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

– Thomas A. Edison

Friday, March 30, 2007

Here are two questions on self inductance:
(1) The inductance between A and D is
(a) 3.66 H (b) 9 H (c) 0.66 H (d) 1 H
This question appeared in AIEEE 2002 question paper. If you examine the arrangement carefully, you will realise that the three iductors, each having value 3 henry are connected in parallel, between the points A and D.
Inductors connected in parallel will produce an effective value given by the reciprocal relation,
1/L = 1/L1 + 1/L2 + 1/L3.
Since all the three inductors have the same value 3H, the effective value of the inductance is 3H/3 = 1 H
[If the three components were capacitors of value 3 μF each, the effective value would be 9 μF because capacitances in parallel get added as C = C1 + C2 + C3 + etc. It will be convenient to remember that Resistors, reactances ( inductive as well as capacitive) and impedances get added when connected in series ( Z = Z1 + Z2 + Z3 + etc.) while their effective value is given by the reciprocal relation when they are connected in parallel (1/Z = 1/Z1 + 1/Z2 + 1/Z3 +…etc.)].
(2) An air cored coil of 600 turns has a self inductance of 90 mH. If the number of turns is reduced to 200 without changing the area of the coil, the self inductance will be
(a) 30 mH (b) 270 mH (c) 180 mH (d) 90 mH (e) 10 mH
The answer to this question is 10 mH since the self inductance of a coil is directly proportional to the square of the numbr of turns. When the number of turns becomes one third, the inductance becomes one nineth.
[The self inductance is equal to the total magnetic flux linked with the coil (because of its own magnetic field) when unit current flows in it. But this magnetic flux is directly proportional to the magnetic field produced by the coil and the total number of turns in the coil. Since the magnetic field produced by the coil is directly proportional to the total number of turns in the coil, the total magnetic flux linked (and hence the self inductance) is directly proportional to the square of the number of turns in the coil]

Saturday, March 24, 2007

Two Questions Involving Gravitational Potential Energy & Escape Velocity

The following question will be simple if you have a clear idea of the potential energy in a gravitational field:
Two bodies of masses m1 and m2 are initially at rest and and at infinite distance apart. When a very feeble momentary push is given to one of them, they move towards each other because of the gravitational force between them. When the separation between them is ‘r’, their relative velocity of approach is (if G is the gravitational constant)
(a) [2Gr(m1+m2)]½ (b) [2Gr(m1–m2)]½ (c) [1/2Gr(m1+m2)]½
(d) [2G (m1+m2)/r]½ (e) [2G(m1 – m2)/r]½
Imagine that initially the masses are at a distance ‘r’ apart. The gravitational potential energy of the system then is –Gm1m2/r. If the separation between the masses is to be increased to infinity, kinetic energy equal to Gm1m2/r is to be supplied to the system. (At infinite separation, kinetic energy as well as potential energy is zero). When the masses are allowed to approach, they lose potential energy (which becomes negative) and gain kinetic energy and at a separation ‘r’, the kinetic energy is +Gm1m2/r.
If v1 and v2 are the velocities of the masses, we have
½ m1v1 2 + ½ m2v2 2 = Gm1m2/r.
Since the momenta are equal (in magnitude), m1v1 = m2v2 so that v2 = m1v1/m2.
Substituting this in the energy equation above, v1 = [2G/(m1 + m2)r]½ ×m2.
Similarly, v2 = [2G/(m1 + m2)r] ½ ×m1.
Since the masses are moving in opposite directions, their relative velocity is
v1 + v2 =[2G/(m1 + m2)r]½ × (m2 + m1) = [2G(m1+m2)/r]½
[The solution to this problem has been edited. Thank you Dr. Ravikrisnan, for pointing out the mistake in the solution which I had posted yesterday].
Now, consider the following question, which will be simple if you have a clear idea of the velocity of escape in a gravitational field:
The surface value of acceleration due to gravity on a planet of radius ‘R’ is ‘g’. If a body of mass ‘m’ at infinite distance from the planet is given a feeble momentary push so that it moves towards the planet, what will be its kinetic energy when it strikes the surface of the planet? (Assume that the gravitational field of the planet only is significant and the effect of atmosphere on the motion of the body is negligible).
(a) Infinite (b) mgR2 (c) ) mgR3 (d) 2mgR (e) mgR

The body will strike the surface of the planet with a velocity equal to the escape velocity which is √(2gR). The kinetic energy with which it will strike the surface is therefore ½ m[√(2gR)]2 = mgR.

Tuesday, March 20, 2007

Power in A.C. Circuits

The power(P) dissipated in an alternating current circuit is given by
P = Vrms Irms cosΦ where cosΦ is the power factor. [You should remember that Φ is the phase difference between current and voltage].
Power in an A.C. circuit is also given by P = (Irms)²R. Note that power is dissipated in resistance only.
Another form of the power relation is P = (Irms)²Z cosΦ, which follows from cosΦ = R/Z

Some of you may be wondering which relation is to be used for calculating power. Well, you may use any one of them which is convenient to you in the context of the problem. Now, consider the following MCQ:
A voltage v = V0 cosωt is applied in series with an inductance L and a resistance R. The average power dissipated per cycle in the circuit is
(a) V02 /R (b) V02/2R (c) V02/(R+L) (d) V02R/[(R2+ L2ω 2)] (e) V02R/[2(R2+ L2ω 2)]
We shall use the relation, P = (Irms)²R. Since Irms = Vrms/Z = (V0/√2)/[√(R2+ L2ω 2)], we obtain P = V02R/[2(R2 + L2ω 2)].
You could have used the other relations (for power) also to obtain the result.
Consider the folowing MCQ which appeared in C.E.T. J&K 2000 question paper:
In an A.C. circuit V and I are given by
V = 100 sin (1000t) volt
I = 1000 sin (1000t + π/3 ) mA.
The power dissipated in the circuit is
(a) 104 W (b) 25 W (c) 10 W (d) 250 W
Here current, voltage and the phase difference (between current and voltage) are given so that it is convenient to use the relation P = Vrms Irms cosΦ. [Note that the current has a phase lead of π/3 and is given in mA].
Therefore, P = (100/√2) × (1000×10 –3/√2) ×cos (π/3) = 25 W

Thursday, March 15, 2007

Magnetic Fields Produced by Current Carrying Conductors

Let us consider the following MCQ which appeared in AIEEE 2004 question paper:
The magnetic field due to a current carrying circular loop of radius 3 cm at a point on the axis at a distance of 4 cm from the centre is 54 μT. What will be its value at the centro of the loop?
(a) 250 μT (b) 150 μT (c) 125 μT (d) 75 μT
The magnetic flux density(B) at a point on the axis at distance ‘x’ from the centre for a single turn loop is given by
B = μ0 r2I / [2(r²+ x²)3/2] where μ0 is the permeabiloty of free space, ‘r’ is the radius of the loop and ‘I’ is the current.
The magnetic field(B') at the centre of the loop is given by
B' = μ0 I/2r
Dividing, B/B' = r3/(r²+ x²)3/2 from which B' = B[(r²+ x²)3/2] /r3.
You may substitute distances in cm itself since their units will get cancelled in the ratio. On substituting B in μT itself, you will get the answer in μT.
Therefore, B' = 54×[(9+16)3/2]/27 = 54×125/27 = 250 μT.
The following multiple choice question appeared in Kerala Medical Entrance 2003 test paper:
A wire of certain length carries a steady current. It is first bent to form a circular coil of one turn. The same wire is next bent to form a circular coil of three turns. The ratio of magnetic induction at the centre of the coil in the two cases is
(a) 9:1 (b) 1:9 (c) 1:3 (d) 3:1 (e) 1:1
Since the magnetic induction (flux density) at the centre of a circular coil is given by
B = μ0 nI/(2r) where μ0 is the permeability of free space, ‘n’ is the number of turns in the coil, I is the current and ‘r’ is the radius of the coil, we have, B α n/r for a given current.
The ratio of magnetic inductions at the centre is therefore B/B' = (n/n') ×(r'/r) = (1/3) ×(1/3) = 1/9, since 2πr = 3×2πr' so that r'/r = 1/3. The correct option therefore is (b).
Here is a simple question (but be careful):
An infinitely long aluminium pipe of radius ‘r’ carries a current. The magnetic flux density out side the pipe at a distance 3r/2 from the axis is 0.04 T. The magnetic flux density inside the pipe at a distance r/2 from the axis will be
(a) 0.01 T (b) 0.02 T (c) 0.04 T (d) zero (e) infinite
Don’t be distracted by the value of the field out side. You should remember that the magnetic flux density at any point inside a long current carrying pipe is zero. The correct option therefore is (d).

Sunday, March 11, 2007

Questions involving Force of Buoyancy

A wooden cylinder is floating (with its axis horizontal) at the interface of water and oil, with a quarter of its volume in water. If the density of oil is 800 kgm–3 , the density of wood is
(a) 820 kgm–3 (b) 830 kgm–3 (c) 840 kgm–3 (d) 850 kgm–3 (e) 860 kgm–3
Even if the cylinder is floating upright, a quarter of its volume will be in water. The cylinder is in equilibrium because the real weight of the cylinder acting vertically downwards is balanced by the forces of buoyancy exerted by water and oil, acting vertically upwards. Therefore we can write
vρg = v1 ρ1g + v2ρ2g where ‘v’ and ‘ρ’ are the volume and the density of the wooden cylinder, ρ2 and ρ2 are the densities of water and oil and v1 and v2 are the volumes of the cylinder within water and oil respectively. [Note that the force of buoyancy is the weight of the displaced fluid]. Thus, we have
vρ = (v/4) ×1000 + (3v/4) ×800 from which ρ = 850 kgm–3
The same question could have been asked as follows:
A wooden cylinder of height 6 cm and radius 2 cm is floating upright at the interface of water and oil, with 1.5 cm of its length in water. If the density of oil is 800 kgm–3, the density of wood is
(a) 820 kgm–3 (b) 830 kgm–3 (c) 840 kgm–3 (d) 850 kgm–3 (e) 860 kgm–3
Since the cylinder is upright and a quarter of its height is within water, a quarter of its volume is within water as in the above case and you will obtain the same answer.
Now consider the following MCQ:
A small piece of wood of density 900 kgm–3 is released from a depth of 20 cm inside water. If the viscous forces are ignored, the height above the surface of water up to which the ball will jump is
(a) 7.5 cm (b)12.5 cm (c) 16.5 cm (d)20.5 cm (e) 25 cm
The apparent weight of the wooden piece inside water is v(ρ-σ)g where v is its volume, ρ is its density and σ is the density of water. Since ρ is less than σ the apparent weight of the ball is negative which means it is directed upwards. It is this force which drives the wooden piece upwards. The magnitude of this force is v(σ- ρ)g.
Since the mass of the wooden piece is vρ, its upward acceleration inside water is, a = v(σ- ρ)g/vρ = (σ- ρ)g/ρ. The velocity ‘v’ of the ball on reaching the water surface is given by the usual equation of uniformly accelerated motion, v2=u2+2as so that
v2 = 0+2[(σ- ρ)g/ρ]s, where ‘s’ is the distance traveled within water (0.2 m ).
Let us use the energy equation for finding the height ‘h’ up to which the ball will jump outside water:
½ mv2 = mgh, from which h=v2/2g=[(σ-ρ)/ρ]s = [(1000-900)/900]×0.2 =0.022 m = 2.2 cm [Option (e)].

Thursday, March 08, 2007

Admission to M.B.B.S. Course -Session 2007-2008- at JIPMER (Jawaharlal Institute of Post-graduate Medical Education & Research)

Jawaharlal Institute of Post-graduate Medical Education and Rrsearch (JIPMER) has invited applications for admission to the first year MBBS course (Session 2007-2008).
Request for supply of application form and prospectus by post should reach the REGISTRAR (ACADEMIC) JIPMER, PUDUCHERRY - 605 006 on or before 14th March, 2007, along with a Crossed Demand Draft, drawn in favour of ‘ACCOUNTS OFFICER, JIPMER’, PAYABLE AT PUDUCHERRY (PONDICHERRY-605 006) and self addressed stamped envelope for Rs.50/- & of size 26 cm × 32 cm (to send the prospectus with application to the candidates). The Bank draft should be Rs. 350/- for General candidates and Rs.250/- for Scheduled Caste / Scheduled Tribe candidates.
Filled in Application Form
should be sent to the REGISTRAR (ACADEMIC), JIPMER, PUDUCHERRY-605006 so as to reach him on or before 28th March 2007, 4.30 P.M.


Candidates can also download the Prospectus and application from the web site
www.jipmer.edu and submit the application ‘ONLINE’. However, they should take a print out in A4 Size Paper and affix the Photograph and sign the application and send the same along with the Demand Draft (Rs.350/- for General Candidates and Rs.250/- for SC/ST Candidates drawn in favour of the Accounts Officer, JIPMER, Puducherry-6 (Pondicherry-605 006). The D.D. should be drawn on any Nationalized Bank) payable at Puducherry (Pondicherry - 605 006) and attested copy of Community Certificate in case of SC/ST candidates and Medical Certificate in case of Physically Handicapped candidates (if applicable), so as to reach the REGISTRAR (ACADEMIC), JIPMER, PUDUCHERRY-605006 on or before 28th March 2007 (4.30 PM) (Wednesday).
The Entrance Examination will be conducted on Sunday the 27th May, 2007 from 10.00 a.m. to 12.30 p.m. at the following centres:
(1) Puducherry (Pondicherry) (2) Chennai, (3) Hyderabad, (4) Delhi, (5) Kolkata and (6) Thiruvananthapuram.
For further details and information updates, visit the site

Monday, March 05, 2007

Optics-Questions on Interference

The following MCQ appeared in Kerala Engineering Entrance 2005 test paper:

In the Young’s double slit experiment, the intensity of the central maximum is observed to be I0. If one of the slits is covered, the intensity at the central maximum will become

(a) I0/2 (b) I0/√2 (c) I0/4 (d) I0 (e) I02

If the resultant amplitude (due to the two interfering waves) at the central maximum is ‘a’. we can write

I0 α a2, since the intensity is proportional to the square of the amplitude.

When one of the slits is covered, the amplitude is reduced to a/2. If ‘I’ is the intensity at the position of the central maximum now, we can write

I α (a/2)2.

From the above, we obtain I = I0/4 [Option (c)].

Now, consider the following question:

In a double slit interference pattern, the intensity at the centre of a bright fringe is I. The intensity at a point one quarter of the distance to the next bright fringe is

(a) I/2 (b) I/4 (c) I/8 (d) I (a) zero

At the centre of a bright fringe the waves arrive in phase. You may imagine that the photons starting from the two slits are in the same state of vibration when they reach the position of the centre of a bright fringe and that is why their amplitudes get added to produce maximum intensity. At the centre of the next bright fringe, the photons will have an extra phase difference of 2π, but this too is ‘in phase’ condition (for the same state of vibration).

At a point one quarter of the distance to the next bright fringe, the phase difference between the interfering photons will be 2π/4 = π/2.

If ‘a’ is the amplitude of each interfering wave, the resultant amplitude at the centre of a bright fringe is 2a and the intensity I is given by

I α 4a2

At a point one quarter of the distance to the next bright fringe, the amplitudes are added with a phase difference of π/2 and the resultant amplitude is √(a2 + a2) = √2 a. The intensity (I') in this case is given by

I' α 2a2

From the above expressions, we obtain I' = I/2 [Option (a)].

[Note that the intensity (I) produced by two interfering waves of the same amplitude ‘a’ is given by I α 4a2cos2(δ/2) where ‘δ’ is the phase difference].

You will find more multiple choice questions with solution in this section at physicsplus: Multiple Choice Questions on Wave Optics and at physicsplus: Questions on Polarisation