## Monday, December 31, 2007

### Two Questions on Specific Heat of Gases

Questions involving specific heats of gases have been discussed in some of the posts on this site. By clicking on the label ‘molar specific heat’ below this post, you will find some of the useful posts in this context. In particular, go through the post dated 1st September 2006.

Let us discuss two more questions involving specific heats of gases:

(1) An ideal diatomic gas in a container is heated so that half of the gas molecules dissociate into atoms. The molar specific heats (at constant volume) of the sample of the gas in the container before and after heating are C1 and C2. Then C1/C2 is

(a) 3/7 (b) 5/7 (c) 7/9

(d) 9/10 (e) 15/11

The important point to note here is that on dissociation, each particle (diatomic molecule) with 5 degrees of freedom produces two particles (individual atoms) with 3 degrees of freedom. Therefore you have to use the value Cv = (5/2)R for the undissociated molecule and the value Cv = (3/2)R for the atoms formed on dissociation (Cv is the molar specific heat at constant volume and R is the universal gas constant).

Assuming that there are ‘n’ moles of the diatomic gas initially, the number of moles after dissociation is (3/2)n, with (n/2)×2 = n moles of atoms and n/2 moles of molecules.

The molar specific heat (at constant volume) before dissociation,

C1 = (5/2)R, appropriate for a diatomic gas.

The molar specific heat (at constant volume) after dissociation,

C2 = (Heat supplied for increasing the temperature through 1 K) /Number of moles

= [n×(3/2)R + (n/2)×(5/2)R] / (3/2)n = (11/6)R

Therefore, C1/C2 = (5/2)R/(11/6)R = 15/11.

(2) An ideal diatomic gas is heated at constant pressure. What is the fraction of the heat energy supplied, which increases the internal energy of the gas?

(a) 2/3 (b) 3/5 (c) 5/7

(d) 7/9 (e) 1/2

This is a simple question but it has appeared in various entrance test papers.

When you heat a gas at constant pressure, the gas expands, thereby doing work. When one mole of a diatomic gas is considered, the increase in the internal energy on heating it through 1 K is equal to its molar specific heat (molar heat capacity) at constant volume, which is (5/2) R where R is the universal gas constant. The total heat energy supplied in this case for increasing the internal energy and for doing the external work is the molar specific heat at constant pressure, which is (7/2) R. The fraction required in the question is therefore [(5/2)R] /[(7/2)R] = 5/7

Merry Christmas!

## Thursday, December 20, 2007

### Multiple Choice Questions (MCQ) involving Inductance

(1) A long straight solenoid of cross section area10 cm2 has 10 turns per cm. A short 50 turn coaxial coil of cross section area 2 cm2 is fixed inside the solenoid at the middle. The mutual inductance of the solenoid and the coil is (in micro henry)

(a) 400 (b) 40 (c) 4π (d) π (e) ) 0.4π

Mutual inductance between an infinitely long straight solenoid and a short secondary coil placed coaxially inside at the middle is given by

M = μ0nNA where μ0 is the magnetic permeability of free space (or air), n is the number of turns per metre of the solenoid, N is the total number of turns in the secondary coil and A is the cross section area of the secondary coil. (If the secondary coil is outside the solenoid, the cross section area of the solenoid will appear in place of the area of the secondary coil).

[Note that the mutual inductance as given by the above expression is the magnetic flux (linked with the secondary coil) per unit current in the solenoid: M = NBA where B is the magnetic field (μ0n×1) produced by unit current in the solenoid]

Substituting for the known quantities in the expression for mutual inductance, we have

M = 4π×10–7×1000 ×50 ×(2×10–4) = 4π×10–6 H = μH.

(2) The secondary coil in the above question is moved with uniform velocity coaxially over a distance of 5 cm in 2 seconds when a current of 2 ampere flows through the solenoid. The emf induced in the coil during the motion is

(a) 200π μV (b) 100π μV (c) 20π μV (d) 10π μV (e) zero

No emf will be induced in the coil since the same flux is linked with the coil throughout its motion and there is no flux change.

(3) In question No.1, suppose the solenoid carries a current of 4 A. If this current is switched off in 100 ms, the emf induced in the secondary coil will be

(a) 0.2π μV (b) 0.4π μV (c) π mV (d) 0.08π mV (e) 0.16π mV

We have ε = –M(dI/dt) = = 4π×10–6 [(4 – 0)/(100×10–3)] = 160π×10–6 volt = 0.16π mV.

4. A battery of emf V volt is connected in series with a coil of inductance L and resistance R at the instant t = 0. The current in the circuit when t = 2τ where τ is the time constant of the circuit is (base of natural logarithm = e)

(a) (V/R)[1 (1/e2)]

(b) (V/R (1/e2)

(c) (V/R)[1 (1/e)]

(d) (V/R)(1/e)]

(e) 2e(V/R)

The expression for the growth of current (with time) in an LR circuit is

I = I0[1– e–Rt/L]

where I0 is the final steady (maximum possible) current which is equal to V/R.

The time constant of the LR circuit is L/R. After 2 time constants, the current will be

I = I0[1– e–R× 2(L/R)/L] = (V/R) [1– e– 2] = (V/R)[1– (1/e2)]

You will find more questions (with solution) at AP Physics Resources: AP Physics C - Multiple Choice Questions on Inductance

## Friday, December 14, 2007

### Two Questions on Force of Buoyancy

How does a balloon filled with helium rise in air? The weight of the balloon with helium is less than the force of buoyancy on the balloon and hence the balloon rises. Here is a question high lighting this principle:

A balloon of mass 6 kg is to be filled with helium so as to lift an instrument weighing 24 kg. The minimum volume of helium to be filled in the balloon is nearly (Density of air = 1.29 kg m–3, density of helium = 0.18 kg m–3 approximately)

(a) ) 22 m3 (b) 27 m3 (c) 30 m3 (d) ) 133 m3 (e) 166 m3

The balloon will start rising when the force of buoyancy on the balloon just exceeds the weight of the balloon with helium and the instrument. The force of buoyancy is equal to the weight of displaced air. Therefore, the minimum volume V of helium to be filled is given by

(6 + 0.18 V + 24)g = 1.29 Vg.

This gives V = 27 m3 approximately.

[The weight of air displaced by the instrument and the empty balloon is negligible and therefore not considered here].

Now consider the following MCQ:

A block of cork floats on the surface of water (in a container) with 30 cm outside water. If the system is placed on the moon where there is no atmosphere and the value of acceleration due to gravity is approximately one sixths that on the earth, the portion outside water will be

(a) 5 cm

(b) slightly less than 5 cm

(c) slightly less than 30 cm

(d) slightly greater than 30 cm

(e) slightly greater than 5 cm

If the acceleration due to gravity has a non-zero value, the exposed portion will remain unchanged since the weight of the floating body and the weight of displaced fluid are directly proportional to the value of the acceleration due to gravity. So, the exposed portion (or, the immersed portion) will remain the same whether you take the system to the moon or mars or inside a coal mine, when you consider the effect of ‘g’ only. But, you have to consider the effect of the atmosphere also. Since there is no atmosphere on the moon, there is no force of buoyancy due to the atmosphere and the net force of buoyancy (which is due to water in the container and due to the air above) is slightly reduced. Therefore, the block of cork gets immersed a little more and the exposed portion is slightly less than 30 cm.

[Note that if ‘g’ is zero, there is no weight and therefore no question of floatation].

You will find similar questions (with solution) at AP Physics Resources: Fluid Mechanics- Questions on Force of Buoyancy

## Thursday, December 06, 2007

### Two Questions on Oscillations

Today we will discuss two multiple choice questions on simple harmonic motion:
(1) A simple pendulum of period 2 s has a small bob of mass 50 g. The amplitude of oscillation of the bob is 10 cm and it is at a height of 45 cm from the ground in its mean position. While oscillating, the string breaks just when the bob is in its mean position. The horizontal distance R `from the mean position where the bob will strike the ground is nearly (a) 35.2 cm (b) 23 cm (c) 15.3 cm

(d)12.4 cm (e) 9.4 cm

The angular frequency (ω) of oscillation of the pendulum is given by

ω = 2π/T = 2π/2 = π rad/s.

The bob has maximum velocity (vmax) in the mean position and is given by

vmax = ωA where A is the amplitude.

Therefore, vmax = π × 0.1 = 0.1 π ms–1

On getting detached from the string, the bob moves like a projectile shot horizontally from a height of 0.45 m with a velocity of 0.1 π ms–1. Its time of flight (t) is obtained from the vertical displacement of 0.45 m:

0.45 = 0 + ½ gt2. (Note that the initial vertical velocity is zero and the vertical acceleration is g, which we may take as 10 ms–2). This gives t2 = 0.09 so that t = 0.3 s.

The horizontal distance covered by the bob during this time is 0.1 π × 0.3 = 0.094 m = 9.4 cm.

[Note that the mass of the bob does not come into the picture and it just serves as a distraction].

Now, consider the following MCQ :

A large horizontal surface moves up and down simple harmonically with an amplitude of 1 cm. If a mass of 3 kg (which is placed on the surface) is to remain continually in contact with it, the maximum frequency of the SHM will be

(a) 5 Hz (b) 2 Hz (c) 8 Hz (d) 10Hz (e) 15 Hz

The mass will remain in contact with the surface if the maximum acceleration produced in the simple harmonic motion does not exceed the acceleration due to gravity (g).

Therefore, we have

ω2A = g where ω is the angular frequency of the SHM.

From this ω = √(g/A) = √(10/ 0.01) = 10×√10.

The maximum frequency (linear) of oscillations is therefore given by

n = ω/2π = (10×√10)/2π = 5 Hz.

[Note that the mass of the body placed on the surface does not come into the picture and it just serves as a distraction].

## Saturday, December 01, 2007

### All India Engineering/Architecture Entrance Examination 2008 (AIEEE 2008)

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2008 (AIEEE 2008) to be conducted by Central Board of Secondary Education (CBSE) on 27th April 2008 (Sunday) are being distributed from 30.11.2007 (Friday) and will continue till 5.1.2008 (Saturday) through selected Branches of Syndicate Bank, Regional Offices of CBSE and designated institutions.

Important dates in this regard are given below:

 1a. Date of Examination 27.04.2008 b Sale of AIEEE Information Bulletin containing Application Form 30.11.2007 to 05.01.2008 c Online submission of application on website http://www.aieee.nic.in 30.11.2007 to 05.01.2008 2. Last date for a. Receipt of request for Information Bulletin and Application Form by Post atAIEEE Unit,CBSE,PS1-2,Institutional Area,IP Extension,Patparganj,Delhi-110092 15.12.2007 b. Sale of Information Bulletin at designated branches of Syndicate Bank, Regional Offices of the CBSE and designated institutions 05.01.2008 c. Online submission of applications 05.01.2008 d. Receipt of complete applications “by post” including Registration Forms with Bank Draft at AIEEE Unit, CBSE, PS1-2,Institutional Area,IP Extension,Patparganj,Delhi-110092 10.01.2008 3 Date of dispatch of Admit Card 10.03.2008 to 31.03.2008 4 Issue/dispatch of duplicate admit card(or request only with fee of Rs. 50/- + postal charges of Rs. 30/- extra for out station candidate. 11.04.2008 to 27.04.2008 (By Hand) 11.04.2008 to 21.04.2008 (By Post) 5 Dates of Examination PAPER – 1 27.04.2008 (0930-1230 hrs)PAPER – 2 27.04.2008 (1400-1700 hrs)

Paper-1 is for B.E./B.Tech and Paper-2 is for B.Arch/B.Planning.

By visiting the site www.aieee.nic.in complete information in this regard can be obtained. Visit the site without any delay.

## Monday, November 26, 2007

### Two All India Institute of Medical Sciences (AIIMS) Questions on Optics

The following questions which appeared in All India Institute of Medical Sciences (AIIMS) 2005 entrance question paper for admitting students to the MBBS Degree course are simple as usual. They are meant for checking your knowledge and understanding of fundamentals.

(1) The apparent depth of water in a cylindrical water tank of diameter 2R cm is reducing at the rate of x cm/minute when water is being drained out at a constant rate. The amount of water drained in c.c. per minute is (n1= refractive index of air, n2 = refractive index of water)

(a) x π R2 n1/n2 (b) x π R2 n2/n1 (c) 2 π R n1/n2 (d) π R2x

Since the refractive index is the ratio of real depth to the apparent depth, we have

Real depth = Apparent depth × refractive index.

Therefore, the rate at which the real depth is decreasing = xn2/n1 cm per minute.

The amount of water drained in c.c. per minute is therefore equal to x π R2 n2/n1, given in option (b).

(2) A telescope has an objective lens of focal length 200 cm and an eye piece with focal length 2 cm. If the telescope is used to see a 50 m tall building at a distance of 2 km, what is the length of the image of the building formed by the objective lens?

(a) 5 cm (b) 10 cm (c) 1 cm (d) 2cm

At the first glance this question may seem to be one involving the magnification produced by a telescope; but, this is quite simple since you are asked to consider the objective only.

The objective will produce the image of the building at the focus (which is at 2 m from the lens) and hence from the expression for magnification (M) we have

M = Distance of image/ Distance of object = Height of image/ Height of object

so that 2/ 2000 = x/50 where ‘x’ is the height of image in metre.

Therefore, x = 2×50/2000 = 0.05 m = 5 cm.

## Thursday, November 15, 2007

### AP Physics Exam Resources- Two Questions (MCQ) on Moment of Inertia

Multiple choice questions discussed on this site will be useful for entrance examinations for admission to various degree courses including professional courses. They will be suitable for those preparing for AP Physics Examination, as can be judged by working out the following two questions:

(1) Three circular discs of radii R, R and 2R are cut from a metallic sheet of uniform thickness and the smaller discs are placed symmetrically on the larger disc as shown in the figure. If the mass of a smaller disc is M, the moment of inertia of the system about an axis at right angles to the plane of the discs and passing through the centre of the larger disc is (a) 5MR2 (b) 7MR2 (c) 9MR2

(d) 11MR2 (e) 12MR2

The mass of the larger disc is 4M (since its radius is twice that of the smaller disc) and its moment of inertia is (4M)×(2R)2/2 = 8MR2.

The moment of inertia of each smaller disc about the axis passing through the centre of the larger disc (as given by the parallel axis theorem) is MR2/2 + MR2 = 3MR2/2.

Note that the moment of inertia is a scalar quantity. Therefore, the total moment of inertia of the system of three discs is 8MR2 + 2×3MR2/2 = 11MR2.

(2) A small body of regular shape made of iron rolls up with an initial velocity ‘v’ along an inclined plane. It reaches a maximum height of 7v2/10g where ‘g’ is the acceleration due to gravity. The body is a

(a) ring (b) disc (c) solid sphere

(d) hollow sphere (e) cylindrical rod

The initial kinetic energy of the body is ½ Mv2 + ½ I ω2 where M is its mass and I is its moment of inertia about its axis (of rolling). The first term is its translational kinetic energy and the second term is its rotational kinetic energy.

Since the entire kinetic energy is used in gaining gravitational potential energy, we have

½ Mv2 + ½ I ω2 = Mgh where ‘h’ is the maximum height reached.

Therefore, ½ Mv2 + ½ I v2/R2 = Mg×7v2/10g, from which

I = (2/5)MR2.

The body is therefore a solid sphere.

## Sunday, November 11, 2007

### Joint Entrance (2008) Examination for Admission to IITs and other Institutions—(IIT-JEE 2008)

Application form and Prospectus of the Joint Entrance Examination 2008 (JEE-2008) for admission to the seven IITs at Bombay, Delhi, Guwahati, Kanpur, Kharagpur, Madras and Roorkee as well as to Institute of Technology, Banaras Hindu University, Varanasi and the Indian School of Mines University, Dhanbad, will be issued with effect from November 23, 2007 (Friday). On-line submission of Application will also commence on the same day.

Application form along with the Information Brochure can be purchased from any one of the selected branches of Canara Bank/State Bank of India/ Union Bank/ Punjab National Bank (visit site: http://www.iitkgp.ernet.in/jee/advt.html for the list of branches) or from any of the IITs between 23.11.2007 (Friday) and 4.1.2008 (Friday). by paying Rs. 500/- in case of SC/ST/Female candidates and Rs. 1000/- in case of all other candidates by way of cash. SC/ST/Female candidates will get the materials in a WHITE coloured envelope while the other candidates will get the materials in a BLUE coloured envelope.

Application Form and Prospectus by Post from IITs:

The request for Application Form and Prospectus by Post from any of the IITs will also be accepted from November 23, 2007. Application Form and Prospectus can be obtained by post from any of the IITs by sending a request along with two self- addressed slips and a Demand Draft for Rs.500/- (in case of SC/ST/Female applicants) and for RS.1000/- in case of other applicants, payable to the “CHAIRMAN, JEE” of the respective IIT, at the corresponding city.[For example, those applying to IIT, Madras, should take the DD in favour of “Chairman, JEE, IIT Madras” and payable at Chennai]. Such requests will be accepted from 23.11.2007(Friday) to 21.12.2007 (Friday).

Submission of filled up Application Forms: Duly completed Application Form, refolded only along the original fold should be inserted in the envelope supplied, along with the attested copy of the 10th Class Pass, or Equivalent Examination Certificate, and the Acknowledgement Card. These items should not be stapled or pasted with the Application form. Irrespective of the Bank/Institute from where the Application has been obtained, they should be re-submitted along with the contents by Registered Post/Speed Post only to the IIT located in the Zone where the centre of the examination chosen by the applicant is located. They may also submitted in person at the JEE office of the IIT concerned.

The last date for receipt of the completed application at the IITs is 17:00 hours on January 4, 2008 (Friday).

Online Submission of Application: This facility will be available between 23.11.2007 (Friday) and 5 pm on 28.12.2007 (Friday) through the JEE websites of the different IITs. The JEE websites of the different IITs are given below:
IIT Bombay: http://www.jee.iitb.ac.in
IIT Delhi: http://www.jee.iitd.ac.in
IIT Guwahati: http://www.iitg.ac.in/jee
IIT Kanpur: http://www.iitk.ac.in/jee
IIT Kharagpur: http://www.iitkgp.ernet.in/jee
IIT Roorkee: http://www.iitr.ac.in/jee

Schedule of the Examination: The examination will be held on April 13, 2008 (Sunday) as per the following schedule:
09:00 – 12:00 hrs Paper - 1
14:00 – 17:00 hrs Paper – 2

## Thursday, November 08, 2007

### Multiple Choice Questions on Elasticity

Here are three questions (MCQ) on elasticity which appeared in Kerala Engineering and Medical Entrance 2004 Examination question papers:

(1) Wires A and B are made from the same material. A has twice the diameter and three times the length of B. If the elastic limits are not reached when each is stretched by the same tension, the ratio of energy stored in A to that in B is

(a) 2:3 (b) 3:4 (c) 3:2 (d) 6:1 (e) 12:1

The work (W) done in increasing the length of a wire or rod by ‘l’ by applying a force ‘F’ is given by

W = ½ Fl

[ Here is the proof for the above: The work dW done for increasing the length by dl is

F×dl. The total work done for increasing the length by ‘l’ is ∫F×dl where the limits of

integration are zero and ‘l’. Since the Young’s modulus, Y = (F/A)(L/l) where A is the

area of cross section and L is the length of the wire, we have F = YAl/L. The total work

done is therefore 0l (Yal/L)dl = ½ (Yal2/L) = ½ (YAl/L)×l = ½ Fl ]

The energy stored (which is equal to the work done) in a wire is therefore directly proportional to the increase in length. The ratio of energy stored is therefore W1/W2 = l1/l2 where l1 and l2 are the increases in length of the wires. But l1 = FL1/A1Y and l2 = FL2/A2Y so that

W1/W2 = l1/l2 = (L1/L2)×(A2/A1) = (3/1)×(1/4) = 3/4 [Option (b)].

(2) A wire of cross section 4 mm2 is stretched by 0.1 mm. How far will a wire of the

same material and length but of area 8 mm2 stretch under the action of the same

force?

(a) 0.05 mm (b) 0.01 mm (c) 0.15 mm

(d) 0.2 mm (e) 0.25 mm

This question as well as the previous one appeared in Kerala Medical Entrance 2004 question paper.

Since the increase in length is inversely proportional to the area of cross section of the wire, the correct option is 0.05 mm.

(3) Compressibility of water is 5×10–10 m2/N. The change in volume of 100 ml of water subjected to 15×10–6 Pa pressure will be

(a) no change (b) increase by 0.75 ml (c) increase by 1.5 ml

(d) decrease by 1.5 ml (e) decrease by 0.75 ml

This question appeared in Kerala Engineering Entrance 2004 question paper. You must definitely remember the expression for bulk modulus ‘B’ as

B = P/(dV/V) where P is the pressure which produces a change in volume dV in a volume V. The negative sign indicates that an increase in pressure will produce a decrease in volume.

Compressibility is the reciprocal of bulk modulus. Therefore we have

1/(5×10–10) = (15×10–10 × 100×10–6)/dV, from which

dV = 0.75×10–6 m3 = 0.75 ml.

Since an increase in pressure produces a decrease in volume, the correct option is (e).

## Saturday, October 20, 2007

### Multiple Choice Questions on Thermoelectric Effect

Questions involving neutral temperature and temperature of inversion often appear in entrance examination question papers. You have to remember that the neutral temperature is a constant for a given thermo couple where as the temperature of inversion is not a constant. The temperature of inversion is dependent on the temperature of the cold junction and is always as much above the neutral temperature as the cold junction is below it.

Now, consider the following MCQ:

The temperature of the cold junction of a thermocouple is 0º C and its neutral temperature is 275º C. If the temperature of the cold is changed to 20º C, the neutral temperature and the temperature of inversion will be respectively

(a) 265º C and 550º C (b) 265º C and 530º C (c) 275º C and 530º C

(d) 275º C and 550º C (e) 275º C and 510º C

The neutral temperature will be unchanged (275º C). Since the cold junction is 255º C below the neutral temperature, the temperature of inversion has to be 255º C above the neutral temperature and will be 530º C. So, the correct option is (c).

Consider now the following simple question:

The metal which does not exhibit Thomson effect is

(a) iron (b) nickel (c) copper

The correct option is (d). In determining thermo electric quantities, lead is often used as one member of the couple because of the absence of Thomson effect in it.

What is the unit of thermo electric power?

The term ‘thermoelectric power’ is a misnomer. Thermoelectric power is dV/dT where V is the thermo emf and T is the temperature of the hot junction. There is no ‘power’ involved in it and the unit is volt per Kelvin.

## Thursday, October 11, 2007

### All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008

Central Board of Secondary Education (CBSE), Delhi has announced the dates of All India Pre-Medical/Pre-Dental Entrance Examination (AIPMT) 2008 for admission to 15% of the total seats for Medical/Dental Courses in all Medical/Dental colleges run by the Union of India, State Governments and Municipal or other local authorities in India except in the States of ANDHRA PRADESH AND JAMMU & KASHMIR. The dates of the Examination are:

(1) Preliminary Examination : 6th April 2008 (Sunday)

(2) Final Examination : 11th May 2008 (Sunday)

Candidates can apply for the All India Pre-Medical/Pre-Dental Entrance Examination in the following two ways:-

(i) Online

Online submission of application may be made by accessing the Board’s website www.cbse.nic.in from 16.10.2007 (10.00 AM) to 26.11.2007 (5.00 PM). Candidates are required to take a print of the Online Application after successful submission of data. The print out of the computer generated application, complete in all respect as applicable for Offline submission should be sent to the Deputy Secretary (AIPMT), Central Board of Secondary Education, Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 301 by Speed Post/Registered Post only. Candidate should pay the examination fee of Rupees 400/- for General Category and Rupees 200/- for SC/ST Category through a Demand Draft in favour of the Secretary, Central Board of Secondary Education, Delhi drawn on any nationalized bank payable at Delhi. Instructions for Online submission of Application Form will be made available on the website www.cbse.nic.in.

(ii) Offline

Offline submission of Application Form may be made on the prescribed Application Form. The Information Bulletin and Application Form costing Rs.400/- for General Category candidates and Rs.200/- for SC/ST candidates inclusive of examination fee can be obtained against Cash Payment from designated branches of Canara Bank/ Regional Offices of the CBSE from 16-10-2007 to 26-11-2007. The details of Banks are given in the Admission Notice which is available on CBSE website www.cbse.nic.in.

Designated branches of Canara Bank in Kerala are:

 KOTTAYAM P.B. No.122, K.K.Road, Kottayam-686 001 QUILANDY Fasila Buidling, Main Road, Quilandy-673 305 TRIVANDRUM Ist Floor, Ibrahim Co. Bldg., Challai, Trivandrum-695 023 TRIVANDRUM Plot no.2, PTP Nagar Trivandrum-695 038 TRIVANDRUM TC No.25/1647, Devaswom Board Bldg. M.G. Road, Trivandrum-695 001 ERNAKULAM Shenoy’s Chamber, Shanmugam Road, Ernakulam, Cochin-682 031 CALICUT 9/367-A, Cherooty Road, Calicut-673 001 TRICHUR Trichur Main Ramaray Building, Round South, Trichur-680001 QUILON Maheshwari Mansion, Tamarakulam, Quilon-691 001 PALGHAT Market Road, Big Bazar, 20/68, Ist floor,Palghat-678 014.

The Information Bulletin and Application Form can also be obtained by Speed Post/Registered Post by sending a written request with a Bank Draft/Demand Draft for Rs.450/- for General Category and Rs.250/- for SC/ST Category payable to the Secretary, Central Board of Secondary Education, Delhi along with a Self Addressed Envelope of size 12” x 10”. The request must reach the Deputy Secretary (AIPMT), CBSE, 2, Community Centre, Preet Vihar, Delhi-110 301 on or before 15-11-2007. The request should be super scribed as Request for Information Bulletin and Application Form for AIPMT, 2008”.

Completed Application Form is to be dispatched by Registered Post/Speed Post only. Last Date for receipt of completed Application Forms for both Offline and Online in CBSE is 28-11-2007.

You can obtain complete information at www.cbse.nic.in

Make it a point to visit the site for information updates.