The density of the metal used to make the cylinder is greater than the density of seawater.Explain why the cylinder floats.(a)...................................................................................................................................................................................................................................................................................... [1]The cylinder has a length of 1.8 m. It floats with 1.2 m submerged in the sea. The bottom ofthe cylinder has an area of cross-section of 0.80 m2.The density of seawater is 1020 kg / m3.Calculate the force exerted on the bottom of the cylinder due to the depth of the seawater.(b)force = .............................................. [4]Deduce the weight of the cylinder. Explain your answer.(c)weight = ............................................................................................................................explanation .................................................................................................................................................................................................................................................................. [2][Total: 7]All the sides of a plastic cube are 8.0 cm long. The diagram shows the cube, (not to scale).2114Downloaded by Saqib Sabir ([email protected])lOMoARcPSD|4272195621

(a) The cylinder floats because the overall density of the cylinder is less than the density of the seawater. Even though the metal itself has a greater density, the cylinder is hollow, meaning it contains air. The combined density of the metal and the air inside the cylinder is less than the density of the seawater, allowing it to float.

(b) The force exerted on the bottom of the cylinder due to the depth of the seawater can be calculated using the formula for pressure in a fluid column: P = ρgh, where ρ is the fluid density, g is the acceleration due to gravity, and h is the height of the fluid column. The force is then given by F = PA, where A is the area of the bottom of the cylinder.

So, first calculate the pressure: P = ρgh = 1020 kg/m³ * 9.8 m/s² * 1.2 m = 11983.2 Pa.

Then calculate the force: F = PA = 11983.2 Pa * 0.80 m² = 9586.56 N.

(c) The weight of the cylinder can be deduced from the fact that it is in equilibrium, i.e., the weight is balanced by the buoyant force. The buoyant force is equal to the weight of the displaced fluid, which is the seawater in this case. The volume of the displaced seawater is equal to the volume of the submerged part of the cylinder, which is 1.2 m * 0.80 m² = 0.96 m³. The weight of the displaced seawater is then its volume times its density times g, which is 0.96 m³ * 1020 kg/m³ * 9.8 m/s² = 9580.8 N. This is approximately equal to the force calculated in part (b), so the weight of the cylinder is also approximately 9580.8 N.

The slight discrepancy between the calculated force and the weight is due to the approximation in the value of g and the rounding in the calculations.