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Asked 6 days ago IIT JEE

Today in these days quality faculty for iit jee are mostly given qualitative basic concept.

Lesson Posted on 12 Jul CBSE/Class 11/Science/Physics IIT JEE/IIT - JEE Advanced/Physics/Modern Physics IIT JEE/IIT - JEE Mains/Physics/Physics and Measurement

This is the foremost important physics topic. Students very often than not forget Conversion of Units of physical quantities to one system while solving numerical questions. This leads to wrong answers and failure in competitive exams. Such errors or failures also cause disaster in real life making the... read more

This is the foremost important physics topic. Students very often than not forget Conversion of Units of physical quantities to one system while solving numerical questions. This leads to wrong answers and failure in competitive exams. Such errors or failures also cause disaster in real life making the society pay a very heavy price.

Physics is fundamentally an empirical science. Measurements, units & uncertainties are at the heart of it. All laws of nature are derived empirically and expressed by mathematical equations. How otherwise scientists can express relationships between measurable physical quantities? It is equally vital to express the measurement in right unit; else invite disaster.

The Mars Climate Orbiter costing about US\$ 125 million came to a catastrophic conclusion because of confusion in UNITS. One engineering team used Metric Units while another team used English Units. As a result in September 1999 the spacecraft entered Martian atmosphere instead of reaching a stable orbit.

Equally important is Uncertainties in measurements e.g. pair of pants of size 34 that varies just 3 percent changes a full inch in waist size. It could end up a 35 and hang on the hips or a 33 making you wonder how you gained all that weight!

Measurements, Units and Uncertainties are crucially important and when properly implemented can create wonders in Life (and also exams); else a disaster. Students should pay utmost importance to Units, Conversion of Units and Uncertainties.

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Lesson Posted on 09 Jul IIT JEE/IIT - JEE Mains/Mathematics/Mathematical Induction

Raj Kumar

I am Six Sigma Black belt certified. I am 2011 pass out in B.tech from NIT Srinagar. I am an experienced,...

The differential equation is that equation which contains. Differntial Independent variable Dependent variable Order of Differential equation: Order of highest order derivative. The degree of the differential equation: Degree of highest order derivative. General Solution: It is a genearlized... read more

The differential equation is that equation which contains.

Differntial

Independent variable

Dependent variable

Order of Differential equation: Order of highest order derivative.

The degree of the differential equation: Degree of highest order derivative.

General Solution:

It is a genearlized solution which contains finite no of the solution.

Particular Solution:

It is a particular solution which contains only one solution of the differntial equation.

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Lesson Posted on 18 Jun IIT JEE/IIT - JEE Mains/Physics/Electrostatics Electromagnetic Waves

Anshuman

I can teach every topic very easily and make it stronger for the students.

This is my lesson plan for the chapter of electrostatics. Electrostatics- The chapter of electrostatics deals with another1 fundamental nature of particles, “Charge”, in particular charges at rest. This Chapter is broadly divide in nine parts, parts of the chapter and time required for... read more

This is my lesson plan for the chapter of electrostatics.

Electrostatics- The chapter of electrostatics deals with another1 fundamental nature of particles, “Charge”, in particular charges at rest.

This Chapter is broadly divide in nine parts, parts of the chapter and time required for each are given in sequence as per lesson plan. The time required for completion of the chapter narrowly varies around 25 hours.

 Serial no. Topics Time required      (Hours) 1 Charge 1 2 Interaction of charges 1 3 Electric field 4 4 Electric potential Energy 2 5 Electric potential 4 6 Relationship between electric field and electric potential 1 7 Electric Dipole 2 8 Gauss’ Law 4 9 Capacitors 6

1. Charge

Definition and  Properties

Definition- It includes formal definiton of charge, the idea of charged bodies, +ve and –ve charges, the transfer of charges, attraction and repulsion of charged bodies.

Properties- The charge carriers, quantisation of charges, conservation of charges. Conductors, free electons and induced charages.

1. Interaction of Charges

Coulomb’s law of electrostatic force, between two point charges. Coulomb’s law in vector form. Principle of superposition, Electrostatic force on a point charge due to surround chage distribution.

1. Electric field

Dfinition of electic field due to a point charge. Calculation of electric field at a point in space using principle of superpositon for descrete distribution and continuous distribution of chages.

Continuous Distribution of charges. (Uniform distribution of chages )

1. Rod of finite length- Calculation of electric field due to uniformly charged rod at point in space. (using calculus)
2. Ring or arc of finite radius- Calculation of elecric field due to uniforly chaged ring at a point on the axis of ring, calculation of electric field at centre of an arc. (using calculus and symmetry) Deriving symmerty betwwen above calculations.
3. Disc or Sector of finite radius -calculation of electric field at a point on the axis of disc or at the centre of a sector.( Using idea of surface chage density and calculus)
4. Hemispherical shell- Calculation of electric field at the centre of hemisphere.( using calculus)
5. Application of all above derivations in calculation of electric field using symmetry for different geometrical symmetries.

Continuous distribution of charges ( Non uniform distribution)Demonstration of concept using some problems, using calculus.

1. Electric Potential Energy

Derivation of electric potentianl energy between two point charges using work-energy theorem.  Electric potential energy of system of charges. Elecric potential energy or self energy of continuous charged bodies. Some typical problems relating work energy theorem.

1. Electric Potential

Dfinition of electic potential due to a point charge. Calculation of electric potential at a point in space using principle of superpositon for descrete distribution and continuous distribution of chages.

Continuous Distribution of charges. (Uniform distribution of chages )

1. Rod of finite length- Calculation of electric potential due to uniformly charged rod at point in space. (using calculus)
2. Ring or arc of finite radius- Calculation of elecric potential due to uniforly chaged ring at a point on the axis of ring, calculation of electric potential at centre of an arc. (using calculus)
3. Disc or Sector of finite radius -calculation of electric potential at a point on the axis of disc and on the edge of disc. Calculation of electric potential due to sector at the centre of a sector.( Using idea of surface chage density and calculus)

The relationship between change in  electric potential energy and electric potential. Some problems on the path independency of  change in potential energy.

1. Relationship between Electric field and Electric potential

Equipotential surfaces, the relationship betweeen jumping from one equipotential surface to other and electric field. Rough mathematical derivation beween electric field and electric potential. DIFFERENTIAL relationship between  and V using  (nabla)

Some problems on euipotential surfaces and electric field.

1. Electric Dipole

Definition and properties of an electric dipole, calculation of net dipole moment i.e vector addition of dipole moments, electric potential at a point in space due to a dipole. The change in position vector i.e.  along  and along . Derivation of electric field at a point in space using differential relationship betweeen  and V, along directions  and . Effects on a electric dipole placed in a uniform electric field, the net torque acting on a dipole due to external electric field, potential energy of a dipole in an external electric field. Net force and torque acting on an electric dipole placed in non uniform external electric field.

1. Gauss’ Law

Electric field line, definiton and properties. Area vector, flux , elecrtic flux. Gauss law, some problems on direct gauss law, Gauss’ law and Coulomb’s law similarities.calculation of electric flux passing through a surface for different geometrical symmetries.

Application of Gauss’ law.

Discussion of properties of a charged conductor. Calculation of electric field due  to a hollow sphere and drawing properties of spherical distribution,  calculation of electric field due to a charged solid sphere, calculation of electric field due to infinite charge distribution i.e cylindrical symmetry, planar symmetry. Calculation of electric field inside a cavity. Application of above mentioned derivations for different geometrical arrangements.

1. Capacitors

Definition of a capacitor, geometrical, and theoretical (capacitance, potential drop, charge) parameters of a capacitor. Different types of capacitors depending on their geometrical arrangements, and calculation of their capacitance. The idea of polarization, permittivity constant of space, Coulomb’s law and Gauss’ law revised. Calculation of capacitance of a capacitor including the concept of permittivity of space. Circuits of capacitors, calculation of equivalent capacitance for an arrangement of capacitors. Energy stored in the capacitors.

1 D      2 B    3 B    4 B    5 D         6 B        7 A      8 B      9 D     10 A

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Answered on 17 Jun CBSE/Class 11/Science/Physics IIT JEE/IIT - JEE Mains/Physics/Physics and Measurement Tuition/BTech Tuition/Engineering Physics +1 IIT JEE less

Most of us would understand the concept of torque. However, I am going to ask a question which is slightly... read more

Most of us would understand the concept of torque. However, I am going to ask a question which is slightly more involved.

I shall spin a rod of length 'L' with an angular velocity 'ω' about its centre of mass and place it on a ground with a kinetic friction coeficcient of ηk.  Without using the torque equation and only applying Newton's 2nd law of motion (F=ma) , could you derive the rod's 'ω' as function of time, 't'.

This is a very interesting exercise to clearly understanding the various forces acting on and inside a rigid body, the direction of those forces and the associated unknowns.

Many students fail to understand the constraints of a rigid body motion. They also do not easily appreciate how using the torque euqation so much simplifies rigid body dynamics.

I hope this exercise will be fun. I will post the solution in a few days though.

Kashish Solanki

Tutor

Most of us would understand the concept of torque. However, I am going to ask a question which is slightly more involved. I shall spin a rod of length 'L' with an angular velocity 'ω' about its centre of mass and place it on a ground with a kinetic friction coeficcient of ηk. Without using... read more

Most of us would understand the concept of torque. However, I am going to ask a question which is slightly more involved.

I shall spin a rod of length 'L' with an angular velocity 'ω' about its centre of mass and place it on a ground with a kinetic friction coeficcient of ηk.  Without using the torque equation and only applying Newton's 2nd law of motion (F=ma) , could you derive the rod's 'ω' as function of time, 't'.

This is a very interesting exercise to clearly understanding the various forces acting on and inside a rigid body, the direction of those forces and the associated unknowns.

Many students fail to understand the constraints of a rigid body motion. They also do not easily appreciate how using the torque euqation so much simplifies rigid body dynamics.

I hope this exercise will be fun. I will post the solution in a few days though.

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Asked on 09 Jun CBSE/Class 11/Science/Physics IIT JEE/IIT - JEE Mains/Physics/Physics and Measurement Tuition/BTech Tuition/Engineering Physics +1 IIT JEE less

Most of us would understand the concept of torque. However, I am going to ask a question which is slightly... read more

Most of us would understand the concept of torque. However, I am going to ask a question which is slightly more involved.

I shall spin a rod of length 'L' with an angular velocity 'ω' about its centre of mass and place it on a ground with a kinetic friction coeficcient of ηk.  Without using the torque equation and only applying Newton's 2nd law of motion (F=ma) , could you derive the rod's 'ω' as function of time, 't'.

This is a very interesting exercise to clearly understanding the various forces acting on and inside a rigid body, the direction of those forces and the associated unknowns.

Many students fail to understand the constraints of a rigid body motion. They also do not easily appreciate how using the torque euqation so much simplifies rigid body dynamics.

I hope this exercise will be fun. I will post the solution in a few days though.

Asked on 09 Jun IIT JEE

Most of us would understand the concept of torque. However, I am going to ask a question which is slightly... read more

Most of us would understand the concept of torque. However, I am going to ask a question which is slightly more involved.

I shall spin a rod of length 'L' with an angular velocity 'ω' about its centre of mass and place it on a ground with a kinetic friction coeficcient of ηk.  Without using the torque equation and only applying Newton's 2nd law of motion (F=ma) , could you derive the rod's 'ω' as function of time, 't'.

This is a very interesting exercise to clearly understanding the various forces acting on and inside a rigid body, the direction of those forces and the associated unknowns.

Many students fail to understand the constraints of a rigid body motion. They also do not easily appreciate how using the torque euqation so much simplifies rigid body dynamics.

I hope this exercise will be fun. I will post the solution in a few days though.

Lesson Posted on 05 Jun CBSE/Class 11/Science/Physics Tuition/Class XI-XII Tuition (PUC)/Physics Exam Coaching/Engineering Entrance Coaching/IIT JEE Coaching +1 IIT JEE/IIT - JEE Advanced/Physics/Modern Physics less

Vishesh Nigam

I am a Chemical Engineer and Khan Academy Talent Hunt Finalist (2017)(please watch my winning video below...

Linear Motion also called rectilinear motion is a one-dimensional motion along a straight line, and can, therefore, be described mathematically using only one dimension. The linear motion can be of two types: uniform linear motion with constant velocity or zero acceleration; nonuniform linear motion... read more

Linear Motion also called rectilinear motion is a one-dimensional motion along a straight line, and can, therefore, be described mathematically using only one dimension. The linear motion can be of two types: uniform linear motion with constant velocity or zero acceleration; nonuniform linear motion with non-zero acceleration or variable velocity. The motion of a particle along a line can be described by its position x, which varies with t (time). An example of linear motion is an athlete running 100m along a straight track.

Check out the webinar video, to understand in-depth.

Have a question? Post your question or comment below.

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Lesson Posted on 04 Jun CBSE/Class 10/Mathematics IIT JEE/IIT - JEE Advanced/Mathematics/Integral calculus IIT JEE/IIT - JEE Mains/Mathematics/Integral Calculus +2 CBSE/Class 9/Mathematics Integral Calculus less

Indrajit

I am successfully providing tuition of Mathematics in Class IX-XII of all Boards near Kavi Najrul Metro...

Before starting the discussion I would like to mention this problem The conventional way to solve this problem is to deal with trigonometric identities and to mange by parts, but this type of problem can be easily solved by applying results of... read more

Before starting the discussion I would like to mention this problem

$\int_0^{\pi/2}\cos^4x\sin^2x=?$

The conventional way to solve this problem is to deal with trigonometric identities and to mange by parts, but this type of problem can be easily solved by applying results of Beta and Gamma Functions.

———————————————————————————————————————————————————

Gamma Function:

The gamma function denoted by  $\Gamma&space;(n)$ is defined for positive values of $n$ by the integral

$\Gamma(n)=\int_0^{\infty}e^{-x}x^{n-1}dx,n>0$                                                          .......(1)

Now,

1. For any $a>0.$

$\int_0^\infty&space;e^{-ax}x^{n-1}dx=\frac{&space;\Gamma(n)}{a^n},n>0$

2.

${\color{Red}&space;\Gamma(n+1)=n\Gamma(n),n>0}$

Integration by parts gives

$\int&space;_{\epsilon}^{B}e^{-x}.x^{n-1}dx=&space;[e^{-x}&space;.\frac{x^n}{n}]_{\epsilon}^B+\frac{1}{n}\int_\epsilon^Be^{-x}x^ndx$

As $B\to&space;\infty$ and $\epsilon&space;\to&space;0+$, the integreted part vanishes at both limits and therefore,

$\int_0^\infty&space;e^{-x}x^{n-1}dx=\frac{1}{n}\int_0^{\infty}e^{-x}x^ndx$

i.e.

$\Gamma(n)=\frac{1}{n}\Gamma(n+1)$

3.     $\Gamma(1)=1.$

By direct computation   $\Gamma(1)$ converges

$\Gamma(1)=\int_0^{\infty}e^{-x}dx=\lim_{B\to\infty}\int_0^Be^{-x}dx=\lim_{B\to\infty}(1-e^{-B})=1$

4. $\Gamma(n+1)=n!$

Combining the above reletions,

$\Gamma(n+1)=n\Gamma(n)=n(n-1)\Gamma(n-1)=\cdots$

$=n(n-1)(n-2)\cdots3.2.1..\Gamma(1)=n!$

Beta Function:

The beta function denoted by  $B(m,n)$ is defined for positive values of m and n by the integral

$B(m,n)=\int_0^{1}x^{m-1}(1-x)^{n-1}dx,&space;m,n>0$

1.      $B(m,n)=B(n,m)$

This reletion can be established by giving the transformation  $x=1-y$

2.

$B(m,n)=\int_0^{\infty}&space;\frac{x^{m-1}}{(1+x)^{m+n}}dx=\int_0^{\infty}&space;\frac{x^{n-1}}{(1+x)^{m+n}}dx=B(n,m)$

Substituting

$x=\frac{1}{1+y}$

,then

$\int_{\epsilon}^{1-\delta}x^{m-1}(1-x)^{n-1}dx&space;=&space;\int_{\frac{1}{\epsilon}-1}^{\frac{\delta}{1-\delta}}\frac{1}{(1+y)^{m-1}}.\frac{y^{n-1}}{(1+y)^{n-1}}.(-1).\frac{1}{(1+y)^2}dy$

Now letting    $\epsilon\to\0+,\delta\to0+$, we have

$\int_0^1x^{m-1}(1-x)^{n-1}dx=B(m,n)=\int_0^{\infty}\frac{y^{n-1}}{(1+y)^{m+n}}dy$

also,

$B(m,n)=B(n,m)$.

3.

$B(m,n)=2\int_0^{\pi/2}&space;\sin^{2m-1}\theta&space;\cos^{2n-1}\theta&space;d\theta;&space;m,n>0$

This reletion can be established by by giving the transformation $x=\sin^2\theta$ in the definition of Beta Function.

4.

$B(\frac{1}{2},\frac{1}{2})=\pi$

This can be established by putting

$m=n=\frac{1}{2}$

in

$B(m,n)=2\int_0^{\pi/2}&space;\sin^{2m-1}\theta&space;\cos^{2n-1}\theta&space;d\theta;&space;m,n>0$

5.

$B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)};&space;m,n>0$

so,

$\Gamma(\frac{1}{2})=\sqrt{\pi}$

—————————————————————————————————————————————————————

Now Lets solve the problem:

$\int_0^{\pi/2}\cos^4x\sin^2x=?$

Now to solve this problem plug

$m=\frac{3}{2},n=\frac{5}{2}$

in

$B(m,n)=2\int_0^{\pi/2}&space;\sin^{2m-1}\theta&space;\cos^{2n-1}\theta&space;d\theta;&space;m,n>0$

$\int_0^{\pi/2}\cos^4x\sin^2x=\frac{1}{2}B(\frac{3}{2},\frac{5}{2})=\frac{\Gamma(\frac{3}{2}).\Gamma(\frac{5}{2})}{\Gamma(\frac{3}{2}+\frac{5}{2})}$

=

$\frac{\frac{1}{2}.\Gamma(\frac{1}{2}).\frac{3}{2}.\Gamma(\frac{3}{2})}{\Gamma(4)}=\frac{\pi}{32}$

As

$\Gamma(\frac{1}{2})=\sqrt{\pi}$

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Lesson Posted on 30 May IIT JEE/IIT - JEE Mains/Mathematics/Sequences and Series CBSE/Class 10/Mathematics/Arithmetic Progression

Hemant P.

1. I have taught at various prestigious institutes of repute in India and abroad. 2. Worked at (i)...

Sequences Mathematics is the science of patterns and homogeneity. Sequences are all about them. A sequence has a traditional meaning in mathematics similar to that in ordinary language. A sequence is any regular pattern exhibiting some characteristic property throughout. To be precise, any sequence... read more

Sequences

Mathematics is the science of patterns and homogeneity. Sequences are all about them.

A sequence has a traditional meaning in mathematics similar to that in ordinary language.

A sequence is any regular pattern exhibiting some characteristic property throughout.

To be precise, any sequence is characterised by a property.

Let us take a simple example of a common most sequence of all, the series of natural numbers, 1, 2, 3 …n. This sequence is characterised by a property that every term is one more than the previous term. Therefore, the sequence is bounded by a property. We may find an infinite number of sequences. You may note that to generate a sequence we need two things. We begin with a starting term, and a rule defined to get the next term or a general term. This general term is often called generating term.

There are of course sequences that are unbounded and may run infinitely on either side. These may be termed as infinite sequences. The series, which we are dealing with, are called finite sequences, for the apparent reason that they are bounded, i.e. they have the last term. The other types of sequences are called infinite series, which we would take afterwards.

We are now in a position to define a sequence, a finite series. We start with the bound of a series. The bound of a series are the values which the terms or members of a sequence can take. You may understand them as the limits of the series. For example, if we have specified that write all natural numbers less than 10, then we have clarified that our sequence has a limit of n varying from 1 to 9 since we have to find all natural numbers less than 10. A series may be generated by a rule alone along with the set or bound in which it is defined. We call this generating term as the general term of a sequence. You may now guess the general or generate the name of our series of natural numbers. Yes, it would be

t­n=n; here t­n is the symbol of the general term. A sequence is, therefore, a function of a general term. For each value of n, we get a particular term of the series. We are now in a position to define few finite sequences ourselves.

0 < n < 10

You can yourself get few values by putting n = 1, 2, 3 etc.

t1 = 2⋅1+1 = 3

t2 = 2⋅2+1 = 5

and t3 = 2⋅3+1 = 7 etc.

Note that this sequence is a contiguous finite sequence hence we would substitute only values in the range of 0 < n < 10; i.e. we have just nine terms in the series.

Till now we have an idea of a finite sequence. Discussion done till now summarises that a finite sequence has following properties.

Starting term or the initial term
The generating term or general term or the rule to get the next term

Please note that a finite sequence is always bounded, but the reverse is not always true. There may be a sequence which is bounded but may not be finite. Consider the sequence 1,.1,.01,.001… .Here the general term is given by tn =. You can see that all the values of this sequence lie in the range 0 and 1 (This sequence is a continuously decreasing sequence, and it will never reach zero as exponent are always positive, but they can reach as near to zero as possible. We call this phenomenon as converging which you will learn in higher grades). Hence this sequence is bounded in the semi-open set (0, 1], i.e. greater than zero and less than or equal to 1.

Let us take a few examples of finite sequences with general term given. You are required to find the first few terms.

tn = 3n+1 ; 0 < n < 3
tn= 2n-1 ; 0 < n < 13

Note that n can only take the value of positive integers, although it may have range.

You can quickly get terms of the sequences defined above, by putting the values of

n = 1, 2, 3, etc. You can check your answers from these:

The sequence is 4, 7, 10
The sequence is 1, 3, 7 … ,

Find more notes here.

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