Saturday, December 27, 2008

IIT-JEE 2008 Reasoning Type Question on Rotation

The following question appeared in IIT-JEE 2008 question paper under Reasoning Type Questions.

STATEMENT -1

Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.

and

STATEMENT -2

By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT -2 is a correct explanation for Statement-1

(B) STATEMENT -1 is True, STATEMENT -2 is True; STATEMENT -2 is NOT a correct explanation for STATEMENT -1

(C) STATEMENT -1 is True, STATEMENT -2 is False

(D) STATEMENT -1 is False, STATEMENT -2 is True

The acceleration (a) of a body of mass M rolling down an inclined plane is given by

a = g sinθ/[1 + (I/MR2)]

where θ is the angle of the plane (with respect to the horizontal), I is the moment of inertia (about the axis of rolling) and R is the radius of the body.

Since the moments of inertia of solid cylinder and hollow cylinder are respectively MR2/2 and MR2 the acceleration a is greater for the solid cylinder. Therefore, the solid cylinder will reach the bottom of the inclined plane first.

Since the cylinders have the same mass and are at the same height they have the same initial gravitational potential energy Mgh. This potential energy gets converted into translational and rotational kinetic energies (obeying the law of conservation of energy) when the cylinders roll down the incline and the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.

Let us consider an ordinary type of multiple choice question now:

A, B and C are three equal point masses (m each) rigidly connected by massless rods of length L forming an equilateral triangle as shown in the adjoining figure. The system is first rotated with constant angular velocity ω about an axis perpendicular to the plane of the triangle and passing through A. Next it is rotated with the same constant angular velocity ω about the side AB of the triangle. If K1 and K2 are the kinetic energies of the system in the first and the second cases respectively, the ratio K1/K2 is

(a) 2/3

(b) 4/3

(c) 8/3

(d) 10/3

(e) 16/3

The rotational kinetic energy (K) is given by

K = ½ 2

Since the angular velocity is the same in the two cases, the ratio of kinetic energies must be equal to the ratio of moments of inertia.

Thereore, K1/K2 = I1/I2 = 2mL2/m(Lsin60º)2

[Note that in the second case the moment of inertia of the system is due to the single mass at C which is at distance Lsin60º from the axis AB].

Thus K1/K2 = 2/(√3/2)2 = 8/3.

Friday, December 19, 2008

AIPMT 2008 Question on Feed back Amplifier

The following question on negative feed back amplifier is simple even though Plus Two students will normally be unprepared to answer it:

The voltage gain of an amplifier with 9% negative feed back is 10. The voltage gain without feed back will be

(1) 10

(2) 1.25

(3) 100

(4) 90

If the voltage gain without feed back is Av and the feed back factor (fraction of output voltage fed back to the input) is β, the voltage gain (Afb) with feed back is given by

Afb = Av/(1 βAv)

In the case of negative feed back the sign of the feed back factor β is negative so that the voltage gain with feed back is given by

Afb = Av/(1+ βAv)

Since Afb = 10 and the magnitude of β is 9% = 0.09, we have on substituting,

10 = Av/(1+ 0.09Av)

This gives Av = 100.

Here is another question on feed back amplifiers:

Pick out the wrong statement:

When negative feed back is applied in a transistor amplifier

(1) its voltage gain is decreased

(2) its band width is decreased

(3) distortion produced by the amplifier is decreased

(4) the transistor current gain is unchanged

The second option alone is incorrect. The band width of a negative feed back amplifier will be greater than that of the same amplifier without the feed back. Since there is a reduction in the voltage gain consequent on the negative feed back, the 3 dB down frequency on the lower side will be shifted towards lower frequency and the 3 dB down frequency on the upper side will be shifted towards higher frequency.

Friday, December 05, 2008

All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009)

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2009 (AIEEE 2009) to be conducted on 26-4-2009 are being distributed from 5.12.2008 and will continue till 5.1.2009. Candidates can apply for AIEEE 2009 either on the prescribed Application Form or make application ‘Online’. Visit the site http://aieee.nic.in immediately for details. Apply for the exam without delay.

You will find many old AIEEE questions (with solution) on this site. You can access all of them by typing ‘AIEEE’ in the search box at the top left of this page and clicking on the adjacent ‘search blog’ box.

Wednesday, November 26, 2008

Electromagnetic Induction- Two Multiple Choice Questions

You should be able to find the answers for the following questions without difficulty. For each question you should take about one minute at the most (if you have understood basic principles in this section thoroughly). (1) A long straight conductor carries a current that increases at a constant rate. The current induced in the circular conducting loop placed near the straight conductor (fig.) is

(a) zero

(b) constant and in the clockwise direction

(c) constant and in the anticlockwise direction

(d) increasing and in the clockwise direction

(e) increasing and in the anticlockwise direction

A current flows in the circular loop because of the increasing magnetic flux linked with it. This increase in the magnetic flux is to be opposed in accordance with Lenz’s law. The magnetic field lines produced by the straight conductor are directed normally into the plane of the circular coil (away from the reader). Therefore, the direction of the magnetic field lines produced by the induced current in the circular coil must be directed normal to the plane of the circular coil and towards the reader. The induced current should therefore flow in the anticlockwise sense.

The magnitude of the induced current is constant since the rate of increase of flux is constant. The correct option is (c).

(2) In the circuit shown a coil of inductance 0.5 H and negligible resistance is connected to a 12 V battery of negligible internal resistance using a resistive network. If I1 is the current delivered by the battery immediately after closing the switch and I2 is the current (delivered by the battery) long time after closing the switch, (a) I1 = 0.5 A and I2 = 0.6 A

(b) I1 = 0.6 A and I2 = 0.6 A

(c) I1 = 0.5 A and I2 = 0.5 A

(d) I1 = 0 A and I2 = 0.5 A

(e) I1 = 0.6 A and I2 = 0.5 A

The current through the inductance cannot rise abruptly on closing the switch (because of the time constant L/R of the LR circuit). At the moment of switching the circuit on, the circuit behaves as though the inductance branch is absent. The initial current is therefore given by

I1 = 12 V/ (16+8) Ω = 0.5 A

The current through the inductance rises exponentially with time t from zero to the final steady value I [in accordance with the equation I = I0(1– eRt/L) with usual notations]. After a long time, the current through the inductance becomes steady and there is no voltage drop across the inductance. The inductance is in effect short circuited and the final steady current is given by

I2 = 12 v/(16+4) Ω = 0.6 A

[Note that the parallel combined value of 8 Ω and 8 Ω is 4 Ω]

The correct option is (a).

You will find many useful posts on electromagnetic induction on this site. You can access them by clicking on the label ‘electromagnetic induction’ below this post.

Tuesday, November 18, 2008

Joint Entrance Examination for Admission to IITs and other Institutions- (IIT-JEE 2009)

The Joint Entrance (2009) Examination for Admission to IITs and other Institutions (IIT-JEE 2009) will be held on April 12, 2009 (Sunday) as per the following schedule:
09:00 – 12:00 hrs Paper – 1
14:00 – 17:00 hrs Paper – 2
Application materials
of the Joint Entrance Examination 2009 will be issued with effect from 19th November, 2008. On-line submission of Application will also commence on the same day.
Application materials can be purchased from designated branches of Banks and all IITs between 19.11.2008 and 24.12.2008 by paying Rs. 500/- in case of SC/ST/PD/Female candidates, Rs.1000/- in case of male general/OBC candidates and US\$ 100 in the case of candidates appearing in Dubai centre.
The last date for postal request of application materials is 16-12-2008.
The last date for receipt of the completed application at the IITs is 24th December, 2008.
Online Submission of Application: This facility will be available between 19.11.2008 and 24.12.2008 between 8:00 hrs and 17:00 hrs. 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://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
Visit one of these web sites (http://jee.iitm.ac.in) for more details.

Thursday, November 13, 2008

AIEEE 2008 Question Involving Kirchoff’s Laws

Occasionally you will find questions requiring the application of Kirchoff’s laws. Here is a question that appeared in AIEEE 2008 question paper which I give here for clearing your doubts regarding the direction of the current you will have to mark in closed loops in direct current networks:

A 5 V battery with internal resistance 2 Ω and a 2 V battery with internal resistance 1 Ω are connected to a 10 Ω resistor as shown in the figure. The current in the 10 Ω resistor is

(a) 0.27 A, P1 to P2

(b) 0.27 A, P2 to P1

(c) 0.03 A, P1 to P2

(d) 0.03 A, P2 to P1

The circuit is redrawn, indicating the currents in the three branches. We have marked the directions of the currents I1 and I2 in the directions we normally expect the 5 V and the 2 V batteries to drive their currents. [Note that there can be situations in which the direction we mark is wrong. There is nothing to be worried about such situations since you will obtain a negative current as the answer and you will understand that the real direction is opposite to what you have marked in the circuit].

Applying Kirchoff’s voltage law (loop law) to the loops ABP2P1 and P1P2CD we have respectively,

5 = 2×I1 + 10×(I1 I2) and

2 = 1×I2 10×(I1 I2)

The above equations can be rewritten as

12 I1 – 10 I2 = 5 and

10 I1 + 11 I2 = 2

These equations can be easily solved to give I1 = 2.34 A (nearly) and I2 = 2.31 A (nearly) so that the current (I1 I2) = 0.03 A.

Since the currents I1 and I2 are obtained as positive, the directions we marked are correct and the current flowing in the 10 Ω resistor is 0.03 A, flowing from P2 to P1 [Option (4)].

[Suppose we had marked the current I2 as flowing in the opposite direction. The current flowing in the 10 Ω resistor will then be (I1 + I2). We will then obtain I1 = 2.34 A and I2 = – 2.31 A, the negative sign indicating that the real direction of I2 is opposite to what we marked. The current flowing through the 10 Ω resistor will again be obtained correctly as (I1 + I2) = 2.34 A– 2.31 A = 0.03 A].

Thursday, October 30, 2008

Geometrical Optics- Questions (MCQ) on Refraction at Prisms

Questions involving the refraction produced by prisms often find place in Medical, Engineering and other Degree Entrance Exam question papers. Here are two questions in this section:

(1) The face AC of a glass prism of angle 30º is silvered. A ray of light incident at an angle of 60º on face AB retraces its path on getting reflected from the silvered face AC. If the face AC is not silvered, the deviation that can be produced by the prism will be (a) 0º

(b) 30º

(c) 45º

(d) 60º

(e) 90º

The deviation (d) produced by a prism is given by

d = i1 + i2 A where i1 and i2 are respectively the angle of incidence and the angle of emergence and A is the angle of the prism. Here we have i1 = 60º, i2 = 0º (since the ray falling normally will proceed undeviated from the face AC if it is not silvered) and A = 60º.

Therefore, d = 30º.

(2) In the above question, what is the refractive index of the material of the prism?

(a) 1.732

(b) 1.652

(c) 1.667

(d) 1.5

(e) 1.414

In the triangle ADN angle AND is 60º since angle DAN = 30º and angle DNA = 90º. Therefore, the angle of refraction at D is 30º. The refractive index of the material of the lens (n) is given by

n = sin i1/ sinr1 = sin 60º/ sin 30º = √3 = 1.732

(3) Two thin (small angled) prisms are combined to produce dispersion without deviation. One prism has angle 5º and refractive index 1.56. If the other prism has refractive index 1.7, what is its angle?

(a) 3º

(b) 4º

(c) 5º

(d) 6º

(e)

Since the deviation (d) produced by a small angled prism of angle A and refractive index n is given by

d = (n – 1)A, the condition for dispersion without deviation on combining two prisms of angles A1 and A2 with refractive indices n1 and n2 respectively is

(n1 – 1)A1 = (n2 – 1)A2

Therefore, 0.56×5 = 0.7×A2 so that A2 = 4º

On this site you will find many questions (with solution) on refraction at plane surfaces as well as at curved surfaces. To access all of them type in ‘refraction’ in the search box at the top left side of this page and click on the adjacent ‘search blog’ box.

Saturday, October 25, 2008

All India Pre-Medical / Pre-Dental Entrance Examination -2009 (AIPMT 2009)

Central Board of Secondary Education, Delhi has invited applications in the prescribed form for All India Pre-Medical / Pre-Dental Entrance Examination -2009 as per the following schedule 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, Municipal or other local authorities in India except in the States of Andhra Pradesh and Jammu & Kashmir:-

1. Preliminary Examination - 5th April, 2009 (Sunday)

2. Final Examination - 10th May, 2009 (Sunday)

Candidate can apply for the All India Pre-Medical/Pre-Dental Entrance Examination ither offline or online as explained below:

Offline

Offline submission of Application Form may be made using the prescribed application form. The Information Bulletin and Application Form costing Rs.600/- (including Rs.100/- as counselling fee) for General & OBC Category Candidates and Rs.350/- (including Rs.100/- as counselling fee) for SC/ST Category Candidates inclusive of counseling fee can be obtained against cash payment from 22-10-2008 to 01-12-2008 from any of the branches of Canara Bank/ Regional Offices of the CBSE. Find details at http://www.aipmt.nic.in/.

Online

Online submission of application may be made by accessing the Board’s website http://www.aipmt.nic.in/. from 22-10-2008 (10.00 A.M.) to 01-12-2008 (5.00 P.M.). 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. Fee of Rs.600/-(including Rs.100/- as counseling fee) for General and OBC Category Candidates and Rs.350/- (including Rs.100/- as counseling fee) for SC/ST category candidates may be remitted in the following ways :

1. By credit card or

2. Through 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 is available on the website http://www.aipmt.nic.in/. Application Form along with original Demand Draft should reach the Board on or before 04-12-2008 which is the last date stipulated.

Visit the web site http://www.aipmt.nic.in/. for all details and information updates.

Tuesday, October 14, 2008

Questions (MCQ) on Newton’s Laws

The following simple questions may prompt you to pick out the wrong option if you are in a hurry. So, be cautious and don’t overlook basic points. Here are the questions:

(1) Two identical frictionless pulleys carry the same mass 2m at the left ends of the light inextensible strings passing over them. The right end of the string carries a mass 3m in the case of arrangement (i) where as a force of 3mg is applied in the case of arrangement (ii) as shown in the adjoining figure. The ratio of the acceleration of mass 2m in case (i) to the acceleration of mass 2m in case (ii) is (a) 2:3

(b) 1:1

(c) 5:3

(d) 3:5

(e) 2:5

In both cases the net driving force is 3mg – 2mg = mg. But in case (i) the total mass moved is 5m where as in case (ii) the total mass moved is 2m. The acceleration in case (i) is mg/5m = g/5 where as the acceleration in case (ii) is mg/2m = g/2.

The ratio of accelerations = (g/5)/ (g/2) = 2/5 [Option (e)].

(2) An object of mass 4 kg moving along a horizontal surface with an initial velocity of 2 ms–1 comes to rest after 4 seconds. If you want to keep it moving with the velocity of 2 ms–1, the force required is

(a) zero

(b) 1 N

(c) 2 N

(d) 4 N

(e) 8 N

The acceleration ‘a’ of the body is given by

0 = 2 + a×4, on using the equation, vt = v0 + at

Therefore, a = – 0.5 ms–2

The body is retarded (as indicated by the negative sign) because of forces opposing the motion. The opposing force has magnitude ma = 4×0.5 = 2 N. Therefore, a force of 2 N has to be applied opposite to the opposing forces to keep the body moving with the velocity of 2 ms–1.

You will find some useful multiple choice questions (with solution) in this section here as well as here

Sunday, September 28, 2008

IIT-JEE 2008 Question (MCQ) on Young’s Double Slit

Today we will discuss two questions on Young’s double slit. The following multiple correct answers type question appeared in IIT-JEE 2008 question paper:

In a Young’s double slit experiment, the separation between the two slits is d and the wave length of the light is λ. The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2. Choose the correct choice(s).

(A) If d = λ, the screen will contain only one maximum.

(B) If λ < d < 2λ, at least one more maximum (besides the central maximum) will be observed on the screen.

(C) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase.

(D) If the intensity of light falling on slit 2 is reduced so that it becomes equal to that of slit 1, the intensities of the observed dark and bright fringes will increase The path difference [S2N = (S2P S1P) in fig.] between the interfering beams is d sinθ. Therefore for obtaining a maximum at the point P on the screen the condition to be satisfied is

dsinθ =

When d = λ, sinθ = n

Since sinθ ≤ 1, n ≤ 1

The possible values of n are 0 and 1. But the maximum corresponding to n = 1 cannot be observed since θ cannot be 90º. Therefore, the central maximum of order zero corresponding to zero path difference (n = 0) alone can be observed. Option A is therefore correct

If λ < d < 2λ, the limiting values of d are λ and 2λ. In addition to the usual central maximum, fringes of order up to n satisfying the equation, dsinθ = will be obtained, where d < 2λ.

If d = 2λ, we have 2λ = nλ/sinθ so that n = 2 sinθ.

Since sinθ should be less than 1 for observing the fringe, the maximum value of n must be 1. Therefore, the central maximum and the maximum of order n = 1 will be observed. Option B too is therefore correct

Options C and D are obviously incorrect since the intensity of the dark fringes will become zero when the light beams passing through the slits are of equal intensity.

Now consider another question on Young’s double slit:

Suppose the wave length λ and the double slit separation d in a Young’s double slit experiment are such that the 6th dark fringe is obtained at point P shown in the above figure. The path difference (S2P – S1P) will be

(a) 5 λ

(b) 5 λ/2

(c) 6 λ

(d) 3 λ

(e) 11 λ/2

The dark fringe of order 1 (1st dark fringe) is formed when the path difference is λ/2. The dark fringe of order 2 is formed when the path difference is 3λ/2 (and not 2λ/2) and the dark fringe of order 3 is formed when the path difference is 5λ/2 (and not 3λ/2). Generally, the dark fringe of order n is formed when the path difference is (2n 1)λ/2. The 6th dark fringe is therefore formed when the path difference is 11λ/2.

By clicking on the label ‘wave optics’ below this post, you can access all related posts on this site.

You will find useful posts on wave optics here at apphysicsresources.