If we did all things we are capable of, we would literally astound ourselves.

– Thomas A. Edison

Thursday, November 29, 2007

KEAM2007 Questions on Electromagnetic Waves

Questions on electromagnetic waves are simple at the 12th class level and you can easily score full marks in this section. Here are the three questions which appeared in the Kerala Medical (Question No.1) and Engineering (Question Nos.2 & 3) Entrance 2007 question papers:

(1) An electromagnetic radiation has energy of 13.2 keV. Then the radiation belongs to the region of

(a) visible light (b) ultraviolet (c) infrared

(d) X-ray (e) microwave

If an electron accelerated by a potential difference of 13.2 kilo volt strikes a target, the maximum energy of the photon generated will be 13.2 kilo electron volt. You know that an accelerating voltage of 13.2 kV in an X-ray tube can produce X-rays. So, the correct option is (d).

If the energy given were in the electron volt (eV) range, you will be in confusion. To be fool proof, it will be better to calculate the wave length of the radiation, noting that the product of the wave length in angstrom (Ǻ) and the energy in electron volt (in the case of photons) is approximately 12400. The wave length of the photon in the present case is therefore given by

λ = (12400/13200) Ǻ = 0.94 Ǻ, which is in the X-ray region.

(2) The dielectric constant of air is 1.006. The speed of electromagnetic wave traveling in air is a×108 ms–1 where ‘a’ is about

(a) 3 (b) 3.88 (c) 2.5 (d) 3.2 (e) 2.8

This is an unusually simple question which the question setter might have deliberately included! You know that the speed of electromagnetic waves in air is almost equal to that in free space and hence the answer is 3 [Option (a)].

Generally, in any medium, the speed (v) of electromagnetic waves is given by

v = c/√(μrεr) where c is the speed in free space (which is 3×108 ms–1), μr is the relative permeability and εr is the relative permittivity (dielectric constant) of the medium.

(3) a. The wave length of microwaves is greater than that of UV-rays

b. The wave length of IR rays is less than that of UV-rays

c. The wave length of microwaves is less than that of IR rays

d. Gamma ray has the shortest wave length in the electromagnetic spectrum

Of the above statements

(A) a and b are true (B) b and c are true

(C) c and d are true (D) a and c are true

(E) a and d are true

This too is a simple knowledge based question and the correct option is (E).

Friday, November 16, 2007

IIT-JEE Matrix Match Type Question on Electromagnetism

The following Matrix Match type question which appeared in the IIT-JEE 2007 question paper aims at checking your understanding of basic principles of electricity and magnetism. Here is the question:

Column I gives certain situations in which a straight metallic wire of resistance R is used and Column II gives some resulting effects. Match the statements in Column I with the statements in Column II and indicate your answer by darkening appropriate bubbles in the 4×4 matrix given in the OR

Column I------------------------------ Column II

(A) A charged capacitor is---------------(p) A constant current flows

connected to the ends of---------------- through the wire

the wire

(B) The wire is moved -----------------(q) Thermal energy is generated

perpendicular to its length with -------- in the wire

a constant velocity in a uniform

magnetic field perpendicular to

the plane of motion

(C) The wire is placed in a constant ---(r) A constant potential

electric field that has a direction ------difference develops between

along the length of the wire-----------the ends of the wire

(D) A battery of constant emf is ------(s) Charges of constant

connected to the ends of the----------magnitude appear at the ends

wire ---------------------------------of the wire
(A) When a charged capacitor is connected to the ends of a wire, a discharge current flows through the wire and the wire is heated. This is given in statement (q) under column II so that (A) is to be matched with (q).

(B) When the wire is moved perpendicular to the magnetic field, free electrons in the wire experience a magnetic (Lorentz) force and shift towards one end. Consequently, a potential difference develops between the ends of the wire. Since the wire is moving with constant velocity in a constant magnetic field, the charges developed at the ends as well as the potential difference between the ends are constant. Therefore, (B) is to be matched with (r) and (s).

(C) When the wire is placed in the electric field, free electrons shift towards the end of the wire which is at higher potential. This shifting of free electrons results in the accumulation of charges at the ends of the wire. But there is no potential difference between the ends of the wire since the shifting of free electrons just nullifies the potential difference caused by the external electric field. Therefore, (C) is to be matched with (s) only.

(D) When a battery of constant emf is connected to the ends of the wire, the effects given at (p) (q) and (r) under Column II occur. Therefore, (D) is to be matched with (p), (q) and (r).

These are to be marked in the 4×4 matrix as shown:

Tuesday, November 13, 2007

IIT-JEE 2008 Application from 23-11-2007

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, Dhanbad, 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

Application form and Prospectus of the Joint Entrance Examination 2008 (IIT-JEE-2008) 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 the 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 the case of SC/ST/Female candidates and Rs. 1000/- in the case of all other candidates by cash. SC/ST/Female candidates will get the materials in a WHITE envelope while the other candidates will get the materials in a BLUE envelope.

Application Material 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: Facility for online submission of applications 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 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 Madras: http://jee.iitm.ac.in
IIT Roorkee: http://www.iitr.ac.in/jee

Visit the website http://www.iitkgp.ernet.in/jee/ for more details.

Friday, November 09, 2007

Two Questions (MCQ) on Elasticity

Have you ever thought why a tree does not grow beyond a limiting height? It is the breaking stress of the tree that limits its height. Here is a question that high lights this point:

A tree has a breaking stress of 4.5×105 Nm–2. If the density of the tree is 900 kg m–3, the maximum height up to which it can grow is (assuming that the tree is cylindrical in shape)

(a) 100 m (b) 75 m (c) 50 m (d) 30 m (e) 25 m

The weight of the tree is AHρg where A is the area of cross section, H is the height, ρ is the density and ‘g’ is the acceleration due to gravity.

Maximum stress will be developed at the base of the tree and will be equal to AHρg/A = Hρg.

Since the breaking stress is 4.5×105 Nm–2, we can write

Hρg = 4.5×105 Nm–2 so that H = 4.5×105 Nm–2/ρg = 4.5×105 Nm–2/9000 nearly.

This works out to 50 m.

Now, consider the following MCQ involving compressibility:

A solid sphere made of copper having compressibility K is placed in an air chamber and the pressure of air is reduced by P. The fractional change in the radius of the sphere will be

(a) PK (b) 3PK (c) P/3K (d) K/3P (e) PK/3

Compressibility (K) is the reciprocal of bulk modulus (B) and we have

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

In terms of compressibility, the above equation can be written as

B = 1/K = –P/(dV/V) so that the fractional change in volume is given by

dV/V = PK.

But V = (4/3) πR3 where R is the radius of the sphere so that dV/V = 3 dR/R. The fractional change in radius is therefore given by dR/R = (dV/V)/3 = PK/3.

You may find all posts on elasticity on this site by clicking on the label ‘elasticity’ below this post or on the side of this page.