Question

a ball of diameter 10 cm and mass 10 grams is dropped in a container of water. the cross sectional area of the container is 100 cm2.. what is the change in the height of the water column

Answer 1

Answer:

h = 9.83 cm

Explanation:

Let's analyze this interesting exercise a bit, let's start by comparing the density of the ball with that of water

       

let's reduce the magnitudes to the SI system

         r = 10 cm = 0.10 m

         m = 10 g = 0.010 kg

         A = 100 cm² = 0.01 m²

the definition of density is

          ρ = m / V

the volume of a sphere

         V = $\frac{4}{3} \ \pi r^{3}$

          V = $\frac{4}{3}$ π 0.1³

          V = 4.189 10⁻³ m³

let's calculate the density of the ball

           ρ = $\frac{0.010}{4.189 \ 10^{-3} }$

           ρ = 2.387 kg / m³

the tabulated density of water is

         ρ_water = 997 kg / m³

we can see that the density of the body is less than the density of water. Consequently the body floats in the water, therefore the water level that rises corresponds to the submerged part of the body. Let's write the equilibrium equation

            B - W = 0

            B = W

             

where B is the thrust that is given by Archimedes' principle

           ρ_liquid  g V_submerged = m g

           V_submerged = m / ρ_liquid

we calculate

            V _submerged = 0.10 9.8 / 997

             V_submerged = 9.83 10⁻⁴ m³

The volume increassed of the water container

           V = A h

            h = V / A

let's calculate

            h = 9.83 10⁻⁴ / 0.01

            h = 0.0983  m

this is equal to h = 9.83 cm