Joint Entrance Examination (JEE) 2013 for Admission to Undergraduate Engineering Programmes in IITs, NITs and other Technical Institutions

Central Board of Secondary Education has notified that admission to Undergraduate Engineering Programmes in IITs, NITs and other Centrally Funded Technical Institutions will be based on a Joint Entrance Examination (JEE) with effect from 2013.

JEE will be conducted in two parts – JEE (Main) and JEE (Advanced).

JEE (Main) will replace AIEEE and JEE (Advanced) will replace IIT-JEE, with the important criterion that only the top 150,000 candidates performing in JEE (Main) will qualify for appearing in JEE (Advanced). Admission to IITs, IT-BHU and ISM Dhanbad will be based only on category-wise All India Rank in JEE (Advanced), subject to the additional condition that such candidates are in the top 20 percentile of successful candiates in Class XII/equivalent examination conducted by their Boards.

For admission to Undergraduate Engineering Programmes in NITs and other Centrally Funded Technical Institutions the rank list will be prepared based on 40% weightage to school board marks in Class XII and 60% weightage to JEE (Main) marks.

JEE (Main) 2013 Paper I (for admission to B.E./B.Tech courses) which replaces Paper I of earlier AIEEE, will be of 3 hours, consisting of objective type questions from Physics, Chemistry and Mathematics.

Paper II for admission to B.Arch/B.Planning courses also will be of 3 hours and will consist of Mathematics, Aptitude Test and Drawing Test as per past practice of AIEEE.

The JEE (Main) (Paper I of earlier AIEEE) for B.E./B.Tech will be held in two modes, viz., offline and online (CBT). The offline examination for JEE (Main) will be conducted on 7^th April, 2013 and the online examinations will be conducted thereafter in April, 2013.

For details you may visit http://jeemain.nic.in 

Apply for All India Engineering Entrance Examination 2012 (AIEEE 2012)

It’s time to apply for AIEEE 2012. Applications for the examination are to be submitted online only, from 15-11-2011 to 31-12-2011.

There are two modes of examination viz., offline examination (Pen & Paper Mode) and online examination (Computer Based Test) with dates as given below:

Offline Examination - 29th April, 2012

Online Examination- 7th May, 2012 to 26th May, 2012

You may visit the site http://aieee.nic.in for details and information updates.

You will find many old AIEEE questions (with solution) on this blog. You can access all of them by trying a search for ‘AIEEE’ making use of the search box on this page or by clicking on the label ‘AIEEE’ below this post. Make use of the ‘older posts’ button to navigate through older posts related to AIEEE.

AIEEE 2011 Question (MCQ) on Magnetic Field due to Current Carrying Conductors

"Seven Deadly Sins:

Wealth without work

Pleasure without conscience

Science without humanity

Knowledge without character

Politics without principle

Commerce without morality

Worship without sacrifice."

– Mahatma Gandhi

The following multiple choice question included in the All India Engineering/Architecture Entrance Examination (AIEEE) 2011 will be interesting and useful to you:

A current I flows in an infinitely long wire with cross section in the form of a semi-circular ring of radius R. The magnitude of the magnetic induction along its axis is

(1) μ₀ I/4πR

(2) μ₀ I/π²R

(3) μ₀ I/2π²R

(4) μ₀ I/2πR

It would have been better if the magnitude of the magnetic induction at the axis was asked for since the words along the axis usually means directed along the axis. [The component of field directed along the axis is zero].

Forget about it. The question setter requires you to calculate the magnetic flux density at points on the axis (of the semicircular ring shaped cross section of the infinitely long wire). In the adjoining figure we have shown the cross section (of the given wire) lying in the XY plane. The length of the wire is along the Z-axis and the current in the wire is supposed to flow along the negative Z-direction. The broad infinitely long wire can be imagined to be made of a large number of infinitely long straight wire strips, each of small width dℓ.

With reference to the figure, we have dℓ = Rdθ.

The magnetic flux density due to the above strip is shown as dB₁ in the figure. It has an X-component dB₁ sinθ and a Y-component dB₁cosθ. When we consider a similar strip of the same with dℓ located symmetrically with respect to the Y-axis, we obtain a contribution dB₂ to the flux density. The flux density dB₂ has the same magnitude as dB₁. It has X-component dB₂ sinθ and Y-component dB₂cosθ. The X-components of dB₁ and dB₁ are of the same magnitude and direction and they add up. But the Y-components of dB₁ and dB₁ are in opposite directions and have the same magnitude. Therefore they get canceled. The entire conductor therefore produces a resultant magnetic field along the negative X-direction.

The wire strip of width dℓ cn be imagined to be an ordinary thin straight infinitely long wire carrying current Idℓ/πR since the total current I flows through the semicircular cross section of perimeter πR. Putting dB₁ = dB₂ = dB we have

dB = μ₀ (Idℓ/πR)/2πR = μ₀ (IRdθ/πR)/2πR = μ₀Idθ/2π²R

The X-component of the above field is (μ₀I/2π²R) sinθ dθ

The field due to the entire conductor is B = ₀∫^π [(μ₀I/2π²R) sinθ]dθ

Or, B = μ₀I/π²R since ₀∫^π sinθ dθ = 2

Apply for All India Engineering/Architecture Entrance Examination 2011 (AIEEE 2011)

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2011 (AIEEE 2011) to be conducted on 24-4-2011 will be distributed from 15.12.2010 and will continue till 14.1.2011. Candidates can apply for AIEEE 2011 either on the prescribed Application Form or make application ‘Online’.

Online submission of the application is possible from 23-11-2010 to 14-1-2011 at the website http://aieee.nic.in

Visit the site http://aieee.nic.in for all details and information updates.

You will find many old AIEEE questions (with solution) on this site. You may search for ‘AIEEE’, using the search box at the top of this page, to access all posts in this context.

AIEEE 2010 Questions (MCQ) from Nuclear Physics

“I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend”

– Sir Isaac Newton

Multiple choice questions based on a given paragraph are seen in many entrance exam question papers. Here are two such questions on nuclear physics (which appeared in AIEEE 2010 question paper):

Directions : Questions (i) and (ii) are based on the following paragraph:

A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.

(i) The speed of daughter nucleus is

(1) c√[Δm/(M + Δm)]

(2) c [Δm/(M + Δm)]

(3) c√(2Δm/M)

(4) c√(Δm/M)

The energy released in the process is Δmc² in accordance with Einstein’s mass-energy relation. This energy is carried by the daughter nuclei (as kinetic energy).

Therefore we have

½ (M/2)v² + ½ (M/2)v² = Δmc² where v is the speed of each daughter nucleus.

Or, Mv²/2 = Δmc² from which v = c√(2Δm/M), as given in option (3).

[Note that the momentum is to be conserved in the process. Therefore, the daughter nuclei have to travel in opposite directions with the same speed since they have the same mass. If you were asked to give the velocities of the daughters, the answer would be v and – v]

(ii) The binding energy per nucleon for the parent nucleus is E₁ and that for the daughter nuclei is E₂. Then

(1) E₁ = 2 E₂

(2) E₂ = 2 E₁

(3) E₁ > E₂

(4) E₂ > E₁

The binding energy per nucleon for the daughter nuclei is greater than that for the parent nucleus since energy is released in the process[Option (4)].

The following question (MCQ) also was included in the AIEEE 2010 question paper:

A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

(1) (A – Z – 4)/(Z – 2)

(2) (A – Z – 8)/(Z – 4)

(3) (A – Z – 4)/(Z – 8)

(4) (A – Z – 12)/(Z – 4)

When an α-particle is emitted, the nucleus loses two protons and two neutrons. When a positron is emitted by the nucleus, a proton in the nucleus gets converted into a neutron. Since the number of neutrons in the original neucleus is A – Z, the number of neutrons in the final nucleus will be (A – Z) – (3×2) + 2 = A – Z – 4.

The protons in the final nucleus will be Z – (3×2) – 2 = Z – 8

Therefore, the ratio of number of neutrons to that of protons in the final nucleus will be(A – Z – 4)/(Z – 8), as given in option (3).

Alternating Current Circuits – Two Questions (MCQ) on Resonance

-->

"We have to do the best we can. This is our sacred human responsibility."

–Albert Einstein

The phenomenon of resonance in alternating current circuits is an interesting one with many practical applications and questions involving resonance are often seen in entrance question papers. You may try searching for ‘resonance’ on this blog (making use of the search box provided on this page) to access all previous posts related to resonance.

Today we will discuss two multiple choice questions meant for high lighting the behaviour of AC circuits at resonance:

-->

(1) An alternating current source of negligible impedance has an emf 6 volt. It is connected in series with a series LCR circuit consisting of a 72 Ω resistor and an inductor-capacitor combination. The inductor offers a reactance of 120 Ω at the operating frequency and the power dissipated in the LCR circuit is 0.5 W. What is the reactance offered by the capacitor?

(a) 60 Ω

(b) 72 Ω

(c) 120 Ω

(d) 192 Ω

(e) Zero

It is always a good idea to check whether the circuit is at resonance (Question setters have a craze to set questions involving resonance).

Since the power is dissipated in the resistor alone, we have

V²/72 = 0.5 where V is the voltage across the resistor.

[Remember the power equation, P = V²/R]

We obtain V² = 36 from which V = 6 volt.

The circuit is indeed at resonance since the entire supply voltage appears across the resistance.

At resonance the inductive reactance is equal to the capacitive reactance and hence the correct option is 120 Ω.

(2) In a series LCR circuit R = 200 Ω and the voltage and the frequency of the main supply are 220 V and 50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltage by 30º. On taking out the inductor from the circuit the current leads the voltage by 30º. The power dissipated in the LCR circuit is

(1) 242 W

(2) 305 W

(3) 210 W

(4) 0 W

This question appeared in AIEEE 2010 question paper.

The phase lag of current produced by the inductance is equal in value to the phase lead of current produced by the capacitor. This means that in the LCR circuit connected to the 220 V, 50 Hz supply, there is no phase difference between the supply voltage and the current and the circuit behaves as a pure resistor. Indeed the circuit is at resonance and the entire supply voltage appears across the resistor.

Therefore, the power P dissipated in the LCR circuit is given by

P = V²/R = 220²/200 = 242 W.

All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010) Questions Involving Electromagnetic Induction

The following interesting multiple choice questions involving electromagnetic induction were included in the AIEEE 2010 question paper:

(1) A rectangular loop has a sliding connector PQ of length l and resistance R Ω and it is moving with a speed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper. The three currents I₁, I₂ and I are

(1) I₁ = I₂ = Blv/6R, I = Blv/3R

(2) I₁ = – I₂ = Blv/R, I = 2Blv/R

(3) I₁ = I₂ = Blv/3R, I = 2Blv/3R

(4) I₁ = I₂ = I = Blv/R

The emf induced in the sliding connector PQ is ε = Blv, which is shown as a battery of internal resistance R in the adjoining figure.

The two external resistances R and R appear in parallel with this battery. The effective external resistance is therefore R/2 and the battery has to drive the current I through a total resistance R + (R/2), which is 3R/2

Therefore, I = ε/(3R/2) = 2Blv/3R.

The current I gets divided equally between the two identical branches and hence I₁ = I₂ = Blv/3R.

(2) In the circuit shown below, the key K is closed at t = 0. The current through the battery is

(1) V(R₁ + R₂)/R₁R₂ at t = 0 and V/R₂ at t = ∞

(2) VR₁R₂/√(R₁²+ R₂²) at t = 0 and V/R₂ at t = ∞

(3) V/R₂ at t = 0 and V(R₁ + R₂)/R₁R₂ at t = ∞

(4) V/R₂ at t = 0 and VR₁R₂/√(R₁²+ R₂²) at t = ∞

The current through the inductance branch is zero at the instant the circuit is switched on. [Remember the exponential growth of current in an LR circuit]

Therefore, at the instant t = 0 the current is limited by R₂ alone and is equal to V/R₂.

At t = ∞ the current has become steady, there is no induced voltage to oppose the flow of current and the inductor functions as a piece of conductor. The current is now limited by R₁ and R₂ which are in parallel with the battery. The current in this case is V/[R₁R₂/(R₁ + R₂)] = V(R₁ + R₂)/R₁R₂, as given in option (3).

You will find similar useful multiple choice question (with solution) here.

Multiple Choice Questions on Diodes [Including EAMCET 2009 (Engineering) and AIEEE 2006 Questions]

Questions from electronics, especially those involving the use of diodes are generally simple at the level expected of you. Here are a few questions on diodes:

(1) Currents flowing in each of the following circuits A and B respectively are

(1) 1A, 2 A

(2) 2 A, 1 A

(3) 4 A, 2 A

(4) 2 A, 4 A

This question appeared in EAMCET 2009 (Engineering) question paper.

In circuit A both diodes are forward biased and hence the circuit reduces to two 4 Ω resistors connected across the 8 V battery. Since the parallel combined value of the two resistors is 2 Ω, the current delivered by the battery is 8 V/2 Ω = 4 A.

In circuit B one diode is forward biased and the other diode is reverse biased and hence the circuit reduces to just one 4 Ω resistor connected across the 8 V battery. The current delivered by the battery is therefore 8 V/4 Ω = 2 A. The correct option is (3).

The following questions [No. (2) and (No. (3)] were included in AIEEE 2006 question paper:

(2) In the following, which one of the diodes is reverse biased?

You should note that all potentials are with respect to the ground. Therefore the diode in circuit (1) is reverse biased.

[In circuit (2) the anode of the diode is at a higher positive potential compared to its cathode and hence it is forward biased. In circuit (3) the cathode of the diode is at a higher negative potential and hence it is forward biased. In circuit (4) the cathode of the diode is at a negative potential and hence it is forward biased].

(3) The circuit has two oppositely connected ideal diodes in parallel. What is the current flowing in the circuit?

(1) 1.33 A

(2) 1.71 A

(3) 2.00 A

(4) 2.31 A

Since the diode D₁ is reverse biased, no current will flow through D₁ and the 3 Ω resistor. The current delivered by the battery is limited by the 4 Ω and the 2 Ω resistors only and is equal to 12 V/(4+2)Ω = 2 A.

The following question is meant for checking your grasp of the behaviour of semiconductor diodes:

(4) In the circuit shown the diodes used are silicon rectifier diodes which require a forward bias of 0.7 volt for appreciable conduction. Their leakage current is negligible. The internal resistance of the battery is insignificant. The potential difference between the terminals A and B is very nearly equal to

(a) 6 V

(b) 3 V

(c) 5.3 V

(d) 0.7 V

(e) 0 V

The upper diode is reverse biased and can be ignored. The lower diode which is connected across the terminals A and B is forward biased and hence keeps the voltage across A and B at 0.7 V [Option (d)].

[You can use the voltage drop across a forward biased diode as a small reference voltage in electronic circuits just as you use the relatively larger breakdown voltages of reverse biased zener diodes].