Showing posts with label Potential Energy. Show all posts
Showing posts with label Potential Energy. Show all posts

Surface Tension Problems with Solutions Three

We are solving series of problems on the property of a liquid called surface tension. This is due to molecular force of attraction among the liquid molecules. If the molecules belongs to same liquid the force is called cohesive force of attraction and if they are due to different liquid molecules then it is called adhesive force of attraction. Which force among this two is dominating depends on the nature of that liquids and the body in contact. Water sticks to our body as adhesive force is dominating and mercury won’t as cohesive force is dominating. This can be even understood in terms of angle of contact and capillary rise of the liquid when a thin tube is placed in it. Surface tension depends on the nature of the liquid as well as temperature of the liquid. If temperature is raised, molecular force of attraction decreases and hence surface tension also decreases.

Problem

A liquid of known density and surface tension is placed in a capillary tube and it has raised to a certain height. We need to find the potential energy of the system and the problem is as shown in the diagram below.



Solution

We know the formula for the gravitational potential energy in terms of the height of the liquid but it is spread over the entire length and not focused at the top. Thus we shall consider the concept of center of mass ans we can assume that the entire mass is consecrated at the center of mass of the liquid which is at geometrically half of the total height. By subsisting the value of the capillary rise in terms of the surface tension, we can solve the problem as shown in the diagram below.


Problem

Several liquid drops of known radius and density are combined to foam a big drop of known radius. If all the energy released in this process is converted into kinetic energy, we need to find the velocity acquired by the drop.


Solution

Energy released can be expressed as the difference between work done for both the radius basing on the definition that surface tension is work done per unit change in the area of cross section of the liquid. We shall equate it to the kinetic energy where mass can be expressed as the product of density and volume. Problem can be further simplified as shown in the diagram below.


Problem

A glass rod of known radius is immersed into a capillary tube of known radius as shown in the problem below. If the arrangement is immersed in the water, we need to find the rise in the liquid level.


Solution

Liquid will rise in the system until the force in the upward direction due to surface tension is balanced by the weight of the raised liquid in the capillary tube system. By writing the mass as the product of volume and density and further volume as the product of area and length, problem can be solved as shown in the diagram below.


Problem

Under isothermal conditions two bubbles of known radius are combined to foam a single drop of know radius. If the external pressure is given to us, we need to measure the surface tension of the system and the problem is as shown in the diagram below.


Solution

We can apply Boyle’s law as the temperature of the system is constant as shown in the diagram below. By simplifying it further we can solve the problem as shown here.


Problem

A soap bubble is blown at the end of very narrow tube as shown in the problem below. Air flows into the tube with a known velocity and it comes to rest inside it. After some time the bubble get some radius and comes out of the tube. We need to find the radius of the bubble so that the air strikes the bubble surface normally.


Solution

We need to equate the force due to pressure to the force due to surface tension and the problem can be solved as shown in the diagram below.




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Surface Tension Problems with Solutions Two

We are solving series of problems based on the concept surface tension. It is the property of the liquid because of which the liquid surface behaves like a stretched elastic membrane. Because of surface tension, liquids always try to acquire minimum surface area. This is the reason why water drops are spherical in shape as the sphere has minimum surface area among all possible three dimensional shapes. Angle of contact is a parameter that measures weather cohesive or adhesive forces are dominating in a given system. If adhesive forces are dominating, angle of contact is less than ninety  degree and vice versa.Angle of contact is the angle drawn between two tangents drawn at the point of contact where one tangent is drawn to the liquid surface and the other tangent is drawn to the wall of the capillary tube into the liquid. Basing on that there will be capillarity rise or fall.


Problem

When a capillary tube is immersed in water the mass of water that raised in the tube is of 5 gram. If the radius of the tube is doubled we need to find the raise of the water in the new tube and the problem is as shown in the diagram below.


Solution

We know that when a capillary tube is placed in a liquid, a component of surface tension generates a force that pulls the liquid upward. Simentaniouly we have liquids weight acting in the downward direction. When this downward force compensate the upward force due to surface tension, liquid stops rising further. By equating this two forces, we can solve the problem as shown in the diagram below.


Problem

We need to find the excess pressure inside an air bubble when the bubble radius is known to us and surface tension of the liquid is also given to us. Problem is as shown in the diagram below.


Solution

When the drop is acquiring spherical shape, there is pressure acting towards its center and it generate excess pressure inside the spherical bubble. Basing on the formula that we have derived, we can solve the problem as shown in the diagram below.



Problem

Surface area and surface tension of the liquid is given to us in the problem as shown in the diagram below. We need to find the excess pressure inside the drop.


Solution

We know the surface area of the sphere and hence we can express the radius of the drop in terms of area of cross section as shown in the diagram below. Substituting that data in the excess pressure formula, we can solve the problem as shown in the diagram below.


Problem

Two soap bubbles are combined to form a single bubble. In this process change in volume and area is given to us in the problem. Pressure and surface tension are given to us and we need to find the relation between them.


Solution

We know that under isothermal conditions, Boyle's law is valid and we can equate product of pressure and volume is conserved as shown in the diagram below. By applying the formula for the pressure and the volume of the sphere we can solve the problem as shown in the diagram below.


Problem

A thread of length L is placed on a soap film of known surface tension. If the film is pierced with a needle we need to measure the tension in that thread and the problem is as shown in the diagram below.


Solution

We can equate the force due to surface tension to the force and centripetal force and solve the problem as shown in the diagram below.



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Surface Tension Problems with Solutions One

We are going to solve series of problems with detailed solutions about a topic called surface tension. Surface tension is the property of liquid due to which the liquid surface experiences a tension and they tend to acquire minimum surface area. It is because of this surface tension, small insects are able to float on the surface of water. It is defined as the force acting on the tangential surface of the liquid normal to the surface of contact per unit length. Surface tension can be explained basing on molecular theory. Every molecule can influence the surrounding and attract the neighboring molecules up to some extend and that distance is called molecular range. Taking the molecule as the center, molecular range as the radius, if we draw a sphere, it is called sphere of influence and within the sphere of influence, core molecule can attract the other molecules.

Problem

The length and thickness of a glass plate is given to us as shown in the diagram below. If this edge is in contact with a liquid of known surface tension, we need to know the force acting on the glass plate due to the surface tension of the liquid.



Solution

We know that surface tension is mathematically force acting on it per unit length. Here length means the length of free surface of the body that is in contact with the liquid. The glass plates both inner and outer surface are in contact with the liquid and hence two lengths has to be taken into count. The problem is solved as shown in the diagram below.



Problem

A drop of water of known volume is pressed between two glass plates so as to spread across a known area. If surface tension of the liquid is known to us, we need to know the force required separating the glass plate and the problem is as shown in the diagram below.


Solution

We can write the volume as the product area of cross section with the length of the liquid. We also know that the surface tension can also be expressed in terms of work done per unit area. Intern work done can be expressed as the product of force and displacement. Taking this into consideration, we can solve the problem as shown in the diagram below.


Problem

A big liquid drop of known radius splits into identical drops of same size in large number and we don’t know the radius of the small drop. We need to measure the work done in this process and the problem is as shown in the diagram below.


Solution

We know that the volume of the liquid is conserved. It means the volume of the big drop is the sum of the volumes of all small drops together and basing on that we can find the relation between smaller and bigger radius as shown in the diagram below. We can write the equation for the work done as the product of surface tension and the change in the area. 


Problem

Work done in blowing a soap bubble of radius R is given to us as W. We need to measure the work done in blowing the same bubble to a different radius and the problem is as shown in the diagram below.


Solution

As discussed in the previous problem, we can define the work done as the product of surface tension and the change in the area of cross section. By applying that data, we can solve problem as shown in the diagram below.




Problem

We need to find the capillary rise of a liquid in a capillary tube when it is dipped in that liquid where surface tension and density of the liquid is given to us. We can treat angle of contact as zero and the problem is as shown in the diagram below.


Solution

We know that when angle of contact is less than ninety degree, the liquid raises above the normal level of the beaker and that property is called capillarity. The capillary rise depends on the radius of the tube, density and surface tension of the liquid. We can apply the formula and solve the problem as shown in the diagram below.



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Mechanical Properties of Fluids Problems and Solutions Five

We are solving series of problems in Mechanical properties of Fluids.Viscosity is the property of the fluid due to which its relative motion is opposed. When a fluid is moving, it moves like different set of layers, each layer moving in the forward direction opposed the lower layers motion in the forward direction and thus viscous force is generated. It is directly proportional to the area of cross section of the fluid, velocity of the fluid flow and it is inversely proportional to the width of the layer. We can find coefficient of viscosity in this way. When a spherical body is moving in a fluid due to opposing viscous force and upward upthrust and weight in downward direction, some where the resultant force becomes zero and the body acquires a constant velocity called terminal velocity.

Problem

A plate of area 100 square centimeter is placed on the upper surface of castor oil having only 2 mm thickness. Coefficient of thickness is given to us and we need to measure the horizontal force required to move the plate with a certain velocity and the problem is as shown in the diagram below.


Solution

We can solve the problem using the very definition of coefficient of viscosity as shown in the diagram below. We know that the viscous force is directly proportional to the area of cross section,velocity of fluid flow and inversely proportional to the distance between the layers. Solution is as shown in the diagram below.


Problem

A vessel has a height of 40 meter. It has three horizontal tubes each of same diameter and length at different heights from the base as shown in the problem below. We need to find out the length of the single pipe of the same diameter that has to be replaced instead of three pipes so that fluid flow is same.


Solution

The sum of rate of flow through each pipe has to be the fluid flow in the single pipe. We can use Poisellie’s equation and find the length of the new pipe as shown in the diagram below.


Problem

A cylindrical tank has a hole of known area at the bottom. If the water is allowed to flow into the tank from a tube above it with a known rate, we need to find the maximum height of the fluid in the cylinder. Problem is as shown in the diagram below.


Solution

At the maximum possible height of the liquid in the cylinder, the rate of fluid in is equal to the rate of the fluid out. Rate of fluid out can be written in terms of equation of continuity and we also know that the velocity of the fluid coming out of the hole is similar to velocity of the freely falling body. Taking this into consideration, we can solve the problem as shown in the diagram below.



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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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