Neutrino Maths

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Kedarnath

A math teacher

My math journey started when I was a kid. Those were the days when I used to see nightmares about math. My fear of math was increasing day by day, because of which I wasn't confident enough in the exams. But when I started scoring less in the exams, I decided that "My fear shouldn't let me down". I took it as a challenge and started working hard. In a few days, I realized Hard-work and consistency can take you to your goal.

And now I teach Math


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Chemistry

11th

For JEE, NEET, and boards

Chapters uploaded:

2/20

Physics

11th

For JEE, NEET, and boards

Chapters uploaded:

3/16

 Sankhyaka shastram: Notes

LECTURE - 1

  • Parichayam 
  • Vargeekruta dattamsham 
  • Avargeekruta dattamsham
          a)Sagatu b)Madhyagatham c) Baahulakam

LECTURE - 2

  • sagatu kanugone paddatulu 
  1. Pratyaksha paddati
  2. Oohinchina sagatu paddati leda vichalana paddati
  3. sankshipta vichalana paddati

 Rendu chara rashulalo rekhiya sameekaranala Jatha: Notes

LECTURE - 1

  • Rendu chara rashulalo rekhiya sameekaranala Jatha - Sadhana
  • Sangatha sameekaranalu 
 a) Paraspara swatantra  sameekaranalu 
b) Paraspara adharita sameekaranaalu
  •  Asangatha sameekaranalu

LECTURE - 2

  • Model paddathi 
  • Jyaamitheeya paddathi 
  • Bheejeeya paddathulu

LECTURE - 3

  • Ivi cheyandi (Pg no. 83) 
  • Udaharana 2 , 3  

Chapter summary: in english

1)Two linear equations within the same two variables are called a pair of linear equations in two variables
a1x+b1y+c1=0(a12+b12≠0)
a2x+b2y+c2=0(a22+b22≠0)
where a1,a2,b1,b2,c1,c2 are real numbers.

2) A pair of linear equations in two variables may be solved by using many methods.

3) The graph of a pair of linear equations in two variables is represented by 2 lines.
i. If the lines intersect at a point then the point gives the unique solution of the two equations. In this case, the pair of equations is consistent and independent.
ii. If the lines coincide, then there are infinitely many solutions -where each point on the line could be a solution. In this case, the pair of equations is consistent and dependent.
iii. . If the lines coincide, then there are infinitely many solutions -where each point on the line is a solution. In this case, the pair of equations is consistent and dependent.

4)We have discussed the subsequent methods for locating the solution(s) of a pair of linear equations
i. Model method
ii.Graph method
iii.Algebraic methods - Substitution method and the Elimination method.

5)There is a relation between the coefficients and nature of system of equations.
i. if (a1 / a2) ≠ (b1 / b2) then the pair of linear equations is consistent.
ii. if (a1 / a2) = (b1 / b2) ≠ (c1 / c2) then the pair of linear equations is inconsistent.
iii. if (a1 / a2) = (b1 / b2) = (c1 / c2) then the pair of linear equations is dependent and consistent.

6) There are several conditions which are also mathematically represented by two equations that don't seem to be linear to start out. But we are able to alter them so that they are goinng to be reduced to a pair of linear equations.

Try these questions

1) For what value of k, the subsequent system of equations has a unique solution. x – ky = 2 and 3x + 2y = –5

2) For what values of m, the pair of equations 3x + my = 10 and 9x + 12y = 30 gives a unique solution.

3) In a rectangle ABCD, AB = x + y, BC = x – y, CD = 9 and AD = 3. Find the values of x and y.

4) Show that the pair Linear Equations 7x + y = 10 and x + 7y = 10 are consultant.

5) Write the Condition for the pair of linear equations in two variables to be parallel lines.

6) If we multiply or divide either sides of a linear equation by a non- zero number, then the roots of that linear equation will remain the same’. is it's true? ? If it is true,then justify with an example.

7) If the current ages of A and B are in ratio of 9 : 4 and after 7 years the ratio of the ages are going to be 5 : 3 then find their current ages.

8) Solve the subsequent pair of linear equations by substitution method. 2x – 3y = 19 and 3x – 2y = 21

Some basic concepts of chemistry: Formula sheets

Chapter summary

Chemistry, as we understand it today is not a very old discipline. People in ancient India, already had the knowledge of many scientific phenomenon much before the advent of modern science. They applied the knowledge in various walks of life.

The study of chemistry is very important as its domain encompasses every sphere of life. Chemists study the properties and structure of substances and the changes undergone by them. All substances contain matter, which can exist in three states – solid, liquid or gas. The constituent particles are held in different ways in these states of matter and they exhibit their characteristic properties. Matter can also be classified into elements, compounds or mixtures. An element contains particles of only one type, which may be atoms or molecules. The compounds are formed where atoms of two or more elements combine in a fixed ratio to each other. Mixtures occur widely and many of the substances present around us are mixtures.

When the properties of a substance are studied, measurement is inherent. The quantification of properties requires a system of measurement and units in which the quantities are to be expressed. Many systems of measurement exist, of which the English and the Metric Systems are widely used. The scientific community, however, has agreed to have a uniform and common system throughout the world, which is abbreviated as SI units (International System of Units).

Since measurements involve recording of data, which are always associated with a certain amount of uncertainty, the proper handling of data obtained by measuring the quantities is very important. The measurements of quantities in chemistry are spread over a wide range of 10–31 to 10+23. Hence, a convenient system of expressing the numbers in scientific notation is used. The uncertainty is taken care of by specifying the number of significant figures, in which the observations are reported. The dimensional analysis helps to express the measured quantities in different systems of units. Hence, it is possible to interconvert the results from one system of units to another.

The combination of different atoms is governed by basic laws of chemical combination — these being the Law of Conservation of Mass, Law of Definite Proportions, Law of Multiple Proportions, Gay Lussac’s Law of Gaseous Volumes and Avogadro Law. All these laws led to the Dalton’s atomic theory, which states that atoms are building blocks of matter. The atomic mass of an element is expressed relative to 12C isotope of carbon, which has an exact value of 12u. Usually, the atomic mass used for an element is the average atomic mass obtained by taking into account the natural abundance of different isotopes of that element. The molecular mass of a molecule is obtained by taking sum of the atomic masses of different atoms present in a molecule. The molecular formula can be calculated by determining the mass per cent of different elements present in a compound and its molecular mass. The number of atoms, molecules or any other particles present in a given system are expressed in the terms of Avogadro constant (6.022 × 1023). This is known as 1 mol of the respective particles or entities. Chemical reactions represent the chemical changes undergone by different elements and compounds. A balanced chemical equation provides a lot of information. The coefficients indicate the molar ratios and the respective number of particles taking part in a particular reaction. The quantitative study of the reactants required or the products formed is called stoichiometry. Using stoichiometric calculations, the amount of one or more reactant(s) required to produce a particular amount of product can be determined and vice-versa. The amount of substance present in a given volume of a solution is expressed in number of ways, e.g., mass per cent, mole fraction, molarity and molality.

Try these questions

Q1. What do you mean by significant figures?
A)Significant figures are the meaningful digits which are known with certainty. Significant figures indicate uncertainty in experimented value. e.g.: The result of the experiment is 15.6 mL in that case 15 is certain and 6 is uncertain. The total significant figures are 3. Therefore, “the total number of digits in a number with the Last digit that shows the uncertainty of the result is known as significant figures.”

Q2. In a reaction A + B2 → AB2 Identify the limiting reagent, if any, in the following reaction mixtures.
(i) 300 atoms of A + 200 molecules of B
(ii) 2 mol A + 3 mol B
(iii) 100 atoms of A + 100 molecules of B
(iv) 5 mol A + 2.5 mol B
(v) 2.5 mol A + 5 mol B

A)Limiting reagent: It determines the extent of a reaction. It is the first to get consumed during a reaction, thus causes the reaction to stop and limiting the amt. of products formed.
(i) 300 atoms of A + 200 molecules of B
1 atom of A reacts with 1 molecule of B. Similarly, 200 atoms of A reacts with 200 molecules of B, so 100 atoms of A are unused. Hence, B is the limiting agent.
(ii) 2 mol A + 3 mol B
1 mole of A reacts with 1 mole of B. Similarly, 2 moles of A reacts with 2 moles of B, so 1 mole of B is unused. Hence, A is the limiting agent.
(iii) 100 atoms of A + 100 molecules of Y
1 atom of A reacts with 1 molecule of Y. Similarly, 100 atoms of A reacts with 100 molecules of Y. Hence, it is a stoichiometric mixture where there is no limiting agent.
(iv) 5 mol A + 2.5 mol B
1 mole of A reacts with 1 mole of B. Similarly 2.5 moles of A reacts with 2 moles of B, so 2.5 moles of A is unused. Hence, B is the limiting agent.
(v) 2.5 mol A + 5 mol B
1 mole of A reacts with 1 mole of B. Similarly, 2.5 moles of A reacts with 2.5 moles of B, so 2.5 moles of B is unused. Hence, A is the limiting agent.

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Multiplication tables

Vectors: Formula sheets

Fluid Dynamics: Formula sheets

Fluid mechanics : Formula sheets

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Vectors







 

Elasticity







 

Fluid mechanics







 

Fluid dynamics






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Some basic concepts of chemistry







 

Atomic structure






Formula sheets


 Formula sheets for class 10, JEE, NEET, [Class11, Class12] 

These formula sheets are only for revision purposes, use them to revise after you finish the reading of the chapters at least once. 
Only formulas of the chapters are given in the PDF and no concepts are explained in these sheets as in notes. Student discretion advised. 

Chemistry

11th

For JEE, NEET, and boards

Chapters uploaded:

2/20

Physics

11th

For JEE, NEET, and boards

Chapters uploaded:

3/16

Atomic structure: Formula sheet

Chapter summary

Atoms are the building blocks of elements. They are the smallest parts of an element that chemically react. The first atomic theory, proposed by John Dalton in 1808, regarded atom as the ultimate indivisible particle of matter. Towards the end of the nineteenth century, it was proved experimentally that atoms are divisible and consist of three fundamental particles: electrons, protons and neutrons. The discovery of sub-atomic particles led to the proposal of various atomic models to explain the structure of atom.
Thomson in 1898 proposed that an atom consists of uniform sphere of positive electricity with electrons embedded into it. This model in which mass of the atom is considered to be evenly spread over the atom was proved wrong by Rutherford’s famous alpha-particle scattering experiment in 1909. Rutherford concluded that atom is made of a tiny positively charged nucleus, at its centre with electrons revolving around it in circular orbits. Rutherford model, which resembles the solar system, was no doubt an improvement over Thomson model but it could not account for the stability of the atom i.e., why the electron does not fall into the nucleus. Further, it was also silent about the electronic structure of atoms i.e., about the distribution and relative energies of electrons around the nucleus. The difficulties of the Rutherford model were overcome by Niels Bohr in 1913 in his model of the hydrogen atom. Bohr postulated that electron moves around the nucleus in circular orbits. Only certain orbits can exist and each orbit corresponds to a specific energy. Bohr calculated the energy of electron in various orbits and for each orbit predicted the distance between the electron and nucleus. Bohr model, though offering a satisfactory model for explaining the spectra of the hydrogen atom, could not explain the spectra of multi-electron atoms. The reason for this was soon discovered. In Bohr model, an electron is regarded as a charged particle moving in a well defined circular orbit about the nucleus. The wave character of the electron is ignored in Bohr’s theory. An orbit is a clearly defined path and this path can completely be defined only if both the exact position and the exact velocity of the electron at the same time are known. This is not possible according to the Heisenberg uncertainty principle. Bohr model of the hydrogen atom, therefore, not only ignores the dual behaviour of electron but also contradicts Heisenberg uncertainty principle.
Erwin Schrödinger, in 1926, proposed an equation called Schrödinger equation to describe the electron distributions in space and the allowed energy levels in atoms. This equation incorporates de Broglie’s concept of wave-particle duality and is consistent with Heisenberg uncertainty principle. When Schrödinger equation is solved for the electron in a hydrogen atom, the solution gives the possible energy states the electron can occupy [and the corresponding wave function(s) (ψ) (which in fact are the mathematical functions) of the electron associated with each energy state]. These quantized energy states and corresponding wave functions which are characterized by a set of three quantum numbers (principal quantum number n, azimuthal quantum number l and magnetic quantum number ml ) arise as a natural consequence in the solution of the Schrödinger equation. The restrictions on the values of these three quantum numbers also come naturally from this solution. The quantum mechanical model of the hydrogen atom successfully predicts all aspects of the hydrogen atom spectrum including some phenomena that could not be explained by the Bohr model.
According to the quantum mechanical model of the atom, the electron distribution of an atom containing a number of electrons is divided into shells. The shells, in turn, are thought to consist of one or more subshells and subshells are assumed to be composed of one or more orbitals, which the electrons occupy. While for hydrogen and hydrogen like systems (such as He+ , Li2+ etc.) all the orbitals within a given shell have same energy, the energy of the orbitals in a multi-electron atom depends upon the values of n and l: The lower the value of (n + l ) for an orbital, the lower is its energy. If two orbitals have the same (n + l ) value, the orbital with lower value of n has the lower energy. In an atom many such orbitals are possible and electrons are filled in those orbitals in order of increasing energy in accordance with Pauli exclusion principle (no two electrons in an atom can have the same set of four quantum numbers) and Hund’s rule of maximum multiplicity (pairing of electrons in the orbitals belonging to the same subshell does not take place until each orbital belonging to that subshell has got one electron each, i.e., is singly occupied). This forms the basis of the electronic structure of atoms.

Try these questions

1)How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom? ( energy required to remove the electron from n =1 orbit).

2)What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

3)Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

4)What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?.

5)Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1 .

6)A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57µm. Calculate the rate of emission of quanta per second.

7)Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν0 ) and work function (W0 ) of the metal.

8)Using s, p, d notations, describe the orbital with the following quantum numbers. (a) n=1, l=0; (b) n = 3; l=1 (c) n = 4; l =2; (d) n=4; l=3.

10th Maths Telugu medium model-paper [2021]- with solutions

Model paper




 Varga sameekaranalu: Lectures


Lecture-1

  • Varga sameekaranalu - parichayam
  • Varga sameekaranalu - moolalu kanugonuta



Lecture-2

  • Varga sameekaranalu - moolalu kanugonuta part -2



 Varga sameekaranalu: Notes

LECTURE - 1

  • Parichayam 
  • varga sameekarana mulalu kanugonuta - paddatulu (Part - 1)

LECTURE - 2

  • varga sameekaranala yokka mulalu kanugone paddatulu (Part - 2)

LECTURE - 3

  • Abhyaasam 5.1

LECTURE - 4

  • Udaaharanalu (3 to 5) 
  • Ivi cheyandi

LECTURE - 5

  • Abhyaasam 5.2 (Que No. 1 to 5)

LECTURE - 6

  • Abhy aasam 5.2 (Que No. 6 to 10)

LECTURE - 7

  • Vargamu purthi cheyuta dwara varga sameekaranamu sadinchuta
  •  Udaharanalu : 6,7 
  •  Ivi cheyandi

LECTURE - 8

  • Udaharanalu : 8 to 13 
  •  Page number 118 to 122

LECTURE - 9

  • Abhyasam 5.3
  •  Question 1 to 3

Shredulu: NOTES

LECTURE - 1

  • Anka shredi - Parichayam 
  • Udaharanalu

LECTURE - 2

  • Abhyasam 6.1 Question no. 1 to Question no. 3

LECTURE - 3

  • Abhyaasam 6.1 Question no. 4 -(i to xiii)

LECTURE - 4

  • 6.3 AnkaShredi yokka "n"va padam 
  •  Udaharana lekkalu (Udaharana 3 nundi 10 varaku)
  •  Page 140 to 143

LECTURE - 5

  • Abhyasam - 6.2 
  • Question number 1 to Question number 5

LECTURE - 6

  • Abhyasam - 6.2 
  • Question number 6 to Question number 10

LECTURE - 7

  • Abhyasam - 6.2 
  • Question number 11 to Question number 17

LECTURE - 8

  •  Anka shredi lo modati n padala mottam
  • ivi cheyandi 
  • udaharana 11 nundi 15 varaku

LECTURE - 9

  • Abhyaasam 6.3
  • 1st and 2nd bit

LECTURE - 10

  • Abhyaasam 6.3
  • 3rd and 5th bit

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Shredulu: Lectures

Lecture - 1 

  • Anka shredi - Parichayam
  • Udaharanalu


Lecture - 2


  • Abhyasam 6.1 ( Question no. 1 to Question no. 3)



Lecture - 3


  • Abhyasam 6.1 Question no. 4 (i - xiii)



Lecture - 4


  • 6.3 AnkaShredi yokka "n"va padam
  •  Udaharana lekkalu (Udaharana 3 nundi 10 varaku)
  •  Page 140 to 143


Trikonamithi - anuvarthanalu : Notes


LECTURE - 1

  • Parichayam 
  • Udaharanalu

LECTURE - 2

  • Abhyasam 12.1 ( Que No 1 to 5)

LECTURE - 3

  • Abhyasam 12.1 ( Que No 6 to 10)

LECTURE - 4

  • Rendu Lamba-Kona tribujalatho kudina samasyalu 
  •  udaharana 6 mariyu 7

LECTURE - 5

  • Abhyasam 12.2 ( Question 1 to Question 4)

LECTURE - 6

  • Abhyasam 12.2 ( Question 5 to Question 10 )

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Trikonamithi - anuvarthanalu: Lectures


Lecture - 1 

  • Parichayam
  • Udaharanalu


Lecture - 2

  • Abhyasam 12.1 ( Que No  1 to  5)

Lecture - 3


  • Abhyasam 12.1 ( Que No 6 to  10)

Lecture - 4

  • Rendu Lamba-Kona tribujalatho kudina samasyalu  
  • udaharana 6 mariyu 7

Lecture - 5
  • Abhyasam 12.2 ( 1 to  4)


Lecture - 6

  • Abhyasam 12.2 ( 5 to 10 )

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