The following is an excerpt of the AIEEE 2010 notification. More details can be found on the official website http://aieee.nic.in/

The All India Engineering Entrance Examination for admissions into National Institutes of Technology (NIT), Deemed Universities, Technical Institutions and other Govt. funded Institutions will be held on April 25 2010

Eligibility Criteria

Nationality: Only Indian nationals are eligible to appear in the All India Engineering Entrance Examination.

Age Limit: The candidates whose age lies between 16-24 years of age on the first day of October of the year of examination. Schedule cast (SC), schedule tribe (ST) and physically handicapped (PH) candidates has been given 5 years of relaxation in the upper age limit. Date of birth as recorded in the Secondary Education Board/University certificate is considered as authentic.

Educational Qualification for appearing in All India Engineering Entrance Examination [AIEEE] is :

a. The 10+2 level examination of any recognized Central/State Board of Secondary Education Examination.
b. Intermediate or two-year Pre-University Examination conducted by a recognized Board/University.
c. Final Examination of the two-year course of the Joint Services Wing of the National Defence Academy.
d. Any Public School/Board/University Examination in India or abroad recognized by the Association of Indian Universities as equivalent to 10+2 system.
e. H.S.C. Vocational Examination.
f. A pass grade in the Senior Secondary School Examination conducted by the National Open School with a minimum of five subjects.
g. 3 or 4-year diploma recognized by All India Council for Technical Education {AICTE} or a State Board of Technical Education.

Other important dates:

To apply online http://aieee.nic.in/aieee2010/aieee/imp.htm

To get important information http://aieee.nic.in/aieee2010/aieee/InfoGlance.htm

To download AIEEE bulletin http://aieee.nic.in/aieee2010/aieee/AIEEE2010.pdf

The following is an excerpt of the IITJEE 2010 notification. More details can be found on the official website www.http://jee.iitm.ac.in/

IIT JEE 2010 Notification
IIT-JEE 2010 Entrance Exam will be held on Sunday April 11, 2010.

IITJEE 2010 Examination Schedule:
April 11, 2010
Paper 1: 09:00 – 12:00 hrs
Paper 2: 14:00 – 17:00 hrs

Eligibility Criteria for IIT JEE:

To be eligible for admission through IIT JEE a candidate should have passed class 12th or equivalent with at least 60% marks in aggregate. There is 5% relaxation for SC/STs. Besides that a candidate should not be more than 25 years as on 1st October 2009. For SC/STs a candidate should not be more than 30 years. Only those candidates who have passed 12th in 2009 or are appearing in 2010 can appear for IIT JEE.

Age Limit: 25 years (30 yrs for SC/ST candidates) on Oct. 1st 2009.

IIT JEE 2010 Exam Pattern:

Paper-1 and Paper-2 will each have separate sections in Chemistry, Mathematics and Physics. Both papers will be of objective type, designed to test comprehension, reasoning and analytical ability of candidates. Exam syllabus will be available on the websites of all IITs.

The Candidate will have the option of submitting application form either on-line or off-line.
Other important dates:

III JEE Offline Application Form 2010:

Sale of OMR IIT JEE 2010 Application Form (Off-line) and Information Brochure will start from 10:00 AM (IST) on November 16, 2009 and close at 5.00 PM (IST) on December 15, 2009.

The IIT JEE 2010 Application Forms material can be obtained from any of the designated bank branches. The cost of IIT JEE Application Form 2010 material including the examination fee is Rs 1000/- for GE/OBC/DS candidates while it is Rs 500/-for Female/SC/ST/PD candidates.

III JEE Online Application Form 2010:

For details about online application process, check out the below link.

http://jee.iitm.ac.in/online2010/main.html

Candidates can apply IIT JEE Online Application Form, if they have access to computer, printer and Internet at home or elsewhere, by logging on to JEE website. The site has been designed in a ‘user friendly’ way to help candidates to apply IIT JEE Online Form in a step-by-step fashion. IIT JEE Online Application portal will be open from 8:00 AM (IST) on November 1, 2009 to 5:00 PM (IST) on December 7, 2009.

The examination fee is Rs 900/- for GE/OBC/DS candidates and Rs 450/- for SC/ST/PD and female candidates. Candidates can make payment On-line with credit/debit cards of major banks. Payment can also be made through challans of designated banks.

The envelope containing the completed OMR IIT JEE 2010 Application Form and enclosures should be sent to the IIT located in the zone where you wish to appear for the examination (irrespective of the institute at which you want to seek admission).

The contact addresses of the zonal IITs to which you should send your envelope with the IIT JEE 2010 Application are given below.

The envelope must be sent by Registered Post/ Speed Post only. Do not send it through ordinary post or private courier services. Please keep the postal receipt for future reference.

One can also submit in person the completed IIT JEE 2010 Application Form at any of the JEE offices at the IITs. IIT JEE Application Forms in IIT Delhi zone may be submitted in person at Union Bank of India, SDA Branch, Hauz Khas, New Delhi – 16.

In IIT Madras Zone, the completed 2010 IIT JEE Application Form can be submitted at the same branch of Indian Bank from where you have purchased the form.

The last date for receipt of the completed Off-line IIT JEE application at the IITs is 5:00 PM on December 19, 2009. Any application received after this date will not be accepted. Office of JEE is not responsible for any postal delay or irregularity or loss during postal transit.

Helpline for IIT JEE 2010 Examination : 044-42925020 & 044-42925021

Concept of friction:

Component of normal reaction

We have seen that a ball rolling on a floor stops after some time. When we switch off the engine of a car, it stops after traveling some distance. Similarly when we apply brakes, our bicycle comes to rest after traveling some distance. The above examples show that some invisible force is opposing the motion of one body over the other. This opposing force is called friction. Friction is an opposing force that comes into play when one body actually moves (slides or rolls) or even tries to move over the surface of another body.

Cause of friction

Roughness of surfaces is the cause of friction.

When two bodies are in contact with each other, the irregularities on the surfaces get interlocked and oppose any relative motion.

Normal reaction

When a body of mass m is lying on a horizontal surface, it presses the surface due its weight. Contact force F_c = R in case no external force is acting on the body. Also in this case friction is zero and F_c is perpendicular to the surface. In the diagram below, R is the normal reaction.

Static and kinetic friction

The opposing force that comes into play when one body tends to move over the surface of another (but the actual motion has not yet started) is called static friction. Limiting friction is the maximum opposing force that comes into play when one body is at the verge of moving over the surface of another body. It is denoted by f_s, and is called maximum force of static friction or limiting friction.

Laws of friction

The following are the laws of limiting friction:

(a) The magnitude of the force of limiting friction f_s, is directly proportional to the normal reaction R.

(b) The direction of the force of limiting friction is always opposite to the applied force.

(c) It is independent of the apparent area of contact.

(d) It depends on the nature and material of the surfaces in contact.

For example, when two polished metal surfaces are in contact,  = 0.2

When these surfaces are lubricated,  gets reduced. Hence, it depends on the nature of the surfaces.

Rolling friction

Rolling friction is always less than dynamic and static friction. When a body rolls on a level track, the area of contact is very small. This causes a depression in the surface below. This causes rolling friction.

The velocity of the point of contact of the wheel with respect to the floor remains zero all the time. Thus, rolling wheel constantly climbs a hill and has to simultaneously get itself detached from the road. Rolling friction is also directly proportional to the normal reaction and is inversely proportional to the radius of rolling body. Combining the two, we get,

F is directly proportional to R/r

F = μx (R/r)

Where μx is called coefficient of rolling friction.

Advantages and Disadvantages of friction

Friction is a necessary evil. It is necessary because we cannot do any work without it. At the same time, it is also an evil because it involves unnecessary wastage of energy.

There are advantages of friction. It can be understood with the help of the following examples:

Walking will not be possible without friction.

Brakes of vehicles will not work without friction.

Writing on a blackboard or a piece of paper is possible only due to friction between the blackboard and the chalk or the paper and the pen. Cleaning with sand paper will not be possible without friction.

There are disadvantages of friction. Friction is an evil and this can be understood with the following   examples:
Extra energy is required to overcome the friction between moving parts.

ü       Friction causes wear and tear of different parts of machinery.

ü       Frictional forces result in the production of heat, which causes damage to the machinery.

Methods of reducing friction

By polishing, by lubrication, By streamlining, By using ball-bearings,

Angle of friction

It is the angle between normal Reaction and the resultant of the normal reaction and limiting frictional force. i.e. we have a body of mass m which is placed on a table and we say that the body and the surface of the table have a coefficient of friction m between them. (fig. 3.7)

If we apply a small force F, the body will not move. Let us gradually increase the force until the body starts moving. At one stage the applied force will be equal to the frictional force. The coefficient of static friction ms  = F/N where F is the applied force and N = mg is the normal force.

Orthocentre is the point of concurrence of the altitudes of a triangle. We denote it by H. The triangle obtained by joining the feet of the altitudes is called pedal triangle. The position vector of the orthocentre is given by

It is possible to find distance of orthocentre from vertices and sides of a triangle.

Let H be the orthocentre

In the figure, we observe that

HK = BK tan ∟HBK = BK tan ∟LBC

HK = BK tan (90°-C) (in ΔBHK)

= AB cos B cot C

=c cos B cos C/ sin C = (c/sin C) cos B cos C

= 2 R cos B cos C

Similarly, we can prove that

HL = 2R cos A cos C

and       HM = 2R cos A cos B

Now, AH = AL sec (90 – C)

= AB cos A cosec C

= (c/sin C) cos A = 2 R cos A

Similarly we get BH = 2R cos B, CH = 2R cos C

Thus, distances of orthocentre from sides a, b, c, are given by
2 R cos B cos C, 2R cos A cos C, 2 R cos A cos B respectively and from vertices A,B,C are given by 2R cos A, 2R cos B and 2R cos C  respectively.

ΔKLM is pedal triangle for ΔABC

In quadrilateral BKHM,

∟BMH +∟ BKH= 90° + 90° = 180°

Þ         BKHM is a cyclic quadrilateral

Þ         ∟BMK = ∟BHK = 90° – ∟HBK

= 90° – ∟LBC

= 90° – (90° – C) = C

=> ∟KMH = 90° – C

Similarly, we can prove that

∟HML = 90° – C

=> ∟KML = 180° – 2 C

Similarly

∟KLM = 180° – 2B and ∟MKL = 180° – 2A

i.e. angles of pedal triangle are supplement of double of opposite angles of the original triangle

Now, in Δ AML,

Similarly, it can be shown that KM = b cos B, KL = c cos C

i.e. sides of pedal triangle are a cos A, b cos B, c cos C.

Illustration

1. MKL is the pedal triangle of ABC; prove that its area is

2 S cos A cos B cos C, where S is area of ΔABC.

Various kinds of substances that constitute matter can be roughly divided into three categories namely- gases, liquids and solids. The existence of matter in either of these three forms is a result of the competition between two opposing intermolecular forces

(1)     The forces of attraction, which hold the molecules together.

(2)     The thermal energy of these molecules which tend to increase the intermolecular distances.

If the thermal energy of the molecules is much greater than the forces of attraction, the state of matter that result is called the gaseous state. On the other hand, if the forces of attraction are greater than the thermal energy, we have the matter in the liquid state. When these forces of attraction are much more greater than the thermal energy compared to the liquid state, we have matter in its solid state condition. However, on the application of heat, the thermal energy of the molecules can be increased and as a result the intermolecular forces of attraction would relatively decrease simultaneously.

Molecules in the gaseous state possess high energy and have almost no force of attraction. They are far apart and show a great uniformity in behaviour, irrespective of their chemical nature, colour or odour. They are highly compressible and can also be expanded without limit. They also produce pressure on the walls of any container uniformly in all directions. They diffuse rapidly through one other to form a homogeneous mixture, and their separation is also not very easy.

Gas laws

Boyle’s law

At constant temperature, the volume of a sample of gas of definite mass varies inversely with its pressure.

i.e    when temperature is kept constant for a given mass of gas

where V = volume and  p = pressure

Introducing a constant k, we have

pV = k = constant                                …(1)

The value of the proportionality constant k depends upon the following factors:

(1)           Nature of a gas

(2) Temperature of the gas, and

(3)           The mass of the gas

Hence, at constant temperature, for a given mass of a gas, Boyle’s Law states

p₁V₁ = p₂V₂ = k = constant                  …(2)

Eq. (1) can be represented graphically as shown in figure given below.

The general term isotherm (i.e. at const. Temp.) is used to describe the above curves.

Have you ever realized that whenever you are sitting alone somewhere, actually, you are not alone (surprised). Well, you are always surrounded by air around you. In fact, air surrounds everything like a blanket. Now, what is air? Well, air is nothing but a mixture of gases.  The gaseous state results when the forces of attraction between the particles of matter are very low. In this state the molecules are far apart from one another and their positions are not fixed. Hence gases have neither definite shape nor definite volume but a gas occupies fully its container. It was correctly stated by Robert Boyle “Imagine the air to be such a heap of little bodies, lying one upon another, as may be resembled to fleece of wool.”

OBJECTIVE

Gaseous state is the simplest state of matter and shows the greatest uniformity in behaviour. Let us in this chapter know more about this ever accompanying companion. In this chapter we will study the behaviour of different gases at different physical condition of pressure and temperature. The study includes different laws related to behaviour of gases e.g., Boyle’s law, Charles’ law, Ideal gas law etc. to comprehend the ideal gas equation. We shall also

• know the meaning of absolute scale of temperature
• learn about Dalton’s law of partial pressures and Graham’s law of diffusion
• understand the kinetic gas equation and average kinetic energy of gaseous molecules
• learn the definition of root-mean-square, average and most probable speeds
• learn about the compression factor ?!
• know about van der Waals equation of state

PRE-REQUISITE

Relative atomic mass of an element

The ratio of average mass per atom of the natural isotopic composition of the element to of the mass of an atom of nuclide ¹²C is known as the relative atomic mass of the element.

Relative molecular mass of Compound

The ratio of average mass per molecule of the natural isotopic composition of the compound to  mass of an atom of nuclide ¹²C is known as relative molecular mass of the compound.

Mole of a substance

One mole of a substance contains as many particles (atoms, molecules, ions) as there are atoms in exactly 12 gm of the nuclide ¹²C. This number is approximately equal to 6.023 × 10²³.

or

The minimum weight of any chemical species which contains 6.023 × 10²³ molecules is called one mole.

or

At NTP or STP (Normal/standard temperature and pressure are 0°C and 1 atm, respectively) one mole occupies 22.4 litre of volume.

Newton’s third law of motion was discovered and formulated, during the investigation of the fact that in all experiments it appeared that “when ever a body exerts a force on a second body, the second body always exerts a force on the first one”

Let us visualize and understand this phenomena with an experiment.

Suppose, we throw a stone on a surface of good strength; and the surface is made of glass, one finds it broken (the surface). From here one concludes that a force was exerted by stone on the surface and consequently it was broken.

Now, the question is, did that surface also exert a force on the stone. Just to know about it let us change our throwing object from stone to an egg of almost equal mass. Now, one throws this egg on the same surface of good strength with the same throwing force which he used for the stone. What happens? Obviously with your daily experience you know that the egg will be broken (And the damage to the surface will not be visible due to egg’s spoiling the observation).

This is only possible if there was a force acting on the egg at the time it hit the surface. In fact we can now conclude that there is a mutual force acting on the contact point of the surface and the object thrown. The breaking of either one (or may be both) depends on their ability to absorb forces without getting damaged (that is their strength) so in precise words:

Þ  To every action there is always opposite and equal reaction, it is equivalent to say that mutual actions of two bodies upon each other are always equal and directed to contrary parts.

Þ  The most important fact to notice here is that these oppositely directed equal action and reaction can never balance or cancel each other because they always act, on two different point (broadly on two different objects) For balancing any two forces the first requirement is that they should act on one and the same object. (or point, if object can be treated as a point mass, which is a common practice).

Illustrations

1.      Suppose in figure 3.2 we put one more block of 5 kg mass adjacent to 10 kg and a force of 150 N acts as shown in the figure 3.3, then find the forces acting on the interface.

Sol.   The combined acceleration of the two bodies when treated as one is

a= F/ (10+5) = 150 / 15 = 10m/sec²

So each one moves with a = 10m/sec² keeping their contact established.

Here you can feel that due to 150N force the body of 5kg feels as if it is being pushed by the 10 kg mass. There is a force acting on 5kg called R₁, to oppose it by third law this body exerts a force R₂ on 10 kg. The interface is as shown in figure 3.4.

Also, third law tells us that R₁ = R₂ in magnitude and is opposite in direction. (figure 3.5)

R₁ = R₂ = R

Here since 150 N force acts on the 10 kg mass and only R acts on the 5kg mass. For motion in 5 kg only R is responsible. We can write the initial equation as:

F = 150 = (10+5) a           i.e., 150 = 10a + 5a

Here 10a is force experienced by 10 kg mass and 5a is experienced by 5 kg mass.

R = 5a      a = 10m/sec² =>           R = 50 N

Net force experienced by 10kg block is (150 – R) = 10a

150 – R = 10 × 10 = 100 N            =>     R = 50 N

Therefore we get R = 50N for both blocks. Hence we find “action and reaction are equal and opposite”. Now net force on the body of 10 kg mass is 100 N & Net force on the body of 5kg mass is 50N and on the interface action and reaction are both equal and also are equal to force experienced by second body.

Circumcircle:

Circumcircle is a circle circumscribing the triangle i.e. passing through the three vertices of the triangle. Its center is naturally found to be at the point of intersection of perpendicular bisectors of the sides of the triangle. Its centre is usually denoted by O and it’s radius by R and called circumradius. It may happen that the circumcentre is lying out side the triangle as any angle may be out use. If O’ denotes the origin, then the position vector of circumcentre O is

Incircle

A circle touching all the three sides of a triangle is said to be incircle. It’s centre lies at the point of intersection of angle bisectors of the angles of the triangle. It’s centre is denoted by I and radius by r and called inradius. This is a circle inscribed in a triangle. Its centre can never lie outside the triangle. Its position vector is given by

Escribed circle

It is a circle touching on of the three sided externally and the rest two extended sides internally. Obviously, every triangle will lead to three escribed circles. The circle opposite to vertex A gets it’s centre denoted by I₁ and radius by r₁ called exradius. Similarly, circles opposite to vertex B and C get their centers and radii denoted by I₂, r₂ and I₃, r₃ respectively. Obviously its centre will lie on the point of concurrence of two external angle bisectors and one internal angle bisector. The position vector of the escribed centre I₁ is given by

It states that rate of change of momentum of a body is equal to the force applied on it, in terms of the magnitude as well as in the sense of direction. Here the momentum is defined as the product of mass and velocity i.e. .

Therefore we can write mathematically.

F = d(mv)/ dt, if ‘m’ remains constant then

Thus acceleration is rate of change of velocity.

Since direction of a is same as F, we can write

F = ma , which is mathematically Newton’s second law of motion.

Here, if F = 0 then we find a = 0. This reminds us of first law of motion. That is, if net external force  is absent, then there will be no change in state of motion, that means its acceleration is zero.

Further we can extend second law of motion, (in fact its decomposition) to three mutually perpendicular directions as per our coordinate system.

If components in x,y, and z directions are Fx, Fy, & Fz respectively, the three accelerations produced when Fx, Fy, & Fz act simultaneously) in the body are ,  Now,

If we add three forces then resultant is called net external force.

Similarly

is called net acceleration produced in the body.

Illustrations

1.             In Figure 3.1. Let us have M = 10 kg and a new net external force in the direction as shown in figure 3.2 is 150 N. Find its acceleration.

Sol.

The discovery of the laws of motion was a dramatic moment in the history of science. Before Newton’s time, the motions of things like the planets were a mystery, but after Newton there was complete understanding.

The contribution of Newton was three laws:

Þ      The first law describes what happens when the body is left alone or is not disturbed.

Þ      The second law gives a specific way of determining how the velocity changes under different influences called forces.

Þ      The third law characterizes the basic nature of force.

OBJECTIVE

In this chapter we will study the three important laws of motion given by Newton. The study includes the dynamics of particle under influence of different forces.

After studying this chapter we will be able to answer the question like Why can’t we drive fast on icy road, in cricket how hard should a player hit the ball for a sixer or how does a small engine pull a large train and many other similar questions.

PREQUISITE

We know by experience that all bodies in nature interact in some way with one another.

Before actually going into details of the three Laws of motion we need to get familiar with  “force”. It was believed and has been accepted that “force” is an external or internal agent present to “influence” the natural state of motion of an object. So this is an influence (force) needed to change the natural state of body; that is of rest or of uniform motion.

Having this idea of force we can now readily move to the Newton’s first law of motion.

Newton’s first law of motion

It states that every object persists in its natural state of motion i.e. continues to be at rest or moves in a straight line with uniform (constant) velocity. (This is what is meant by natural state of motion); In the absence of a net external force acting (impressed) on it.

Mathematically, it is equivalent to say that for producing acceleration (that is for changing velocity) in a body, we need to have a net external force. (By net external force we mean vector sum of all the external forces acting on it)

It can be easily deduced from the statement of change in the state of motion. It is directly related to a frame of reference about which we have discussed earlier. To mark the point here, we can discover that by viewing objects from different frame of references the natural state of motion as perceived by different observers will be obviously different (can only be same if the frames are truly equivalent). Therefore, the change in state will also depend on the choice of reference frame. Finally, the amount of acceleration produced in a body (or change in velocity) will depend on our choice of reference frames.

A mass ‘M’ is lying (figure 3.1) on a table which is at rest (w.r.t. table). Explain its state with the help of Newton’s First Law of motion.

Since ‘M’ is lying on a table, there is no external force acting on it (forget about gravity just for the immediate discussion). As per Newton’s first law of motion it will keep on lying at rest with respect to table for infinite time.

Here, comes out a very important, intrinsic (that is inherent) property of a body which is that it retains its state of motionlessness (as well as of motion, if its in motion) which is termed as INERTIA of an object. This is present in all materialistic bodies in this universe.