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].

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) 7º

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.

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.

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, v_t = 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

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 = nλ

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 = nλ 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 6^th 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 (1^st 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 6^th 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.

Moment of inertia - Two Multiple Choice Questions

The value of a man resides in what he gives and not in what he is capable of receiving

– Albert Einstein

You will find many questions involving moment of inertia on this site and you can access all of them by clicking on the label ‘MOMENT OF INERTIA’ below this post. I give you two more questions here, the first being the one which appeared in the AIEEE 2008 question paper. You will have to apply the parallel axis theorem to answer these questions.

(1) Consider a uniform square plate of side ‘a’ and mass ‘m’. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is

(1) (5/6) ma²

(2) (1/12) ma²

(3) (7/12) ma²

(4) (2/3) ma²

The moment of inertia (I_CM) of the square plate about an axis through its centre of mass (CM) passing perpendicular to its plane is given by

I_CM = m(a² + a²)/12 = ma²/6.

[Note that generally for a bar of rectangular cross section, the moment of inertia about an axis passing through the centre, perpendicular to its length is m(l² + b²)12 where l is the length and b is the breadth].

The moment of inertia (I) about a parallel axis through one corner O is given by

I = I_CM + m(a/√2)² since the corner is at a distance a/√2 from the centre of mass (fig).

Thus I = ma²/6 + ma²/2 = (2/3) ma²

(2) A thin uniform rod of mass 6m is bent to form a regular hexagon of side a. What is the moment of inertia of the hexagon about an axis passing through its centre and perpendicular to its plane?

(a) 5 ma²

(b) 5 ma²/6

(c) 5 ma²/3

(d) 6 ma²

(e) 6 ma²/5

Consider one side (such as AB) of the hexagon. Since its mass is m and its length is a, its moment of inertia about an axis through its centre and perpendicular to its length is given by

I_CM = ma²/12.

The moment of inertia of this side about a parallel axis through the centre C of the hexagon is given by

I = I_CM + m[(^√3/₂)a] ² since the side is at a distance (^√3/₂)a from the centre of the hexagon (fig).

Thus I = ma²/12 + 3ma²/4 = 5 ma²/6.

Since there are six sides, the total moment of inertia of the hexagon is 6×5 ma²/6 = 5 ma².

Resistive Networks- AIPMT 2008 Questions

The following questions on resistive networks appeared in All India Pre-Medical/Pre-Dental Entrance Examination 2008 question paper:

(1) In the circuit shown, the current through the 4 Ω resistor is 1 amp when the points P and M are connected to a d.c. voltage source. The potential difference between the points M and N is

(1) 1.0 volt

(2) 0.5 volt

(2) 3.2 volt

(2) 1.5 volt

Since the current through the 4 Ω resistor is 1 ampere, the potential difference between the points P and M is 4 volt. The 1 Ω resistor and the two parallel resistors (each of value 0.5 Ω) make the total resistance in the branch PNM equal to 1.25 Ω. The potential drop across the 1 Ω resistor is therefore equal to 4×1/1.25 = 3.2 volt.

[The p.d. of 4 volt gets divided between 1 Ω and 0.25 Ω]

(2) A current of 3 amp flows through the 2 Ω resistor shown in the circuit. The power dissipated in the 5 Ω resistor is

(1) 2 watt

(2) 1 watt

(3) 5watt

(4) 4watt

The potential difference across the 2 Ω resistor is 2×3 = 6 volt.

The same p.d. appears across the branch containing 1 Ω and 5 Ω. Therefore, the current through 1 Ω and 5 Ω is 6 V/6 Ω = 1 A.

Therefore, the power dissipated in the 5 Ω resistor is I²R = 1²×5 = 5 watt.