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

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 Three

We are dealing with the series of problems based on fluid statics and fluid dynamics. When a fluid is in the state of motion, it has potential energy,kinetic energy and also pressure energy. We know that the total energy of the system is constant. As per Bernoulli's theorem, the sum of potential energy,kinetic energy and pressure energy per unit mass always remains constant. We can also consider the viscous force acting opposite to the motion. It is the force that always opposes the relative motion. It depends on the area of cross section of the fluid, velocity of the fluid flow and the distance between the two layers of the fluid in the motion.

Problem

A vessel is kept on a table and filled with water. From the bottom of the vessel, a orifice is made and its location is given in the problem as shown in the diagram below. We need to measure the horizontal distance at which it is going to strike the floor.


Solution

We know that the velocity of the fluid coming out of orifice is similar to a velocity of a freely falling body and it depends on the height of the fluid above the opening. Once it is out, it is under the influence of the gravity and its horizontal distance distance can be measured with the equations of motion.There is no gravity acting on it so its velocity is uniform along horizontal direction. Further problem can be solved as shown in the diagram below.


Problem

Velocity of air flow on the upper surface of the wing of airplane is 40 meter per second and on the lower surface is 30 meter per second and its area of cross section is given to us and its mass is also given to us. We need to measure the force experienced by the wing and the problem is as shown in the diagram below.


Solution

As the wing is airplane is horizontal, there is no change in potential energy and we can remove that components of Bernoulli's theorem. Applying kinetic energy data, we can get the difference in the pressure at both the cases. We know that pressure is the force per unit area and hence we can measure the force acting on the system as shown in the diagram below.


Problem

Specific gravity of two liquids combined together when they have same mass and volume is given to us and we need to measure the density of each liquid. The problem is as shown in the diagram below.


Solution

We know that the density is the ratio of mass to the volume. We can find the effective density of the system when they have equal volume and equal mass can be found as shown in the diagram below. We have all ready derived equations for both of them and it can be done as shown below.



Problem

A sphere has known density and it is falling through a fluid of known density and we need to measure the acceleration of the body and the problem is as shown in the diagram below.


Solution

When a body is moving in a fluid, its weight acts in the downward direction and the upthrust acts in the upward direction. We need not consider viscous force as there is no data about coefficient of viscosity. We can write equation for the resultant force as the difference between weight and upthrust. We can use Newton’s second law and find as shown in the diagram below.


Problem

When a polar bear jumps on to ice block it just sinks and we need to measure the weights of that ice block and the specific gravity of ice and sea water is given to us. Problem is as shown in the diagram below.


Solution

We know that weight of the liquid displaced is the sum of weights of ice block and polar bear. We can equate the data and solve the problem as shown in the diagram below.



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Mechanical Properties of Fluids Problems and Solutions Two

Fluids are some one which can flow from one place to other and both liquids and gases falls under this category. Fluids in the state of rest and their properties were studied in fluid statics and as we deal more about water, it is also called as hydro statics. Fluids in the state of motion are studied in fluid dynamics. Apart from pressure, here we need to talk about their energies also. They have potential energy,kinetic energy and pressure energy and satisfy law of conservation of energy. That is explained in terms of Bernoullie’s theorem. If the density of the fluid is constant they are called in compressible. We can ignore their viscous property and they are then called non viscous fluids.

Problem

A horizontal pipe of non uniform area of cross section has water flows through it at a point with a speed of 2 meter per second where the pressures is 40 kilo pascal. We need to know the pressure at a point where the velocity is 3 meter per second.


Solution

As the tube is horizontal, there is no potential difference between the two points. We need to apply Bernoullie’s theorem according to which the total energy of the system that is the sum of potential energy,kinetic energy and pressure energy is the constant. If one energy increases, other energy decreases but their sum remains constant.

We can apply this with the given data barring potential energy part as it is same on both the sides and we can solve the problem as shown in the diagram below.


Problem

Two water pipes of diameters 4 and 8 centimeter are connected in series to a main pipe we need to find the ratio of velocities of the water in these two pipes and the problem is as shown in the diagram below.


Solution

We know that as the two pipes are connected in series the rate of flow of water is same in both of them and hence the equation of continuity is very well valid concept here. According to it the volume of the water passes through a given system is always constant. It is nothing but the product of area of cross section of the pipe and the velocity of the pipe is constant. Area can be written as circular and it is proportional to the square of radius or diameter. Problem can be further solved as shown in the diagram below.


Problem

Two equal  drops of water are falling through with a velocity 10 meter per second and that is constant. If these two drops are combined together and a single drop is formed, we need to find the velocity of that combined drop and the problem is as shown in the diagram below.


Solution

The constant velocity acquired by a spherical drop while passing through a medium is called terminal velocity and at that instant the resultant force acting on the system is zero. We have proved earlier in this chapter concepts that the terminal velocity is directly proportional to the square of the radius of the drop.

When two drops are combined as both of them are of same density, its total volume is the sum of the volumes of two drops. Taking that into consideration, we can find the relation between small and large radius and the problem can be solved as shown in the diagram below.


Problem

A large tank is filled up to certain height and we need to know the ratio of time taken by a small hole placed at the bottom to empty first half of the height of the water when compared with the second height of the water. The problem is as shown in the diagram below.


Solution

We can prove and we have proved while handling this chapter that time taken to empty the tank is directly proportional to the difference of height of fluid from where to where they are emptied in terms of heights of fluid under square root. In the first case, it is emptied from full to half and in the second case, it is emptied from half to zero. By writing that data, we can solve the problem as shown in the diagram below.



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Mechanical Properties of Fluids Problems and Solutions One

Fluid statics is a branch of physics that deals with the fluids in the state of rest and its properties like density,pressure and upthrust. Density is the property of a body which measures how its mass is distributed over its volume. If density is more, its more mass is concentrated in a small volume and vice-versa. Pressure is defined as the force experienced by a body per unit surface area. In this case, force is acting on the body in all directions and hence it changes the volume of the body. Every body in a fluid experience a resultant upward force and it is called upthrust or buoyant force.It depends on the volume of the fluid displaced by the body when it is immersed in the fluid, density of the fluid and acceleration due to gravity. If the body is completely immersed in the fluid, volume of the fluid displaced is equal to the volume of the body itself.

Problem

A body is floating in the water in such a way that six parts out of ten parts of its volume is inside the water. We need to measure the density of the water and the problem is as shown in the diagram below.




Solution

When a body is under water, we know that it experience upthrust. As the body is in the equilibrium state, its weight is balanced by the upthrust and they can be equated mathematically. In the place of weight, we can write the product of volume of the body with its density. Volume of the fluid displaced is six parts of volume of the body as only that much is immersed in the fluid. By applying this data, we can solve the problem as shown in the diagram below.



Problem

A brass sphere weights 100 gram weight in air. It is suspended by a thread in a liquid of known specific density. If the specific density of the brass is also known, we need to find the tension in the thread and the problem is as shown in the diagram below.



Solution

Weight is given in gram weight. We need to convert not only gram into kilogram, we also need to multiply it with acceleration due to gravity so that we are converting it into force that is measured in newton. Tension and upthrust acts in upward direction and weight of the body in the downward direction. As the system is in equilibrium position, we can equate this upward and downward force and solve the problem as shown in the diagram below.


Problem

Ninety grams of sulfuric acid is mixed with ninety grams of water and both the densities are known to us. If the specific gravity of the mixture is given to us, we need to find the loss of volume of the system due to mixing and the problem is as shown in the diagram 
below.



Solution

We need to write the total volume initially is the sum of volumes of the first and second ones and it is to found using the definition as the ratio of mass to the density of the each one as shown in the diagram below.



This much volume is supposed to be there but we can also find out the actual volume available using the data like total mass of the system and the effective density of the mixture and as shown below it is different from the supposed to be the volume measured in the first case. Hence the difference between is the loss of the volume. It is further solved as shown in the diagram below.



Problem

Weight of a solid in air, in water is given to us in the problem as shown in the diagram below and we need to measure the weight of that body when it is completely immersed in a liquid of known specific gravity.



Solution

We can find the loss of the weight of the system as the difference between weight of the body in air to the water. Loss of that weight multiplied with the density of the new system, we can find the loss of the weight of the body in that fluid. Actual mass of the body in the fluid is the difference between actual mass of the body in air to the loss of the weight in the fluid. The solution is as shown in the diagram below.


Problem

A liquid is placed in a cylindrical jar and it is rotating about its axis. If radius and angular velocity of the cylinder is know to us, we need to know the difference between the center of the liquid and rise of the liquid at the sides and the problem is as shown in the diagram below.


Solution

The kinetic energy of the fluid is converted into penitential energy and hence as per the law of conservation of energy, we can equate them. We need to replace linear velocity in terms of angular velocity and radius of the system and the problem can be solved as shown in the diagram below.





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Mechanical Properties of Solids Problems with Solutions Four

We are solving series of problems of on the topic called mechanical properties of solids. Solid material has molecules attached very close to  each other and they have strong force of attraction among them. When some external force is applied on the body, its molecules are disturbed and distance between them increases and hence body looses its shape temporarily. When the applied force is withdrawn, internal potential energy pulls the molecules back to the original position and hence body will all most recover its original shape. This property is called elasticity and it is the property of the given solid material and it depends on the nature of the material. Elastic property of the material is studied with a physical quantity called modulus of elasticity.

Problem

A rubber cord of catapult has a known cross section and length. It is stretched by  a length 12 centimeter and a stone of mass 5 gram is released from it.If young's modulus of the rubber is given to us as shown in the diagram below, we need to measure the velocity with which the stone is released from the catapult ?


Solution

We know that when rubber is pulled from its original position, we have used some of our energy and that is stored in the rubber in the form of elastic potential energy. All this energy is given to the stone and it gets kinetic energy. So we can use law of conservation of energy and solve the problem as shown in the diagram below.


Problem

A steel wire of length one meter and known area of cross section is hung from a rigid support with a stone of volume 2000 centimeter square  hanging from the other end. We need to measure the decrease in the elongation of  the wire when the body is immersed in the water.


When the wire is under the load of the sphere the force acting on the wire is nothing but weight of the wire. When that load is immersed in the water, there is a new force called upthrust and hence the resultant force acting on the wire decreases. It is the difference between the weight and the upthrust. As a result the expansion of the wire also decreases.


Problem

A metal wire of length and area of cross section known is available. If the wire breaks due to a applied load, we need to know the rise in the temperature of the system and the problem is as shown in the  diagram below.


Solution

To solve this problem, we need to equate the strain energy to the heat energy generated in the system. Heat energy is in the form of specific heat and it is from that definition itself. By equating them, the problem is solved as shown in the diagram below.


Problem

One end of a long metallic wire of known length is tied to a ceiling and the other end is tied to a mass less spring of known spring constant. If the attached load is slightly pulled down down and released  it will oscillate with a time period and we need to find that time period and the problem is as shown in the diagram below.


Solution

The wire also behaves like a spring and we can find the spring constant equivalent of the wire and we can find it using the definition of Young's modulus. The other spring and this are in series and we can find the effective spring constant of the system as shown in the diagram below. We can also write the equation for the time period as shown in the diagram below.


Problem

A thin uniform metallic rod of mass M and length L is rotated with a angular velocity in a horizontal plane about a vertical axis passing through one of its ends. The tension in the middle of the rod has to be found and the problem is  as shown in the diagram below.


Solution

Let mass per unit length m, so that we can find the total mass of the wire as shown in the diagram below. We can find the centripetal force at any given instant using its definition. To find the total value, we shall integrate the equation as shown in the diagram below.


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Mechanical Properties of Solids Problems with Solutions Two

Mechanical Properties of Solids Problems with Solutions Three