A solid sphere made of plastic density of 1350 kg/m3 has a radius of 35.0 cm. It is suspended by a massless cord. 75% of its volume is in water and 25% of it is in oil. The density of water is 1000 kg/m3 and the density of the oil is 850 kg/m3. What is the total buoyant foce on the sphere

B_total = 1694 N ,  T = - 937.7 N

Explanation:

This is an exercise of the Archimedes principle that establishes that the thrust of a liquid is equal to the weight of the dislodged volume

          B = ρ g V_body

For this case we will assume that the weight of the body is in equilibrium with the thrust of the liquids and the tension of the rope

          B₁ + B₂ - W + T = 0

suppose liquid 1 is water and liquid 2 is oil

          ρ₁ g V₁ + ρ₂ g V₂ = W

for body weight let's use the definition of density

         ρ = m / V

         m = ρ V

we substitute

         T = ρ g V - (ρ₁ g V₁ + ρ₂ g V₂)

In the problem we are told that the volume in the water is 75% and the volume in the oil is 25% of the body's volume.

         T = ρ g V - g (ρ₁ 0.75 V + ρ₂ 0.25 V)

let's calculate the volumes

Body

         V = 4/3 π r³

liquid 1 (water)

         V₁ = 0.75 V = ¾ V

liquid 2 (oil)

         V₂ = 0.25 V = ¼ V

we substitute

        T = ρ g π r³ - g (ρ₁   π r³ + ρ₂   π r³)

        T = ρ g r³ - g π r³ (ρ₁ + ρ₂)

The term of the floating force is

        B_total = g π r³ (ρ₁ + ρ₂)

let's calculate its value

        B_total = 9.8 π 0.35³ (1000 + ⅓ 850)

        B_total = 1694 N

therefore the tension in the string is

         T = 1350 9.8 0.35³ - 1694

         T = 756.3 - 1694

         T = - 937.7 N

The negative sign indicates that the system tends to come out of the liquid, for which a downward force must be applied to keep it in position.

molecules are free to separate from each other and will take on the shape of the container. molecules move completely at random and high speeds. they vibrate in a fixed position and do not move around much. molecules expand to the edge of their container.