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

(d) lead (e) bismuth

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 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

(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

Designated branches of Canara Bank in Kerala are:


P.B. No.122, K.K.Road, Kottayam-686 001


Fasila Buidling, Main Road, Quilandy-673 305


Ist Floor, Ibrahim Co. Bldg., Challai, Trivandrum-695 023


Plot no.2, PTP Nagar Trivandrum-695 038


TC No.25/1647, Devaswom Board Bldg.

M.G. Road, Trivandrum-695 001


Shenoy’s Chamber, Shanmugam Road, Ernakulam,

Cochin-682 031


9/367-A, Cherooty Road, Calicut-673 001


Trichur Main Ramaray Building, Round South,



Maheshwari Mansion, Tamarakulam, Quilon-691 001


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

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

Saturday, October 06, 2007

Rotational Motion of a Reel – Two Multiple Choice Questions

The following question on the acceleration of a reel carrying stitching thread is an interesting one:

A reel of mass ‘m’ in the form of a short solid cylinder carries light inextensible string wound round it. The free end of the string is tied to a hook fixed to the ceiling and the reel is allowed to roll down, unwinding the string. The linear acceleration of the reel is

(a) g (b) g/3 (c) g/2 (d) 3g/2 (e) 2g/3

If ‘a’ is the acceleration of the of the reel we have

Mg – T = ma

where ‘M’ is the mass of the reel and ‘T’ is the tension in the string.

Therefore, T = mg – a

[You can directly write the above value of T if you remember that T is equal in magnitude to the weight of the reel while moving down with acceleration ‘a’ as in a lift]

If the radius of the reel is ‘R’, the torque on the reel is TR.

Therefore, we have

TR = Iα

where ‘I’ is the moment of inertia of the reel ( I = MR2/2) and ‘α’ is its angular acceleration, which is equal to a/R.

The above equation therefore becomes

(Mg – a)R = (MR2/2) (a/R)

This gives a = 2g/3.

Now consider another MCQ:

A solid cylinder having mass M and radius R is free to rotate about its axis and has a light inextensible string wound roun it. A body of mass M/2 is attached to the free end of the string. If the system is released from rest, the mass will move down with a linear acceleration of

(a) g (b) g/3 (c) g/2 (d) 3g/2 (e) 2g/3

The tension in the string is given by

T = (M/2)(g – a).

Since TR = Iα, we have

(M/2)(g – a)R = (MR2/2) (a/R).

This equation yields a = g/2.

[It will be interesting to note that if the mass suspended were M, the acceleration would have been 2g/3, which is the acceleration of the falling reel in the previous question].