Thursday, March 29, 2007

Multiple Choice Questions on Radioactivity

The following Question appeared in Karnataka CET 2003 question paper:
Half life of a radioactive substance is 20 min. The time between 20% and 80% decay will be
(a) 25 min. (b) 30 min. (c) 40 min. (d) 20 min
If the initial activity is A0, the activity ‘A’ after ‘n’ half lives is given by
A = A0/2n.
Let us take the initial activity as 100 units. After 20% decay, the activity becomes 80 units and after 80% decay, the activity becomes 20 units. These two cases can be stated as
80 = 100/2n and
20 = 100/2m
where ‘n’ and ‘m’ are the numbers of half lives required for 20% decay and 80% decay respectively.
Dividing, 80/20 = 2m/2n = 2(m–n). Or, 2(m-n) = 4, from which (m–n) = 2 half lives = 2×20 min. = 40 min.

Now consider the following MCQ which appeared in Karnataka CET 2004:
A count rate meter shows a count of 240 per minute from a given radioactive source. One hour later the meter shows a count rate of 30 per minute. The half life of the source is
(a) 80 min. (b) 120 min. (c) 20 min. (d) 30 min.
From the equation, A = A0/2n, we have 30 = 240/2n so that n = 3. Therefore one hour is equal to 3 half lives which means the half life of the substance is 20 min.

Tuesday, March 27, 2007

MCQ on Oscillation of Magnets

The period (T) of angular oscillations of a magnet suspended freely in a magnetic field of flux density ‘B’ is given by
T = 2π √(I/ mB) where ‘I’ is the moment of inertia of the magnet about the axis of the angular motion and ‘m’ is the magnetic dipole moment. You will find questions based on this relation in your entrance examination question papers. Here is a question which is meant also for checking your grasp of the moment of inertia and the vector property of the magnetic dipole moment:
Two identical magnets are placed one above the other and tied together so that their like poles are in contact. The period of oscillation of this combination (on suspending horizontally using torsionless suspension fibre) in a horizontal magnetic field is 2 s. What will be the period of oscillation if the magnets are placed one above the other such that they are mutually perpendicular and bisect each other?
(a) 2 s (b) √2 s (c) 2√2 s (d) 2¼ s (e) 2×2¼ s

The period of oscillation is given by T = 2π √(I/ mB) with usual notations.
In the both cases, the moment of inertia is twice that of one magnet. In the first case, the net dipole moment is twice that of one magnet since the like poles are pointing in the same direction. In the second case, the net dipole moment is √2 times the moment of one magnet since the moment vectors at right angles get added.
The equations for the period in the two cases are therefore
T1 = 2π √(2I/ 2mB) = 2π √(I/ mB) = 2 s (as given in the question) and
T2 = 2π √(2I/ √2 mB) = 2π √(√2 I/ mB) = T1 √(√2) = 2×2¼ s.
The following MCQ appeared in EAMCET (Med) A.P.2003 question paper:
The period of oscillation of a magnet at a place is 4 seconds. When it is remagnetised so that the pole strength becomes 4 times the initial value, the period of oscillation in seconds is
(a) ½ (b) 1 (c) 2 (d) 4

The period of oscillation is given by T = 2π √(I/ mB) with usual notations. The period is therefore inversely proportional to the square root of the dipole molent ‘m’. When the pole strength is made 4 times the initial value, the dipole moment is made 4 times the initial value so that the period becomes half the initial value. The correct option is 2 seconds.

Thursday, March 22, 2007

Two Questions Involving Gravitation

Most of you may be knowing that the moon does not possess an atmosphere because the thermal velocity acquired by gas molecules on the moon when heated by the solar radiations is significant compared to the escape velocity on the moon’s surface (2.4 km/s). The escape velocity on the earth’s surface is 11.2 km/s which is much greater than the velocity acquired by oxygen and nitrogen gas molecules on getting heated by solar radiations. (In the case of hydrogen molecules, this is not the case). It is enough that the most probable velocity
[√(2RT/M)] of a gas molecule is in excess of about 20% of the escape velocity, for the molecule to escape to outer space.
Now, consider the following question:
The radius of the earth is 6400 km and the acceleration due to gravity on the earth’s surface is 9.8 ms–2. The universal gas constant is 8.4 J mol–1 K–1. The temperature at which the r.m.s. velocity of oxygen gas molecules becomes equal to the velocity of escape from the surface of the earth is
(a) 1.59×106 K (b) 1.59×105 K (c) 1.59×104 K (d) 1.59×103 K (e) 1.59×102 K
The escape velocity is given by ve = √(2gRE) where ‘g’ is the acceleration due to gravity and ‘RE’ is the radius of the earth.
On substituting for ‘g’ and ‘RE’, the escape velocity, ve = 11.2×103 m/s
The molecular velocity (r.m.s.) is given by v = √(3RT/M) where ‘R’ is universal gas constant, ‘T’ is the temperature (in Kelvin) and ‘M’ is the molar mass of the gas (oxygen in the present case).
Therefore, √(3RT/M) = 11.2×103. Substituting for R = 8.4 and M = 0.032 kg, the temperature works out to be 1.59×105 K.
Now, consider the following question which is based on Kepler’s law:
A planet moves around the sun. When it is farthest away from the sun at distance r1, its speed is v1. When it is closest to the sun at distance r2 its speed will be
(a) r1v1/r2 (b) (r1/r2)2 v1 (c) √(r1/r2) ×v1 (d) r2v1/r1 (e) √(r2/r1)× v1

According to Kepler’s law, the line joining the planet to the sun sweeps out equal areas in equal intervals of time. If we consider a very small time interval δt, the areas swept when the planet is at apogee (farthest away) and at perigee (closest to the sun) will be triangles whose areas are directly proportional to v1r1 and v2r2 respectively. [The bases of the triangular areas swept in the time δt are v1δt and v2δt and the altitudes are r1 and r2 respectively].
Therefore, from Kepler’s law, r1 v 1 = r2 v2 so that v2 = r1v1 /r2

Sunday, March 18, 2007

Vibration of Strings

The fundamental frequency of vibration (n) of a string (or wire) is given by
n = (1/2L)√(T/m) where L is the length of the wire, T is the tension and ‘m’ is the linear density (mass per unit length) of the string.
You may get questions based on this relation. See the following MCQ:
A sonometer wire and a tuning fork are excited together. Four beats are heard when the length of the wire is 60 cm as well as 62 cm. The frequency of the tuning fork is ( in Hz)
(a) 512 (b) 488 (c) 384 (d) 256 (e) 244
Since the frequency of vibration of the wire is inversely proportional to its length, we can write, in the two cases,
(n + 4) α 1/60 and
(n – 4) α 1/62 where ‘n’ is the frequency of the tuning fork.
[Note that the frequency of the wire is greater than that of the fork when its length is smaller].
Dividing, (n + 4)/ (n – 4) = 62/60 from which n = 244Hz.
Now, consider the following questionwhich appeared in EAMCET (Med) 2003 question paper:
Two uniform wires are vibrating simultaneously in their fundamental modes. The tensions, lengths, diameters and the densities of the two wires are in thr ratio 8:1, 36:35, 4:1 and 1:2 respectively. If the note of the higher pitch has a frequency 360 Hz, the number of beats produced per second is
(a) 5 (b) 10 (c) 15 (d) 20
Since the frequency of vibration of a stretched wire is given by
n = (1/2L)√(T/m) = (1/2L)√(T/πr2ρ) where L is the length of the wire, T is the tension and ‘m’ is the linear density (mass per unit length) which is πr2ρ where ‘r’ is the radius and ρ is the density of the material of the wire, we have,
n1/n2 = (L2/L1) √[(T1/T2)( r2 /r1)221)]
= (35/36) √[(8/1)( 1 /16) (2/1)]
= (35/36)×1
This means that n2 is the higher frequency. Since the higher frquency is given as 360 Hz, we have n1/360 = 35/36 from which n1 = 350 Hz.
Therefore, beat frequency = 360 – 350 = 10 Hz.

Friday, March 16, 2007

Questions (MCQ) on Direct Current Circuits

Here is a question which you can work out using Ohm’s law only:
Two constantan wires P and Q have their lengths in the ratio 1:2 and radii in the ratio 2:1. They are connected in series and potentials 2V and 20V are applied at the free ends of P and Q respectively. The potential at the junction of the wires is
(a) 2V (b) 4V (c) 9V (d) 12V (e) 16V
The resistances of P and Q are in the ratio 1:8 since the length of Q is twice that of P and the area of cross section of Q is a quarter if that of P. [The resistance is given by R = ρL/A where ρ is the resistivity, L is the length and A is the area of cross section].
The potential drops across P and Q (when current flows in them) are in the ratio 1:8. The potential difference applied across the series combination of P and Q is (20–2) = 18V. Therefore, the potential drop across Q = 18×8/(1+8) = 16V.
The potential at the junction of P and Q is (20V– 16V) = 4V.
Now, consider the following MCQ:
In the circuit shown, the power dissipated in the 2Ω resistor is 9W. What is the power dissipated in the 4Ω resistor?
(a) 18W (b) 12W (c) 9W (d) 3W (b) 2W
If the current through the 2Ω resistor is ‘I’, the current through the 4Ω resistor is I/3 since the total resistance (9Ω) in the 4Ω resistor branch is 3 times the total resistance (3Ω) in the 2Ω resistor branch. The expressions for power dissipation in 2Ω and 4Ω are respectvely
P = I2 ×2 and
P' = (I/3)2 ×4 so that P'/P = 2/9 from which P'= P×(2/9) = 9×2/9 = 2W.

Wednesday, March 14, 2007

Vibration of Air Columns – Resonance Column & Organ Pipe

In Acoustics a closed pipe or organ pipe means a tube closed at one end. An open pipe or organ pipe means a tube open at both ends. When a standing wave (stationary wave) is formed in an organ pipe, the closed end will be a node and the open end will be an antinode. This is why the length of the pipe in the fundamental mode is equal to λ/4 (which is the distance between neighbouring node and antinode) in a closed pipe. In an open pipe, in the fundamental mode, the length of the pipe is equal to λ/2 since the consecutive antinodes are located at the ends of the tube.
You should remember that a closed pipe can support odd harmonics only where as an open pipe can support all harmonics. In other words, the frequencies of vibration of the air column in a closed pipe are in the ratio 1: 3 : 5 : 7 : etc., while those in an open pipe are in the ratio 1 : 2 : 3 : 4 : 5 : etc.
Now, consider the following MCQ:
An open organ pipe and a closed organ pipe have the same length. The ratio of their fundamental frequencies is
(a) 1 (b) 2 (c) 3 (d) 4 (e) 3/4
If ‘L’ is the length of the pipe, we have, L = λ/2 for the open pipe and L = λ'/4 for the closed pipe where λ and λ' are the wave lengths of sound in the two cases (in the fundamental mode).
The corresponding fundamental frequencies are n = v/ λ = v/2L and n' = v/λ' = v/4L, from which n/n' = 2 [Option (b)].
Let us consider another question:
Almost the entire length of an aluminium pipe of length 1.1m is dipped vertically in water contained in a tall jar. An excited tuning fork of frequency 500 Hz is held at the upper end of the pipe and the pipe is gradually raised. How many discrete resonance conditions are possible? (Speed of sound in air = 330 m/s)
(a) 1 (b) 2 (c) 3 (d) 4 (e) 5
The arrangement mentioned in this problem makes a simple resonance column apparatus. The wave length of sound emitted by the fork, λ = v/n = 330/500 = 0.6666 m.
The first resonance (fundamental mode) is obtained when the exposed length of the pipe is λ/4. The second resonance is obtained when the exposed length is 3 λ/4. These two are definitely possible since the length of the pipe is 1.1 m and λ = 0.6666 m.
The third resonance will be obtained when the exposed length is 5λ/4 = 5×0.6666/4 = 0.83 m. This too is possible.
The fourth resonance will be obtained when the exposed length of the pipe is 7λ/4 = 7×0.6666/4 = 1.16 m. This is not possible since the length of the entire pipe is 1.1 m only.
So, the correct option is (c).

Saturday, March 10, 2007

MCQ on Communication Systems

Questions on communication systems at the higher secondary/plus two level are simple. Here is a question on optical communication systems:
An optical communication system operates at a wave length of 750 nm. The available channel band width for optical communications is only 1% of the optical source frequency. How many TV signals can the system accommodate if each signal requires a band width of 5 MHz?
(a) 8×105 (b) 7.5×105 (c) 6×105 (d) 5×105 (a) 4×105

The optical source frequency, f = c/λ = 3×108/(750×10–9)= 4×1014 Hz.
Total band width available in the system = 1% of 4×1014 Hz = 4×1012 Hz.
Therefore, no. of TV signals that can be accommodated = (4×1012 Hz)/ (5×106 Hz) = 8×105.
Now, consider the following MCQ:
A 9 MHz signal is transmitted from a ground transmitter at a height of 300m. The maximum electron density of the ionosphere is 1.44×1012. A receiver at a distance of 60 km can receive the signal by
(a) space wave only (b) sky wave only (c) space wave and sky wave (d) satellite transponder only (e) sky wave and satellite transponder

The maximum line of sight distance possible is given by d = √(2Rh) where ‘R’ is the radius of the earth (6400 km) and ‘h’ is the transmitter height. Therefore, d = √(2×6400×103×300) = 62×103 m = 62 km.
Reception by space wave is possible since the receiver is at 60 km.
The upper frequency limit for ionospheric reflection (critical frequency) is given by
fc = 9√Nmax = 9√(1.414×1012) = 10.8×106 Hz =10.8 MHz.
The transmitter frequency is 9 MHz only so that the waves can be reflected by the ionosphere.
So, reception by sky wave also is possible and the correct option is (c).
[Note that satellite transponder doesn’t come into the picture since the waves cannot penetrate through the ionosphere].
Here is another MCQ:
A photo detector is made using a semiconductor having a band gap of 1.55 eV. The maximum wave length it can detect is nearly
(a) 500 nm (b) 600 nm (c) 750 nm (d) 800 nm (e) 850 nm
If the wave length is too large, the photon energy will be too small and the incident light will not be able to produce charge carriers in the semiconductor. With a given semiconductor therefore, there is an upper limit for the detectable wave length. Note that the product of the wave length in Angstrom and the energy in electron volt of any photon is 12400 (very nearly). So, the 1.55 eV photon has wave length equal to (12400/1.55) Ǻ = 8000 Ǻ = 800 nm. This is the maximum wave length this semiconductor can detect [Option (d)].

Thursday, March 08, 2007


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 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.
Further details can be obtained from the prospectus as well as from the website Make it a point to visit the site for information updates.

Sunday, March 04, 2007

Charged Particles in Magnetic and Electric Fields

Here is a multiple choice question which appeared in AIEEE 2002 question paper:
If an electron and a proton having same momenta enter perpendicular to a magnetic field, then
(a) curved path of electron and proton will be same (ignoring the sense of revolution)
(b) they will move undeflected
(c) curved path of electron is more curved than that of the proton
(d) path of proton is more curved

The radius of the circular path of the electron is obtained by equating the magnetic force to the centripetal force: qvB = mv²/r. The radius ‘r’ is therefore given by r = mv/qB. The radius is therefore directly proportional to the momentum (mv) and inversely proportional to the charge (q) of the particle.
The momenta are given as equal in the problem. Since the proton and the electron have the same charge magnitudes and are moving in the same magnetic field B, they will follow paths of the same radius[Option (a)].
You might have noted that the path of a charged particle in an electric field is generally parabolic. This is because of the fact that in a uniform electric field, the electric force on the particle has the same direction everywhere. The motion is similar to the projectile motion in a gravitational field. Now, consider the following question:
A particle of charge ‘+q’ and mass ‘m’ is projected with a velocity ‘v’at an angle ‘θ’ with respect to the horizontal, in an electric field ‘E’ which is directed vertically downwards. If there are no gravitational or magnetic fields, the horizontal range of the particle is
(a) (v²sin2θ)/E (b) (v²sin2θ)/qE (c) (mv²sin2θ)/E (d) (mv²sin2θ)/qE (e) (qmv²sin2θ)/E
In the case of the motion of a projectile in a gravitational field, the expression for horizontal range is R = (v² sin2θ)/g. In the present case, the gravitaional acceleration ‘g’ is replaced by the acceleration produced by the electric field.
Acceleration produced by the electric field = Force/ Mass = qE/m. The correct option therefore is (d).
The following MCQ appeared in AIIMS 2004 question paper:
The cyclotron frequency of an electron gyrating in a magnetic field of 1 T is approximately
(a) 28 MHz (b) 280 MHz (c) 2.8 GHz (d) 28 GHz
The cyclotron frequency is the frequency with which a charged particle describes circular path in a magnetic field and is given by f =qB/2πm with usual notations. [You can get it this way: qvB = mrω² where ω is the angular frequency. Substituting v = ωr in this, we get ω = qB/m. Frquency f = ω/2π =qB/2πm].
Substituting for the mass and charge of the electron, we have
f = (1.6×10–19 ×1)/ (2π×9.1×10–31 ) = 28×109 Hz = 28 GHz.

Saturday, March 03, 2007

Questions on Polarisation -Brewster’s law

Most of you might have noted that the transverse nature of light wave was proved by the phenomenon of polarisation. You should remember that sound wave cannot be polarised since it is longitudinal.
You will often find questions based on Brewster’s law in the section on polarisation. When unpolarised light proceeding through a rarer medium is incident at an angle ‘i’ on a denser medium, the reflected beam is plane polarised if tan i = n where ‘n’ is the refractive index of the denser medium with respect to the rarer medium.. This is Brewster’s law. [ The transmitted beam will be partially plane polarised].
Here is a simple question based on Brewster’s law:When unpolarised beam of light is incident on a glass slab, the rflected beam is found to be completely plane polarised. The angle between the reflected beam and the transmitted beam is
(a) 30° (b) 45° (c) 60° (d) 90° (e) dependent on the refractive index
The correct option is (d). You can easily prove this as follows:
Since n = sin i/sin r, we have tan i = sin i/ sin r so that sin i/ cos i = sin i/sin r. Therefore, cos i = sin r. Therefore, r = 90° – i so that i + r = 90°. With reference to the figure, the angle BOC ( which is the angle between the reflected and transmitted beams) is therefore equal to 90°.
Now consider the following question:
When unpolarised light proceeding through air is incident at an angle of 60° on a transparent slab, the reflected rays are found to be completely plane polarised. The refractive index of the slab and the angle of refraction into the slab are respectively
(a) 1.7, 30° (b) 1.414, 40° (c) 1.5, 30° (d) 1.3, 45° (e) 1.732, 30°
If you are in too much hurry, you may pick out option (a). But the correct option is (e). Refractive index, n = tan i = tan 60° = √3 = 1.732. Angle of refraction r = 90° – i = 90° – 60° = 30°.