P_2 = \frac{P_1 V_1}{V_2} \]
Where:

• \( P_1 \) = 101.3 kPa
• \( V_1 \) is the initial air volume
• \( V_2 \) is the final air volume.

This final air pressure accounts for the reduced volume due to the water entering the cylinder.