Final answer:
The force on the vessel's side wall is determined by the fluid's density and depth. Since oil's density is half that of water, when the vessel is filled only with oil, the force on the side wall is reduced compared to when it is filled half with water and half with oil. This reduction equates to one fifth, due to the relationship between pressure, force, and density.
Step-by-step explanation:
To show how the force on each side wall of the vessel will be reduced by one fifth when filled with oil compared to when it is filled with an equal proportion of water and oil, we need to understand how pressure and force relate in a liquid and how density affects this relationship.
Since the oil is twice lighter than water, its density (ρ) is half that of water. If we take the density of water as ρw = 1000 kg/m³, then the density of oil will be ρo = 500 kg/m³. The pressure exerted by a fluid at a given depth (h) is given by the equation P = ρgh, where g is the acceleration due to gravity. The force exerted on the side of the container is the pressure multiplied by the area of the side of the container (A): F = PA.
When the vessel is filled with equal volumes of water and oil, the average density of the combination is ρav = (ρw + ρo)/2. Therefore, the force on the wall due to the combined liquids would be Fav = ρavghA. Since the oil is half as dense as the water, the force when the vessel is filled only with oil would be Fo = ρoghA. To find the reduction in force, we can compare Fo and Fav.
On comparing the two, we will see that Fo is one fifth less than Fav, which is due to the oil's reduced density. This change in force is a direct application of an understanding of fluid pressure and density, illustrating that lower density fluids exert less force on the container's walls due to lower pressure at equal depths.