Mechanical Properties of Fluids Problems and Solutions Four

We are solving series of problems on fluid dynamics. Viscous force acts on all bodies that have relative motion. If a spherical ball is falling through a fluid, it experience a viscous force and it can be measured using a law called Stroke’s law. According to this rule, viscous force is directly proportional to the square of the radius of the ball. There is weight of the ball acting in the downward direction and upthrust acting always in the upward direction. Viscous force always acts opposite to the relative motion of the body. After some travel, the resultant force acting on the body becomes zero and the body acquires a constant velocity and that velocity is called as terminal velocity. 

Problem

An open U tube contains mercury. When certain amount of water of known height is filled in one tube over mercury, we would like to measure the difference between the liquid levels in both the sides and the problem is as shown in the diagram below.


Solution

We know that at the bottom of the system both the pipes has one common point whose height is same and hence the pressure is same at both the points. We know that the pressure can be expressed as the product of height of the liquid, density of the liquid and acceleration due to gravity at the given place. Thus by equating the pressure at both the points we can solve the problem as shown in the diagram below.


Problem

A sphere of known radius is having a cavity of half the radius of the sphere. It is found that the sphere is just floating in the water with its highest point in touch with the water. We need to find the specific gravity of the material and the problem is as shown in the diagram below.


Solution

We know that when the sphere is just floating, its weight is balanced by the upthrust acting on it. We know that upthrust is the product of volume of the fluid displaced, density of the fluid and acceleration due to gravity. As the body is completely immersed in the fluid, volume of the fluid displaced is equal to the volume of the body itself. We know the values of density of water and acceleration due to gravity.

Weight acting in the downward direction is only due to the mass present in the system. We can write mass as volume and density product of the body. Volume of the content means we shall count only the part of mass present in the system. Thus we can measure the weight and equate it ot the upthrust as shown in the diagram below.


Problem

The upthrust force acting on the wing of aeroplane is given to us. Velocity of the air on its lower surface and area of cross section is also given to us and we need to find the velocity of air on its upper surface ad the problem is as shown in the diagram below.


Solution

We can assume that the wing is horizontal and hence there is no difference in the gravitational energy. As velocity is different kinetic energy per unit mass and hence pressure energy are going to be different. We can apply Bernoullie’s theorem, we can solve the problem as shown in the diagram below.


Problem

Due to a disease main artery expanded in its area of cross section as per the given data in the problem. Velocity of the blood in non expanded area is given to us and blood density is also given to us. We need to find the excess pressure developed due to this and the problem is as shown in the diagram below.
  

Solution

We can apply equation of continuity and velocity of the other part of artery as shown in the diagram below. Now knowing the velocity, we can apply conservation of energy concept and solve the problem as shown in the diagram below.


Problem

The coefficients of viscosity of two liquids is given to us as 2:3 densities of the fluids is given to us as 4:5. In the equal time we need to know the rate of fluid flow in the tubes and the problem is as shown in the diagram below.


Solution

We need to use a concept called Poisellie’s equation. As per it we can write the equation for the rate of flow as shown in the diagram below and solve the problem. Length and diameter of tubes is given as same and the problem is solved as shown in the diagram below.



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