Showing posts with label units. Show all posts
Showing posts with label units. Show all posts

Uses and Applications of Dimensional Analysis Video Lesson

Dimensional formula is a representation of a physical quantity in terms of fundamental quantities. We represent mass with M, length with L and time with T in dimensional formula. Dimensional analysis can be done basing on principle of homogeneity and according to it the dimensions of left hand side and right hand side of the equation has to be equal. It mean to tell us that we can add or subtract only similar physical quantities but not dissimilar ones. Using the dimensional formula and analysis, we can convert a physical quantity from one system of unit to other. 

 Conversion of physical Quantity from one system to other

For example we can that the energy is measured in the unit joule in SI system and erg in CGS system. Because they are the units of same physical quantity in different system of units, they shall be having some relation between them. We can find that relation using the principle of homogeneity.




 Checking the correctness of given equation

 We can also use the dimensional analysis to check the correctness of a given equation. We have so many equations in physics and before they are correct as per the subject, they have to be correct conceptually. That can be verified using the dimensional analysis. Here we are depending on the concept of principle of homogeneity and according to it LHS and RHS of a equation shall have same dimensions of the physical quantities. If the sum of left hand side physical quantities is velocity then the right hand side sum or difference of the terms shall also be the velocity. In the other sense, we are adding or subtracting physical quantities of the same nature but not different.





 Finding Relation among Physical Quality

 We can also use dimensional analysis to find the relation between physical quantities basing on principle of homogeneity. We will simple equate the dimensions of left hand side of the equation with the right hand side equation so that we will be getting mathematical equations and by solving them, we can get the dimensions of the physical quantities. Here we will be able to find only the relation but we can not find the proportional constant values using this method. This is one limitation of the dimensional analysis.




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Principle of Homogeneity and Limitations of Dimensional Analysis Video Lesson

Dimensional formula is a representation of a physical quantity in terms of fundamental quantities. When we write a physics equation, it shall be in such a way that the dimensions of left hand side of the equation has to be equal to the dimensions of all physical quantities along the right hand side of the equation. This is possible only when you add or subtract similar quantities at either left or right side of the equation. 

It is simple understanding that velocity can be added only with velocity to get another velocity We can not add velocity with force and the summation can not give either velocity or force and the summation is meaning less.This principle is called principle of homogeneity. If any equation is not satisfying the law, then it cannot be correct with respect to physics. But we need to be careful that the dimensionally correct equation can not be correct with respect to physics. Being satisfying the principle of homogeneity is the fundamental condition to be accepted as a physics equation.



Limitations of Dimensional Analysis

 Dimensional formulas helps us to understand the relation among the physical quantities and it helps us also in converting the physical quantity from one system of unit to other system of unit. But they have certain limitations. Trigonometric and exponential functions won’t have any dimensions and if they are involved in any equation, we can not solve them basing on dimensional analysis. If the equation on the right hand side is depending on more fundamental quantities than the left hand side, then we cannot solve the equation basing on dimensional analysis. We also won’t be able to find out the proportionality constants of a science equation using the dimensional analysis.





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Units and Dimensions an Introduction Video Lesson

We are dealing with the basics of concepts units and dimensions in a video lesson. Physical quantity is a way of understanding nature and its physics applications and to measure them, we use units.

Introduction to Units

Physical quantities are helpful in understanding the nature in terms of measurement. We try to understand the world around us with minimum physical quantities and they are the alphabets and words of the physics language. If a physical quantity is independent of any other physical quantities, they are called fundamental physical quantities and they are the irreducible set of quantities to represent other things in the nature around. Length, mass and time are basic examples of fundamental physical quantities.

Other physical quantities are derived basing on this fundamental quantities and they are called derived physical quantities. Velocity, acceleration and force are the some of the examples of derived physical quantities. To measure this physical quantities, we need to use units. Unit is a way of measuring the physical quantities in a standard way that is acceptable to all.

As physical quantities are of two types, the units are also two types and they are fundamental and derived units. To measure the fundamental physical quantities, we have fundamental units with different kinds of systems like FPS,CGS,MKS and SI system of units. We can further define and write units for the derived quantities based on the the fundamental system of units. At present we are using SI system as the standard system to measure the fundamental physical quantities. 



Introduction to Dimensions

Each physical quantity is either a fundamental or derived physical quantity. Derived physical quantity is obtained from fundamental quantity by manipulating the fundamental quantities as per the requirement of the concepts and applications. The representation of a physical quantity in terms of fundamental quantities is called dimensional formula and the powers to which the fundamental quantities are raised are called dimensions. Thus using the dimensional formula, we will be knowing the content of fundamental quantities in it.



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Units and Dimensions Problems and Solutions Six

We would like to start the way of solving problems in the chapter units and dimensions with a problem that deals with the energy and expressing it in another non convention  form of units. We know that, if we have energy, we can do work and vice versa. So the work done and the energy are having the same units and dimensions.

Problem One


Solution

Work or energy can be measured as the dot product of force and displacement. Thus it will have units that are the product of units of force and displacement. We can express its dimensional formula and it is representation of the physical quantity in terms of fundamental quantities.

By converting them into new terms as shown in the diagram below, we can solve the problem as shown in the diagram.


Problem Two

In this problem, we need to express young's modules in terms of velocity, acceleration and force. The problem is as shown in the diagram below.


Solution

Young's modulus is used to measure the elastic nature of the material. It is mathematically defined as the ratio of stress and strain. Stress is further defined as the restoring force acting on a body per unit area and the longitudinal strain is the ratio of change in the length of the body to its original length. Thus young's modulus has the dimensions of stress per strain.

We need to express this in terms of velocity that is defined as the rate of change of displacement. Acceleration is defined as the ratio of rate of change of velocity and force is defined as the rate of change of momentum.

By using principle of  homogeneity, we know that any equation shall have same dimensions on both the sides of  the equation. By using this concept, we can solve the problem as shown in the diagram below.


Problem Three

The problem is about potential energy. It is the energy possessed by the body by virtue of its position. We need to express it in terms of distance and in terms of some constants. We need to find the dimensions of that constants.


Solution

As the left hand side of equation is potential energy, the right hand side shall be the same. In the denominator, we are adding a constant with displacement with a unknown term. As we can add only similar physical quantity, that unknown term is also displacement. With the combination of the numerator, we shall get energy. The problem is solved as shown in the diagram below.


Problem Four

The last  problem of the post is about expressing joule in terms of the new physical quantities that the present in the problem as shown in the diagram below.


Solution

We can express energy in terms of fundamental quantities like mass, length and time with different dimensions. But now we need to express it in different terms different terms of units of length, time and mass. The problem is solved as shown in the diagram below.


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Units and Dimensions Problems and Solutions Five

We would like to solve the first problem regarding the units of energy. Energy is mathematically defined as the product of force and displacement. If the units of length  and force are changed how does the units of energy changes is the question. By writing the dimensional formula for the energy, we can find how energy changes can be determined as shown in the diagram below.

Problem One


Solution

Energy can be defined as the product of force and displacement. Here in the problem as the unit of length and force are multiplied by four times, the energy units will be increased by 16 times as shown in the diagram below. Displacement is similar to length itself.


Problem Two

The second problem is regarding force relation with other physical quantities. It is given that force depends on mass, radius and time period as shown in the diagram below.


Solution

We know that force is defined as the rate of change of momentum. It has the units of momentum per time. Time period is nothing but certain time. By using dimensional analysis, we can solve the problem  as shown in the diagram below. Thus we can measure the value of the unknown power of force.


Problem Three

The next problem is regarding critical velocity. It is the constant velocity acquired by a body when the resultant force acting on it is zero. It depends on the coefficient of viscosity, density and radius. Coefficient of viscosity is the measure of the viscous nature of the medium.


Solution

We know that viscous force depends on the area of cross section, velocity of fluid flow and is inversely proportional to the distance between the layers.

By writing the dimensions on both the sides of the equation, we can find the value for the critical velocity as shown in the diagram below.


Problem Four

The following problem is about measuring the error in identifying the error when we measure of two resistors when they are connected in parallel. We need to find the effective value of the resistance as well as the error in the measurement.


Solution

We know the formula to  find the effective resistance when two resistors are connected in parallel. Each part of the resistance has its own error thus we need to find the effective error in the total value as shown in the diagram below.



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Units and Dimensions Problems and Solutions Four

We would like to solve a problem basing on significant figures. Significant figures are the numbers in represent a physical quantity basing on which we can say the accuracy of a physical quantity. If we don't want that much accuracy, we can round off the given number as per our requirement.

Problem One 


Solution

Volume of the cube can  be determined by the cube of the side of the cube. The given physical quantity has two significant figures and hence even the volume also shall have only two physical significant figures. This is the rule that we shall follow while we are doing multiplication and division.


Problem Two

This problem is regarding measuring the error in a mathematical measurement where some physical quantities are involved. Given physical quantity is density which is defined as the ratio mass to the volume of the body.


Solution

We need to measure the percentage error in mass and volume. It is nothing but the ratio of error in measuring the physical quantity and its original value. The total error in volume is the percentage error of volume and mass. We can measure it as shown in the diagram below.


Problem Three

This problem is also about measuring the error in a physical quantity like area. It is defined as the product of length and breadth of the given body.


Solution

As area is the product of length and breadth, percentage error in area is the sum of fractional error in length and breadth. Once we know the factional error, we can get the error in area, by multiplying the fractional error in area with original area as shown in the diagram below.


Problem Four

The following problem is regarding error measurement while we are using the device vernier calipurse.


Solution

Least count is the least value that we can measure with a device. With the Vernier  caliperse, we can measure a least value of one by hundredth of the centimeter. So it gives better accuracy when compared with the normal scale.


Any way this is also similar to previous problem where we are going to measure the error in the area.

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Units and Dimensions Problems and Solutions Three

Let us solve a problem where we are going to convert a physical quantity from from units of measurement to other. The physical quantity that we are trying to convert is is acceleration to gravity. It is nothing but acceleration and is defined as the rate of change of velocity. 

Problem one


Solution

We need to solve this problem basing on principle of homogeneity. Writing the dimensional analysis  on both the sides and then substituting the appropriate units on both the sides of the equation, we can solve the problem as shown in the diagram below.

Problem Two

The second problem deals with expressing energy not in normal terms of mass, length and time as fundamental quantities but representing it in terms of force, acceleration and time. The problem is as shown in the diagram below.


Solution

We need to express in terms of the above mentioned physical quantities. Time is a common term in both the formats. We need to express length and mass in terms of force and acceleration so that the problem can be solved as shown below.


Problem Three

In this problem we need to solve the dimension of a unknown physical quantity used in the given equation. The left hand side of the equation is the number of particles crossing per unit area and per unit time. Thus shall have dimensions of per area and per time as number won't have any dimensions.


Solution

The other side of the equation has unknown physical quantity D and we need to find it. Numbers are the number of particles crossing the given space per unit volume on the right hand side of the equation and X is displacement itself.

By equating the dimensions on both the sides of the equation, we can solve the problem as shown in the diagram below.


Problem Four

The problem is regarding measuring the mean deviation of the given parameters. Length of the cylinder is measured in multiple times and each time different reading is obtained as shown in the diagram below.


Solution

To measure the mean deviation, first we need to measure the average of the given values. Average is simply sum of the readings by the number of the readings.

We can now measure how much each reading is different from the other readings. That is called deviation and we need to measure the deviation of each reading from the average. We can then measure the average of these deviations and it is called mean deviation. Thus we are able to certify that the given value is the sum of mean value and sum or difference of the mean deviation. Thus though we are not able to say exactly what is the value is, we can at least say that the reading vary between the given range.



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