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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