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Physics Derivations For Class 11 Pdf 3844: A Comprehensive Guide for Aspiring Physicists


Physics Derivations For Class 11 Pdf 3844: A Comprehensive Guide




Physics is one of the most fascinating and challenging subjects for class 11 students. It involves understanding the fundamental principles of nature and applying them to solve various problems. One of the key skills that physics students need to develop is the ability to derive formulas and equations from basic concepts. Physics derivations are not only essential for scoring well in exams, but also for developing a deeper insight into the subject.




Physics Derivations For Class 11 Pdf 3844



However, many students find physics derivations difficult and confusing. They often struggle to remember the steps and logic behind each derivation. They also face problems in finding the right resources and guidance to learn and practice physics derivations. That's why we have created this comprehensive guide on physics derivations for class 11 pdf 3844. In this guide, you will learn:


  • What are physics derivations and why are they important for class 11 students?



  • What are the basic concepts and formulas of physics that you need to know?



  • What are the advanced topics and derivations of physics that you need to master?



  • How to download and use physics derivations for class 11 pdf 3844?



  • What are some common FAQs related to physics derivations for class 11?



By the end of this guide, you will have a clear understanding of physics derivations and how to ace them in your exams. So, let's get started!


Basic Concepts and Formulas of Physics




Before we dive into the advanced topics and derivations of physics, let's review some of the basic concepts and formulas that you need to know. These concepts and formulas form the foundation of physics and will help you in deriving more complex equations later on. Here are some of the important topics that you need to revise:


Units and Dimensions




Units are the standard quantities that are used to measure physical quantities. For example, meter is the unit of length, kilogram is the unit of mass, second is the unit of time, etc. Dimensions are the powers to which the fundamental units are raised to express a physical quantity. For example, speed has the dimension of [L][T], where [L] represents length and [T] represents time.


The main formulas that you need to know in this topic are:


  • The principle of homogeneity: The dimensions of all the terms in a physical equation must be same.



  • The dimensional formula: The expression that shows how a physical quantity depends on the fundamental units.



  • The dimensional analysis: The method of using the dimensions of physical quantities to check the validity of equations, to derive relations, and to convert units.



Here is an example of a derivation using dimensional analysis:


Derivation: Derive the relation between the escape velocity (ve) of a body from the surface of a planet and the gravitational constant (G), the mass (M) and the radius (R) of the planet.


Solution: We know that the escape velocity is the minimum velocity required by a body to escape from the gravitational field of a planet. Therefore, it must depend on the gravitational constant, the mass and the radius of the planet. Let us assume that:


ve α GMR


where a, b and c are constants to be determined. Taking logarithms on both sides, we get:


log ve = log GMR


log ve = a log G + b log M + c log R


Comparing the dimensions of both sides, we get:


[L][T] = [M][L][T][M][L]


Equating the powers of [M], [L] and [T], we get:


-a + b = 0


3a + c = 1


-2a = -1


Solving these equations, we get:


a = 1/2, b = 1/2, c = -1/2


Substituting these values in the original equation, we get:


ve α √(GM/R)


ve = k√(GM/R)


where k is a dimensionless constant. To find the value of k, we use the fact that the escape velocity from the earth's surface is 11.2 km/s, and the values of G, M and R for the earth are 6.67 x 10 Nm/kg, 6 x 10 kg and 6.4 x 10 m respectively. Substituting these values, we get:


k√(GM/R) = 11.2 x 10


k = 11.2 x 10/√(6.67 x 10x 6 x 10/6.4 x 10)


k ≈ 1


Hence, the final relation is:


ve ≈ √(GM/R)


Scalars and Vectors




A scalar is a physical quantity that has only magnitude and no direction. For example, mass, speed, distance, etc. A vector is a physical quantity that has both magnitude and direction. For example, displacement, velocity, acceleration, force, etc. Vectors are represented by arrows with a specific length and direction.


The main formulas that you need to know in this topic are:


  • The addition and subtraction of vectors: The resultant vector of adding or subtracting two or more vectors can be found by using either the parallelogram law or the triangle law.



  • The multiplication of vectors: The product of two vectors can be either a scalar or a vector, depending on whether it is a dot product or a cross product.



  • The dot product: The dot product of two vectors A and B is defined as A.B = AB cos θ, where θ is the angle between them. The dot product is a scalar quantity that gives the projection of one vector on another.



The cross product: The cross product of two vectors A and B is defined as A x B = AB sin θ n̂, where θ is the angle between them and n 71b2f0854b


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