Showing posts with label Dimensional formula. Show all posts
Showing posts with label Dimensional formula. Show all posts

NEET 2025 UNITS AND MEASUREMENTS QUESTION BASED ON priniciple of homogenity

 Mastering the Units and Measurements chapter is a crucial step for any student aiming to crack NEET 2025. In this video, we provide a detailed, step-by-step solution to a high-yield question focused on the Principle of Homogeneity and Dimensional Analysis.

Dimensional analysis is more than just a topic; it’s a powerful tool that helps you verify equations and derive relationships between physical quantities during the exam. Whether you are a beginner or looking to refine your problem-solving speed, this walkthrough is designed for you!

What You Will Learn in This Tutorial:

We take a complex problem involving a balloon's inflation and the speed of escaping gas, breaking it down into simple, manageable steps:

  • Understanding the Problem Statement: We identify how time depends on several physical quantities: surface tension , area , density , and radius  [00:59].

  • The Power of Dimensional Analysis: Learn how to write the dimensional formulas for each quantity:

    • Surface Tension: Force per unit length  [01:30].

    • Area:  [02:02].

    • Density: Mass per unit volume  [02:02].

    • Radius: [02:10].

  • Applying the Principle of Homogeneity: We equate the dimensions of Mass , Length , and Time  from both sides of the proportionality equation to create a system of linear equations [02:19].

  • Step-by-Step Solving: Watch as we solve for the exponents:

    • Finding alpha = -1/2 from the time dimension [03:41].

    • Deriving gamma = 1/2 from the mass dimension [03:51].

  • Smart Exam Strategy: We demonstrate how to use the process of elimination with the options provided to save time and verify the final result [04:09].

  • The Final Verification: See why Option 4 alpha = -1/2, beta = -1, gamma = 1/2, delta = 7/2) satisfies the final length equation [07:14].

Why Watch This Video?

  • Simple & Logical: We avoid unnecessary jargon and focus on the logical flow of the solution.

  • Exam Shortcuts: Learn how to narrow down options early in the calculation.

  • Foundation for Success: The Principle of Homogeneity is frequently tested in NEET and JEE—this video ensures you have a rock-solid understanding of it [08:15].



No comments:
Labels: , , , , ,

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.




Related Posts


No comments:
Labels: , , ,

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.





Related Posts


No comments:
Labels: , , ,

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.



Related Posts



No comments:
Labels: , , ,

Oscillations Problems with Solution Two

We are solving series of problems based on  the concept of oscillations. It is also called vibratory or harmonic motion where there  is to and fro motion that gets repeated at regular intervals of time. The vertical projection of uniform circular motion is simple harmonic and we have derived equations for displacement, velocity and acceleration for the body in simple harmonic motion based on that. The time taken to complete one oscillation is called time period and the number of oscillations per one second is called frequency.

Problem

A body of known mass is connected with two springs and they in tern are connected to rigid support as shown in the diagram below. We need to find the effective time period of the system and the problem is as shown in the diagram below.


Solution

When ever the body is slightly disturbed, it starts oscillating and one spring expands and the other spring contracts by the same magnitude. As the force acting on both of them is same, the two springs behaves as if like they are connected in series and we can find the effective spring constant of the system and time period as shown in the diagram below.


Problem

A spring of spring constant K and length L is cut into two parts and the relation between the lengths of two parts and their ratio is given to us in the problem as shown in the diagram below. We need to measure the spring constant of one part of the spring.


Solution

Spring constant is the measure of nature of spring and it depends on the length of spring in the inverse proportional ratio. Taking that into consideration, we need to solve the problem as shown in the diagram below.


Problem

A piece of wood known dimensions is given to us and its density is also known to us as given in the problem below. It is floating in water with one surface vertical to the surface. It is pushed down and released and it starts oscillating. We need to measure the time period of the system.


Solution

When ever a force is applied on the body, there is buoyant force also and the system starts executing oscillatory motion with a certain restoring force. By comparing that with the standard equation, we can solve the problem as shown in the diagram below.


Problem

A cylindrical piston is used to close a cylinder with a certain gas and when it is slightly distributed, it starts oscillating in simple harmonic motion. We need to find the time period of the system and the problem is as shown in the diagram below.


Solution

By comparing it with the standard equation, we can solve the problem as shown in the diagram below. We need to identity the restoring force and the equation for the acceleration of the body in SHM. The problem is solved as shown in the diagram below. Here in the first case, we get the equation for the force.


We need to equate to the product of mass and acceleration and we further need to substitute the value of acceleration for a body in SHM so that we can compare it with the standard equation. Thus we can get the time period of the system as shown in the diagram below.


Problem

A spherical ball of known mass and radius is rolling with out slipping on a concave surface of known radius as shown in the diagram below. If the oscillations are small, we need to find the time period of the system.


Solution

Let us consider the spherical ball is at a particular position and we can find the angle basing on the definition as shown in the diagram below. As we know the equation for the acceleration of a body sliding on a inclined surface, by writing that equation, we can get that and find the time period as shown in the diagram below.



Related Posts


2 comments:
Labels: , , , , ,

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.


Related Posts

No comments:
Labels: , , , ,
Subscribe to: Comments (Atom)