Fluid Mechanics

The original weight Weight of cube:$ \rho_b $

Cube A Floating Alone:Cube A floats with a fraction of its volume immersed in water.Let the length of the cube be , giving it a volume .The weight of cube A is .The buoyant force acting on cube A equals the weight of the displaced water: .At equilibrium, the weight of cube A equals the buoyant force: Simplifying, we find: Cube B Placed on Top of Cube A:Let the density of cube B be .The total weight of the system (both cubes A and B) must be balanced by the buoyant force when cube A is fully submerged.The total weight is .Since , we can write: The buoyant force when cube A is fully submerged is .At equilibrium, the total weight equals the buoyant force: Dividing both sides by , we obtain: Combining the Equations:From the first equation, .Substituting into the second equation gives: Simplifying, we solve for : To find the ratio : Therefore, the ratio is:

Given:Density of water: Density of the ball: Buoyant Force:According to Archimedes’ principle, the buoyant force is the weight of the displaced water: Weight of the Ball:The weight of the ball is: Net Force on the Ball:The net force acting on the ball when submerged in water is the difference between its weight and the buoyant force: Effective Acceleration:The effective acceleration is the net force per unit mass: Initial Velocity :Using the standard result for vertical circular motion, where the velocity is given by: Substituting and :