## Thursday, December 31, 2009

### Two Questions (MCQ) involving Kinetic Energy and Potential Energy

The following question appeared in Kerala Engineering Entrance 2007 question paper:

A simple pendulum is released from A as shown. If m and l represent the mass of the bob and the length of the pendulum, the gain in kinetic energy at B is

(a) mgl/2

(b) mgl/√2

(c) (mgl√3)/2

(d) 2mgl/√3

(e) mgl

From position A, the bob of the pendulum has fallen through a distance l cos 30º (fig.). Therefore, the loss of potential energy is mg l cos 30º = (mgl√3)/2.

Therefore, the gain in kinetic energy by the bob (by the law of conservation of energy) is (mgl√3)/2.

Here is another question which will be useful in the present context:

The bob of a simple pendulum has mass 0.2 kg. The bob is drawn aside so that the string is horizontal and is released. When it swings through 60º its kinetic energy is 0.5 J. The kinetic energy when the sting becomes vertical will be

(a) √3 J

(b) 1/√3 J

(c) √3/2 J

(d) 2/√3 J

(e) 1 J

You can use the figure used with the previous question for working out the present question as well since the angle turned is 60º. The loss of potential energy by the bob on swinging through 60º is mg l cos30º itself. [Or, you may take it as mg l sin60º].

Since the loss of potential energy is equal to the gain of kinetic energy, we have

mg l cos30º = 0.5 J

Or, (mgl√3)/2 = 0.5 J.

The kinetic energy when the string becomes vertical will be mgl since the bob falls through a distance l and loses its entire initial potential energy mgl.

From the above equation mgl = 1/√3 J [Option (b)].

[Note that the mass of the bob given in the question just serves as a distraction].

You will find some useful questions in this section here.

## Tuesday, December 15, 2009

### IIT-JEE 2009 Multiple Choice Question (Single Answer Type) on Elastic Collision

The following question which appeared in IIT-JEE 2009 question paper is based on the principle that in an elastic collision between two particles of the same mass, the velocities get interchanged:

Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v respectively as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, these two particles will again reach the point A?
(a) 4
(b) 3
(c) 2
(d) 1
Particle P1, after traversing one third (AB) of the circular track in the anticlockwise direction, will collide with particle P2 at position B. This collision occurs when the particle P2  has traversed two thirds (ACB) of the circular path (since the speed of P2 is twice that of P1).

Since the collision is elastic and the particles are of equal masses, their velocities are interchanged. P1 now travels in the clockwise direction with speed 2v and P2 travels in the anticlockwise direction with speed v. The second collision takes place at position C. (BAC is two thirds of the circular path while BC is one third of the path). The second collision at C results in the reversal of the velocities and P1 now travels in the anticlockwise direction with speed v where as P2 travels in the clockwise direction with speed 2v. Therefore, the third collision will occur at A.
Since the third collision at A is not to be counted, the answer is 2 [Option (c)].
*    *    *    *    *    *    *    *    *    *    *    *   *    *    *    *    *

Now suppose we modify the above question as follows:
Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular frictionless track. Their speeds are v and 3v respectively as shown in the figure. After making how many elastic collisions, other than that at A, these two particles will again reach the point A ?
(a) 4
(b) 3
(c) 2
(d) 1
[You may use the figure shown with the question above, replacing the velocity 2v with 3v]

You can easily arrive at the answer which is 3 [Option (b)].

## Saturday, December 05, 2009

### Apply for Entrance Examination for Admission to Medical/ Agriculture/ Veterinary/ Engineering/ Architecture Degree Courses 2010 (KEAM 2010), Kerala

The Commissioner for Entrance Examinations, Govt. of Kerala, has invited applications for the Entrance Examinations for admission to the following Degree Courses in various Professional Colleges in the State for 2010-11.
(a) Medical: (i) MBBS (ii) BDS (iii) BHMS (iv) BAMS (v) BSMS
(b) Agriculture: (i) BSc. Hons. (Agriculture) (ii) BFSc. (Fisheries) (iii) BSc. Hons. (Forestry)
(c) Veterinary: BVSc. & AH
(d) Engineering: B.Tech. [including B.Tech. (Agricultural Engg.)/B.Tech. (Dairy Sc. & Tech.) courses under the Kerala Agricultural University]
(e) Architecture: B.Arch.
Dates of Exam:
Engineering Entrance Examination (For Engineering courses except Architecture)
19.04.2010 Monday 10.00 A.M. to 12.30 P.M. Paper-I : Physics & Chemistry.
20.04.2010 Tuesday 10.00 A.M. to 12.30 P.M. Paper-II: Mathematics.

Medical Entrance Examination (For Medical, Agriculture and Veterinary Courses)
21.04.2010 Wednesday 10.00 A.M. to 12.30 P.M. Paper-I : Chemistry & Physics.
22.04.2010 Thursday 10.00 A.M. to 12.30 P.M. Paper-II: Biology.

Application form and Prospectus will be distributed from 07.12.2009 to 06.01.2010 through selected branches of Post Offices in Kerala and outside the State. The amount towards the fee of application (Rs. 700/- for general candidates and Rs.350 for SC/ST candidates) is to be remitted in cash at the Post Offices. To find the list of post offices, selected as sales centres, and details of additional fee of Rs.8500/- in the case of candidates opting Dubai as the centre of exam, visit the site http://www.cee-kerala.org/.
Online submission of application is possible in the case of candidates claiming no reservation benefit and those belonging to Non-Keralite category. See details at http://www.cee-kerala.org/.
Last Date for receipt of Application by CEE: 06-01-2010 (Wednesday) - 5 PM
Candidates seeking admission to B.Arch. course should also submit their application to the CEE. There is no state level Entrance Examination for this purpose. These candidates should write the National Aptitude Test for Architecture and should forward the NATA score and mark list of the qualifying examination to the CEE on or before 05-06-2010.
You will find complete details and information updates at http://www.cee-kerala.org/

If you want to see earlier KEAM questions discussed on this site, type in ‘Kerala’ in the search box at the top left of this page and strike the enter key or click on the search button.

## Wednesday, December 02, 2009

### 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].

## Saturday, November 21, 2009

### Questions (MCQ) on Nuclear Physics

The following three questions from nuclear physics are simple but are useful in your preparation for entrance tests:
(1) Two radioactive samples S1 and S2 have half lives 3 hours and 7 hours respectively. If they have the same activity at a certain instant t, what is the ratio of the number of atoms of S1 to the number of atoms of S2 at the instant t?
(a) 9 : 49
(b) 49 : 9
(c) 3 : 7
(d) 7 : 3
(e) 1 : 1
If The number of atoms present at the instant t is N, we have
N = N0eλt where N0 is the initial number, e is the base of natural logarithms and λ is the decay constant.
Therefore, activity, dN/dt = λ N0eλt = λN
If N1 and N2 are the number of atoms of S1 and S2 respectively when the activities are the same, we have
λ1N1 = λ2N2 from which N1/N2 = λ2/λ1
But the decay constant λ is related to the half life T as T = 0.693/λ.
Therefore, N1/N2 = λ2/λ1 = T1/T2 = 3/7 [Option (c)].
(2) A nucleus ZXA has mass M kg. If Mp and Mn denote the mass (in kg) of proton and neutron respectively, the binding energy in joule is
(a) [ZMp + (A – Z)MnM]c2
(b) [ZMp + ZMnM]c2
(c) M – ZMp – (A – Z)Mn
(d) [M– ZMp – (A – Z)Mn]c2
(e) [AMn M]c2
Total mass of the Z protons is ZMp. Since the total number of nucleons is A, the total number of neutrons is (A – Z) and the total mass of the neutrons is (A – Z)Mn.
The mass defect M is the difference between the total mass of the nucleons (protons and neutrons together) and the mass of the nucleus: M =[ZMp + (A – Z)MnM].
Therefore, binding energy.= Mc2 where ‘c’ is the speed of light in free space.
Thus binding energy = [ZMp + (A – Z)MnM]c2
(3) If the aluminium nucleus 13Al27 has nuclear radius of about 3.6 fm, then the tellurium nucleus 52Te125 will have radius approximately equal to
(a) 3.6 fm
(b) 16.7 fm
(c) 8.9 fm
(d) 6.0 fm.
(e) 4.6
The nuclear radius R is given by
R = R0A1/3 where R0 is a constant (equal to 1.2×10–15 m, nearly) and A is the mass number of the nucleus.
If Rl and R2 are the radii of the given Al and Te nuclei respectively, we have
Rl = R0 (27)1/3 = 3R0 and
R2 = R0 (125)1/3 = 5R0
Dividing, Rl/R2 = 3/5
Therefore, R2 = 5R1/3 = (5×3.6)/3 fm = 6 fm.
By clicking on the label ‘nuclear physics’ below this post, you can access all posts related to nuclear physics on this site.
You can find useful posts in this section here.

## Tuesday, November 17, 2009

### Apply for All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010)

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2010 (AIEEE 2010) to be conducted on 25-4-2010 will be distributed from 1.12.2009 and will continue till 31.12.2009. Candidates can apply for AIEEE 2010 either on the prescribed Application Form or make application ‘Online’.
Online submission of the application is possible from 16-11-2009 to 31-12-2009 at the website http://aieee.nic.in

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 typing ‘AIEEE’ in the search box at the top left of this page and then hitting the enter key (or clicking the search button).

## Tuesday, November 10, 2009

### IIT-JEE 2010 – Candidates to get Performance Score Cards

Leading news papers have flashed a welcome news item which will be of great interest to candidates appearing for IIT-JEE 2010. Here is the gist of the news item:

The JEE Board will issue performance cards specifying the marks and the ranks secured by candidates who will be appearing for IIT-JEE 2010. The performance cards can be considered as certificates by many other institutions wanting to give admission to JEE candidates. A decision for issuing such performance cards has been taken by the Joint Admission Board (JAB) for IITs in its meeting on August 23. The performance score cards will be issued two weeks after the results are declared. The marks of the students will be published on the IIT-JEE websites just a week after the results are declared. The next JAB meeting in April 2010 will give the final approval to the scheme.

## Thursday, November 05, 2009

### MCQs on Magnetism including EAMCET 2009 (Medical) Question

Some multiple choice questions on magnetism have already been posted on this site. You can access them by clicking on the label ‘magnetism’ below this post. Today we will discuss a few more multiple choice questions on magnetism.

(1) The period of oscillation of a magnetic needle in a magnetic field is T. If an identical bar magnetic needle is tied at right angles to it to form a cross (fig), the period of oscillation in the same magnetic field will be

(a) 21/4T

(b) 21/2T

(c) 2T

(d) T√3

(e) T/2

The period of oscillation (T) of the single magnetic needle is given by

T = 2π√(I/mB) where ‘I’ is the moment of inertia of the magnetic needle about the axis of rotation, ‘m’ is the magnetic dipole moment of the needle and ‘B’ is flux density of the magnetic field.

When two magnetic needles are tied together to form a cross, the moment of inertia becomes 2I and the magnitude of the magnetic dipole moment becomes √(m2 + m2) = m√2.

[Note that magnetic dipole moment is a vector quantity. Two identical vectors (each of magnitude m) at right angles will yield a resultant magnitude m√2].

The resultant magnetic moment will be directed along the bisector of the angle between the axes of the individual magnets since the magnets are identical. In the absence of a deflecting torque, the resultant dipole moment vector will align along the applied magnetic field B. On deflecting from this position, the system will oscillate with period T1 given by

T1 = 2π√(2I/mB√2) = 2π√(I√2/mB) = 21/4T

(2) Three identical magnetic needles each L metre long and of dipole moment m ampere metre are joined as shown without affecting their magnetisation. At points B and C unlike poles are in contact. The dipole moment of this system is

(a) m

(b) 2m

(c) 3m

(d) 3m/2

(e) 5m/2

The distance (AD) between the ends of the compound magnet is 2L. Since the pole strength is m/L, the dipole moment of the compound magnet is (m/L)2L = 2m

(3) A magnet of length L and moment M is cut into two halves (A and B) perpendicular to its axis. One piece A is bent into a semicircle of radiur R and is joined to the other piece at the poles as shown in the figure below:

Assuming that the magnet is in the form of a thin wire initially, the moment of the resulting magnet is given by

(1) M/2π

(2) M/π

(3) M(2 + π)/2π

(4) Mπ/(2 + π)

The above question appeared in EAMCET 2009 (Medicine) question paper.

The distance between the poles of the resulting magnet is (L/2) + 2R

Since the semicircular portion of radius R is made of the magnetised wire of length L/2, we have L/2 = πR so that R = L/2π and 2R = L/π

Therefore, length of the resulting magnet (L/2) + (L/π)

The pole strength (p) of the magnet is given by

p = M/L

Therefore, the dipole moment of the resulting magnet = Pole strength×Length = (M/L)[ (L/2) + (L/π)] = M(2 + π)/2π