Encyclo Entrance

Army Engineering For Boys

2013-11-26T17:56:00.001-08:00

* Courses starting on July 2014

* Registration ending on 29th November 2013

Army engineering courses are for 5 years and could enter in a rank of left and officers. At least 70% mark required in the subjects Physics,chemistry and maths in +2. Going +2 students can apply this.Students those who are between 16.5 & 19.5 age could apply for this exam.

Selection

Eligible students are asked to face the five day service selection board interview.That is not only an interview, they check mental growth and each and every personality.All types of medical tests are required.

Advantages

After 3 year a stiff of RS 21000 per month will be rewarded.After the courses a job of salary RS 27000 and other privileges will be provided.Salary and service all together will award a person RS 65000.There are promotions based on your performance.

Registration

http://www.joinindianarmy.nic.in/ Apply online for TES 31 course.2 copy of application must be printed.Photo and signature must be printed on it.Attested 12th and 10th examination certificate (going 12th student only 10th examination certificate) must be keep.During the interview this must be taken along with you.

JOIN INDIAN ARMY

HOW TO FILL ONLINE APPLICATION FORM
(IMPORTANT INSTRUCTIONS)

	Only Male Candidates to Apply
	TES :-
	Age Eligibility Criteria: “A Candidate must not be below 16 ½ years and above 19 ½ yrs on the first day of the month in which the course is commencing i.e., the candidate should not be born before 01 Jan 1995 and not after 01 Jan 1998 (both days inclusive) only are eligible to apply”.
	TGC :-
	Age Eligibility Criteria: “A Candidate must not be 20 years and above 27 yrs on the first day of the month in which the course is commencing i.e., the candidate should not be born before 02 July 1987 and not after 01 July 1994 (both days inclusive) only are eligible to apply”.
	AEC :-
	Age Eligibility Criteria: “A Candidate must not be below 23 years and above 27 yrs on the first day of the month in which the course is commencing i.e., the candidate should not be born before 02 July 1987 and not after 01 July 1991 (both days inclusive) only are eligible to apply”.
	Education Qualification Eligibility Criteria:
	TES-31 :- Candidates scoring Aggregate % of Marks 70% and above only are eligible.
	TGC-119 :- BE/B.Tech/B.Arch Degree in the notified streams. Application of Cut off % is at the discretion of this office.
	AEC :- MA/M.Sc in 1^st or 2^nd Division in Notified Subjects.

Click on ONLINE APPLICATION Link.
To fill up ONLINE APPLICATION form, the Candidate has to first sign up for Registration.
On Clicking Sing Up tab, Sign Up window appears and candidate is required to fill all the details.
Every Candidate must select own unique User Name and Password (Not more than 10 characters)
After filling all the tabs,Click Submit.
Welcome window appears displaying User Name and Link for Online Application for TES 31 Course.
Click Online Application Form for TES 31 Course and then a new window appears displaying the actual Online Application Form.
Online Application Form comprises of four parts (a) Personal, (b) Communication, (c) SSB, (d) Academic.
Fill in all the fields in respective parts of Online Application and Click Save.
On clicking save Tab, next parts of the Online Application Form appear on the screen and repeat the same procedure to fill all four parts of Online Application Form.
Thereafter First Click Save and then Click on Submit Tab.
In case you want to check or edit any information filled in any part of the application Form, please don’t submit and Log out.
Login again by clicking Online Application Link by using same User Name and Password used at the time of Registration.
Online Application Form as filled in by you will appear on the screen and the same can be edited. After editing, click Save and then Submit. Once submitted the Application can’t be edited.
Roll No will be auto generated and your Online Form in PDF Format with complete details as filled in by you will appear on your screen.
Take printout of the PDF File and in case you want to take print out later then save the same in your PC.

BORON FAMILY

2013-11-08T17:12:00.000-08:00

//Here we are going to study more about boron family in p-block

These are the main elements in Boron family

Now lets study more detailed about boron family.

*Boron is a well known metal and it always form covalent bonds. It has 5 electrons which revolve around nucleus. Its atomic number is 5.

Atomic size

There is always a increase in atomic size due to the increase in number of shells. But as an exception there is a small decrease in size of aluminium to gallium. This is observed due to the poor screening effect of 10 d electrons,and there is a considerable increase in size from In to Tl due to the poor screening effct of 14 f electrons.

Ionisation enthalpy

The Solid State

2013-11-06T08:39:00.000-08:00

General Characteristics of Solid State

(i) They have definite mass, volume and shape.

(ii) Inter molecular distance are short

(iii) Inter molecular forces are strong

(iv) Their constituent particles (atoms, molecules or ions) have fixed positions and can only oscillate about their mean positions

(v) They are incompressible and rigid

Solids can be classified as crystalline or amorphous on the basis of the nature of order present in the arrangement of their constituent particles.

Property	Crystalline Solids	Amorphous Solids
Shape	Definite characteristic geometrical shape.	Irregular shape.
Melting Point	Melt at a sharp and characteristic temperature	Gradually soften over a range of temperature
Cleavage Property	When cut with a sharp edged tool, they split into two pieces and the newly generated surfaces are plain and smooth	When cut with a sharp edged tool, they cut into two pieces with irregular surfaces
Heat of Fusion	They have a definite and characteristic heat of fusion	They do not have definite heat of fusion
Anisotropy	Anisotropic in nature	Isotropic in nature
Nature	True solids	Pseudo solid or super cooled liquids
Order in arrangement of constituent particles	Long range order	Only short range order

Examples

Sodium chloride and Quartz

Glass, Rubber and plastics

Classification of crystalline solids

1) Molecular solids

(i) Non polar :- Dispersion or London force | Ar , CCl4 , H2 , I2 , CO2

(ii) Polar :- Dipole-dipole interactions | HCl , SO2

(iii) Hydrogen bonded :- Hydrogen bonding | H2O (ice)

2) Ionic solids :- Coulombic or electrostatics | NaCl , MgO , ZnS , CaF2

3) Metallic solids :- Metallic bonding | Fe , Cu , Ag , Mg

4) Covalent or network solids :- Covalent bonding | SiO2(quartz) , SiC , C(diamond,graphite) ,Hard Insulators Very high AIN

Crystal Lattices

A regular three dimensional arrangement of points in space

Bravais Lattices

There are only 14 possible three dimensional lattices.These are called Bravais Lattices

Lattice Point or Lattice Site

Each point in a lattice is called Lattice Point or Lattice Site

Magnetic Effect of Current

2013-11-05T07:47:00.001-08:00

Magnetic Effect of Current

Magnetic Field

The space around a current carrying conductor in which its magnetic effect can be experienced is called magnetic field.

Biot-Savart’s Law

The law states that “the magnetic field at any point due to an element of a conductor carrying a current is

(i) Directly proportional to the strength of current

(ii) Directly proportional to the length of the element

(iii) Directly proportional to the sinθ , where 'θ' is the angle between the element and the line joining the element to the point

(iv) Inversely proportional to the square of distance between the element and the point

dB ∝ idlsinθ/r²

dB=(µo)/4π x idlsinθ/r²

Unit: - Wb/m² or tesla

µo Is called permeability of free space

µo = 4π x 10000000

Right hand Grip Rule

It states that “If conductor carrying current is held in the right hand (grasped) such that the thumb points in the direction of the current , then the direction of finger tips round the wire represents the direction of the magnetic field lines.

Maxwell’s right handed screw rule

If a right handed cork screw is rotated so that its tip moves in the direction of flow of current through the conductor, then the rotation of the head of the screw gives the direction of magnetic lines of force.

Right Hand Palm Rule

The rule states that “If the direction of the circular current is represented by the direction of the closed fingers of the right hand, then the stretched thumb will give the direction of the magnetic field”.

Ampere’s circuital Law

Theorem states that “the line integral of the magnetic field (B) around any closed path in free space is equal to µo times the total current enclosed by the path.”

∫B x dl=µo(i net)

Remaining topics will be added soon

Short Notes on Electro Statics

2013-11-04T08:27:00.000-08:00

Short Notes on Electro Statics

Number of electrons =(Total charge)/(Charge of one electron)

n=q/e

q=ne

Coulomb’s Law

“It states that two point charges repel or attract each other with a force which is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between them.”

F=1/4Πεο×q1q2/r²

Linear charge density:- λ=q/L

Surface charge density:- σ=q/A

Volume charge density:- ρ=q/V

In Coulomb’s law the charges if placed in another medium,

Then,

F medium = F air / K = 1/4Πεο×1/K×q1q2/r²

Where K is called dielectric constant of medium

E=F/q

F=qE

Electric field due to a point charge

Consider a single point charge ‘q’ , Let ‘r’ be the vector from q to a general point P where the test charge qο is placed.

By Coulomb’s law,

F=(1/4Πεο)x(qqο/r²)x r^

E=q/qο

E=(1/4Πεο)x(q/r²)x r^

The electric field vector points radially outwards if q > 0 and radially inwards if q < 0.

Electric field inside a conductor is zero.

Dipole moment

→ →

P = q(2a)

Electric field on the axial line of an electric dipole

E=[1/4Πεο]×[(2Pr )/(r²-a²)²]

E=[1/4Πεο]×[(4qra )/(r²-a²)²]

When a<<r

Then ,

E=[1/4Πεο]×[2P/r³]

Electric field on equatorial line

E=[1/4Πεο]×[P/(r²+a²)^3/2]

When a<<r

Then ,

E=[1/4Πεο]×[P/r³]

Torque of the dipole in the uniform electric field,

→ →

ζ=P × E

Electric Flux

φ=∫E.ds

Inverse Trigonometric functions

2013-10-25T05:20:00.002-07:00

Domain and range of inverse functions:

Function Domain Range

Sin^-1x [-1,1] [-π/2,π/2]

Cos^-1x [-1,1] [0,π]

Tan^-1x R [-π/2,π/2]

Cot^-1x R [0,π]

Sec^-1x R-[-1,1] [0,π]-{π/2}

Cosec^-1x R-[-1,1] [-π/2,π/2]-{0}

COMPOSITION OF TRIGONOMETRIC FUNCTIONS AND THEIR INVERSE:

Sin^-1(sinX)=cos^-1(cosX)=tan^-1(tanX)=X

{provided 'X' is in the domain of the function.

SUM OF TWO INVERSE FUNCTIONS:

Sin^-1X+Cos^-1X=π/2

Tan^-1X+Cot^-1X=π/2

Sec^-1X+Cosec^-1X=π/2

RELATION BETWEEN TWO FUNCTIONS

Sin-1(1/x)=Cosec-1(x)

Cos^-1(1/x)=Sec^-1(x)

Tan^-1(1/x)=Cot^-1(x)

Encyclo Entrance

Army Engineering For Boys

BORON FAMILY

Atomic size

Ionisation enthalpy

The Solid State

Magnetic Effect of Current

Short Notes on Electro Statics

Inverse Trigonometric functions

Domain and range of inverse functions:

Function Domain Range

Sin-1x [-1,1] [-π/2,π/2]

Cos-1x [-1,1] [0,π]

Tan-1x R [-π/2,π/2]

Cot-1x R [0,π]

Sec-1x R-[-1,1] [0,π]-{π/2}

Cosec-1x R-[-1,1] [-π/2,π/2]-{0}

COMPOSITION OF TRIGONOMETRIC FUNCTIONS AND THEIR INVERSE:

SUM OF TWO INVERSE FUNCTIONS:

RELATION BETWEEN TWO FUNCTIONS

Function Domain Range

Sin^-1x [-1,1] [-π/2,π/2]

Cos^-1x [-1,1] [0,π]

Tan^-1x R [-π/2,π/2]

Cot^-1x R [0,π]

Sec^-1x R-[-1,1] [0,π]-{π/2}

Cosec^-1x R-[-1,1] [-π/2,π/2]-{0}