A metal ball with a volume of V = 200 cm³ suspended on a thread was lowered into a cylindrical vessel

A metal ball with a volume of V = 200 cm³ suspended on a thread was lowered into a cylindrical vessel with water. Determine the change in water pressure at the bottom of the vessel if the bottom area S = 50.0 cm².

If a metal ball with a volume of ΔV = 200 cm³ = 0.0002 m³ is lowered into a cylindrical vessel with water volume V₀, a metal ball with a volume of ΔV = 200 cm³ = 0.0002 m³, then the liquid level h₀ will increase by Δh and become h = h₀ + Δh, since the liquid with the ball will have a volume V = V₀ + ΔV. In this case, Δh = ΔV / S, where S = 50.0 cm² = 0.005 m² is the bottom area of ​​a cylindrical vessel.
To determine the change in water pressure at the bottom of the vessel, we use the formula to determine the weight pressure of the liquid at the bottom: Δр = ρ g Δh or Δр = ρ g ΔV / S, where the coefficient g = 9.8 N / kg, and the values we find the density of water in the reference tables: ρ = 1000 kg / m³.
Substitute the values ​​of physical quantities into the calculation formula and perform the calculations:
Δp = 1000 kg / m³ · 9.8 N / kg · 0.0002 m³ / 0.005 m²;
Δp = 392 Pa.
Answer: the water pressure at the bottom of the vessel will increase by 392 Pa.