Surface Tension–Questions involving excess of pressure inside a bubble

Here is a question which is popular among question setters:

Two spherical soap bubbles of radii r₁ and r₂ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to

(a) (r₁+r₂)/2 (b) √(r₁+r₂) (c) r₁r₂/(r₁+r₂) (d) (r₁+r₂)/√2 (e) √(r₁²+r₂²)

As the temperature is constant, we have P₁V₁+ P₂V₂ =.PV where P₁ and P₂ are the pressures inside the separate bubbles, V₁ and V₂ are their volumes, P is the pressure inside the combined bubble and V is its volume.

Since the bubbles are located in vacuum, the pressure inside the bubble is equal to the excess of pressure 4T/r, where T is the surface tension and ‘r’ is the radius so that we have (4T/r₁)×[(4/3)πr₁³]+ (4T/r₂)×[(4/3)πr₂³] = (4T/R)×[(4/3)πR³] where R is the radius of the combined bubble.

This yields R = √(r₁²+r₂²).

[You can work out this problem by equating the surface energies: 4πr₁²T+4πr₂²T= 4πR²T, from which R = √(r₁²+r₂²)]

Now, consider the following MCQ:

Excess of pressure inside one soap bubble is four times that inside another. Then the ratio of the volume of the first bubble to that of the second is

(a) 1:16 (b) 1:32 (c) 1:64 (d) 1:2 (e) 1:4

If r₁ and r₂ are the radii of the bubbles and T is the surface tension of soap solution, we have (4T/r₁)/(4T/r₂) = 4.

Therefore, r₁/r₂ = ¼. Since the volume is directly proportional to the cube of the radius, the ratio of volumes V₁/V₂ = (r₁/r₂)³ = (¼)³ = 1/64 [Option (d)].

The following MCQ appeared in AIEEE 2004 question paper. This question is popular among question setters and has appeared in other entrance test papers as well:

If two soap bubbles of different radii are connected by a tube,

(a) air flows from the bigger bubble to the smaller bubble till the sizes become equal

(b) air flows from the bigger bubble to the smaller bubble till the sizes are interchanged

(c) air flows from the smaller bubble to the bigger

(d) there is no flow of air

The excess of pressure inside the smaller bubble is greater than that inside the bigger bubble. (Remember, P = 4T/r and hence excess of pressure ‘P’ is inversely proportional to the radius ‘r’). Therefore, air will flow from the smaller bubble to the bigger bubble [Option (c)].

MCQ on Vibration of Strings & Sonometer

In continuation of the post dated 18th March 2007, let us discuss two more questions involving transverse waves in stretched strings. The following MCQ appeared in Kerala Engineering Entrance Examination 2006 question paper:

Two stretched strings of same material are vibrating under same tension in fundamental mode. The ratio of their frequencies is 1:2 and ratio of the lengths of the vibrating segments is 1:4. Then the ratio of the radii of the strings is
(a) 2:1 (b) 4:1 (c) 3:2 (d) 8:1 (e) 4:5
You know that the frequency (n) of vibration of a stretched string in the fundamental mode is given by
n = (1/2L)√(T/m) = (1/2L)√(T/πr²ρ) where L is the length of the vibrating segment 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.
Therefore, if n₁ and n₂ are the frequencies of the two wires having lengths L₁ and L₂ and radii r₁ and r₂ respectively, we have
n₁/n₂ = [(1/2L₁)√(T/πr₁² ρ)] / [(1/2L₂)√(T/πr₂²ρ) = L₂r₂/L₁r₁. [Note that the strings have the same density since they are of the same material].
Since n₁/n₂ = ½ and L₁/L₂ = ¼ as given in the question,
½ = 4r₂/r₁ from which r₁/r₂ = 8 [Option (d)].

Now, consider the following question:
A sonometer wire is kept stretched by suspending a 60 kg mass from the free end of the wire. The suspended mass has a volume of 0.01 m³. The fundamental frequency of the wire in this condition is 300 Hz. If the suspended mass is completely immersed in water, the fundamental frequency of the wire will be approximately
(a) 285 Hz (b) 273 Hz (c) 300 Hz (d) 330 Hz (e) 355 Hz
The frequency will decrease since the tension is decreased (because of the decreased weight of the mass in water).
When the mass is in air, the tension,T₁ = mg = 60×g where ‘g’ is the acceleration due to gravity.
When the mass is in water, tension, T₂ = mg – up thrust = 60×g – 0.01×1000×g = 50×g. [Note that the up thrust is the weight of 0.01 m³ of water displaced by the suspended mass].
Since the frequency is given by n = (1/2L)√(T/m) with usual notations, n α √T.
If n₁ and n₂ are the frequencies in the two cases, n₁/n₂ = √(T₁/T₂) or,
300/n₂ = √(60/50) = √1.2 = 1.1 (nearly) so that n₂ = 273 (nearly).

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 = 100/2^m
where ‘n’ and ‘m’ are the numbers of half lives required for 20% decay and 80% decay respectively.
Dividing, 80/20 = 2^m/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.

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) 2^¼ s (e) 2×2^¼ 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) = 2×2^¼ 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.

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 v_e = √(2gR_E) where ‘g’ is the acceleration due to gravity and ‘R_E’ is the radius of the earth.
On substituting for ‘g’ and ‘R_E’, the escape velocity, v_e = 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

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)² (ρ₂/ρ₁)]
= (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.

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.

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).

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
f_c = 9√N_max = 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)].

JAWAHARLAL INSTITUTE OF POSTGRADUATE MEDICAL EDUCATION AND RESEARCH (JIPMER)- Admission to M.B.B.S. Course- 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.

PROSPECTUS AND APPLICATION FORM CAN ALSO BE OBTAINED IN PERSON AT THE BANK OF BARODA, EXTENSION COUNTER, JIPMER, PUDUCHERRY-6 ON PAYMENT OF RS.350/- (IN RESPECT OF GENERAL CANDIDATES) AND RS.250/- (IN RESPECT OF SC/ST CANDIDATES) IN CASH DURING OFFICE (BANK) HOURS TILL 28th MARCH, 2007 (WEDNESDAY) ON ALL WORKING DAYS BETWEEN 9.30 AM TO 3.00 PM ( On Saturday between 9.30 am to 12.00 noon).

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

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.

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 indexThe 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°.

Questions (MCQ) on Tansmission of Heat

In this section you have to remember the following:

(1) The quantity (Q) of heat conducted through a rod of thermal conductivity ‘K’ and cross section area ‘A’ in time ‘t’ under a temperature gradient dθ/dx is given by

Q = KA(dθ/dx)t

(2) If ‘n’ conductors of the same area of cross section having lengths d₁, d₂, d₃….. d_n and thermal conductivities K₁, K₁, K₃,….K_n are joined in series, the same quantity of heat will flow through them in a given time. The thermal conductivity of such a compound rod is given by K = (d₁+ d₂+ d₃+….. +d_n)/[(d₁/K₁)+ (d₂/K₂)+ (d₃/K₃)+…...+(d_n/K_n)]

If two rods of the same length and the same area of cross section are joined in series, the equivalent thermal conductivity can be obtained from the above general expression as K = 2K₁K₂/ (K₁+K₂)

(3) Stefan’s law: Energy (E) radiated per second from a perfectly black body of surface area ‘A’ is given by E = AσT⁴ where ‘σ’ is Stefan’s constant equal to 5.57×10^–8Wm^–2K^–4.

If the body is not a perfect black body, E = AeσT⁴ where ‘e’ is a dimensionless fraction called emissivity.

In the case of a perfectly black body at temperature T, with surroundings at temperature T_0, the net rate of loss of radiant energy is E = Aσ(T⁴ – T₀⁴), since the body emits energy AσT⁴ and absorbs energy AσT₀⁴ per second in accordance with Kirchhoff's law.

(4) Wein’s displacement law: λ_mT = constant (= 0.29cmK) where λ_m is the wavelength for which the energy radiated is maximum at temperature T. If the constant is taken as 0.29 cmK, the value of λ_m is to be substituted in cm.

(5) Newton’s law of cooling: The rate of cooling is directly proportional to the excess of temperature of the body over the surroundings if the excess of temperature is small compared to the temperature of the body (T) and the temperature (T_S) of the surroundings: dT/dt α (T– T_S)

Now consider the following MCQ which appeared in IIT screening 2002 question paper:

An ideal black body at room temperature is thrown in to a furnace. It is observed that

(a) initially it is the darkest body and at later times the brightest

(b) it is the darkest body at all times

(c) it cannot be distinguished at all times

(d) initially it is the darkest body and at later times it cannot be distinguished

A black body is a good emitter as well as a good absorber. Initially it will absorb energy and hence will appear dark. Once it attains the temperature of the furnace, it emits better than the other parts of the furnace and hence appears the brightest. So the correct option is (a).

Consider the following question:

Two copper spheres A and B having the same emissivity have radii 12 cm and 3 cm respectively . If they are maintained at temperatures 727°C and 1727°C respectively, the ratio of energy radiated by A and B is

(a) 0.032 (b) 0.12 (c) 0.48 (d) 1 (e) 2

Since the energy radiated per second is given by E = AeσT⁴ and the emissivities are equal, we have E₁/E₂ = (A₁/A₂)(T₁/ T₂ )⁴.

Since the radii of A and B are in the ratio 4:1, the surface areas are in the ratio 16:1. Since the temperatures are 1000K and 2000K, the ratio (T₁/ T₂ )⁴ = 1/16. Therefore, E₁/E₂ = 1.

Here is a simple question involving Wein’s displacement law:
On examining the spectrum of a star, it is found that mximum energy is emitted at a wave length of 5800 Ǻ. The surface temperature of the star is
(a) 4500 K (b) 5000 K (c) 5500 K (d) 6000 K (e) 6500 K
This question indicates how one can determine the temperature of a distant star from spectroscopic data. According to Wien’s law, we have λ_mT = 0.29cmK, from which T =0.29/(5800×10^–8) = 5000K. [Note that we have substituted the wave length in cm].

Consider the following question involving heat conduction:

Three identical iron rods are welded together to form the shape of Y. The top ends of the ‘Y’ are maintained at 0°C and the bottom end is maintained at 600°C. The temperature of the junction of the three rods is

(a) 100°C (b) 200°C (c) 250°C (d) 300°C (e) 400°C

The quantity of heat conducted per second through the bottom rod making the ‘Y’ gets divided equally at the junction of the three rods. If ‘θ’ is the temperature of the junction, we have

KA(θ–0)/L = 2KA(600 – θ)/L where K is the thermal conductivity, A is the area of cross section and L is the length of the identical rods.

[ Note that the L.H.S. is the quantity of heat conducted through the lower single rod making the ‘Y’ and the R.H.S. is the sum of the quantities of heat conducted through the upper two rods].

The above equation yields θ = 400°C

Heat engine and Refrigerator

From the section Thermodynamics, you will usually get questions on heat engines or refrigerators. You will definitely remember the expression for the efficiency (η) of a Carnot engine: η = (Q₁ –Q₂)/Q₁ = (T₁–T₂)/T₁ where Q₁ is the heat absorbed from the source at temperature T₁ and T₂ is the heat rejected to the sink at the lower temperature T₂.
The coefficient of performance (β) of a refrigerator (heat pump) is given by
β = (Heat removed from the cold body) / (Work done by the pump)
= Q₂/W = Q₂/(Q₁– Q₂) = T₂/(T₁–T₂)Now consider the following simple question which appeared in AIEEE 2002 question paper:
Even Carnot engine cannot give 100% efficiency because we cannot
(a) prevent radiation (b) find ideal source
(c) reach absolute zero temperature (d) eliminate frictionThe correct option is (c). The efficiency is given by the expression, η = (T₁–T₂)/T₁. The percentage efficiency is [(T₁–T₂)/T₁] ×100. This shows that the efficiency is 100% only if either the source temperature T₁ is infinite or the sink temperature T₂ is zero. Both are impossibilities.
Now see the following MCQ:
In a Carnot engine 800 J of heat is absorbed from a source at 400 K and 640 J of heat is rejected to the sink. The temperature of the sink is
(a) 320 K (b) 100 K (c) 273 K (d)250 K (e) 200 KIn a Carnot engine, Q₁/T₁ = Q₂/T₂ so that the temperature of the sink, T₂ = T₁Q₂/Q₁ = 400×640/800 = 320 K.
The following question appeared in Kerala Engineering Entrance 2000 question paper:
A heat engine undergoes a process in which its internal energy decreases by 400 J and it gives out 150 J of heat. During the process
(a) it does 250 J of work and its temperature rises
(b) it does 250 J of work and its temperature falls
(c) it does 550 J of work and its temperature rises
(d) it does 550 J of work and its temperature falls
(e) 250 J of work is done on the systemThe internal energy of the system will decrease when the system does work and/or gives off heat. Since the heat given out is 150 J and the reduction in internal energy is 400 J, the work done by the engine is 400– 150 = 250 J.
When the internal energy is reduced, the system is cooled. So, the correct option is (b).
Now, consider the following question:
The temperature inside a refrigerator is 4°C and the room temperature is 27°C. How many joules of heat will be delivered to the room for each joule of electricity consumed by the refrigerator?( Treat the refrigerator as ideal).
(a) 1 J (b) 12 J (c) 8.3 J (d) 13 J (e) 6 JThe coefficient of performance of the refrigerator, β = Q₂/W = Q₂/(Q₁– Q₂) = T₂/(T₁–T₂) = 277/(300–277) = 12. [Note that we have converted the temperature to the Kelvin scale].
Therefore, Q₂ =12 W. Heat delivered to the room is Q1 = Q₂+W = 12W+W = 13W. Here W is the work done by the pump. So for each joule of work done (for each joule of electricity consumed), the quantity of heat pumped out in to the room will be 13 joule.

Given below is a question of the type which often finds a place in Entrance test papers:

An ideal gas is taken through a cycle of operations shown by the indicator diagram. The net work done by the gas at the end of the cycle is

(a) 6P0V0 (b) 4P0V0 (c) 15P0V0 (d) 10P0V0 (e) 3P0V0

The work done in a cyclic process indicated by a PV diagram is the area enclosed by the closed curve. The area under the slanting curve showing the expansion of the gas from volume 2V0 to volume 5V0 gives the work done by the gas. This is greater than the area under the (horizontal) curve showing the compression of the gas (from volume 5V0 to volume 2V0), which gives the work done on the gas. The vertical portion of the curve is an isochoric (volume constant) change which involves no work since the area under it is zero. The area enclosed by the closed curve gives the net work done by the gas. The triangular area enclosed is ½ ×3V0×2P0 = 3P0V0 [Option (e)].
[ Note that in problems of the above type, the work is done by the gas if the arrow showing the cycle is clockwise. If the arrow is anticlockwise, work is done on the gas. In either case, the work done is the area enclosed by the curve].

Questions on Isothermal and Adiabatic Changes

You can expect questions involving isothermal and adiabatic processes in most entrance examinations. Here is a typical question:
A gas at a temperature of 27°C inside a container is suddenly compressed to one sixteenths of its initial volume. The temperature of the gas immediately after the compression is (Ratio of specific heats of the gas, γ = 1.5)
(a) 19200 K (b) 1200°C (c) 927°C (d) 108°C (e) 19200°C
This is an adiabatic change since the compression is sudden so that the volume(V) and the temperature (T) are related as TV^γ–1 = constant.
Therefore we have 300 V^0.5 = T(V/16)^0.5. Note that the temperature is to be substituted in Kelvin. The final temperature is given by T = 300×16^0.5 = 1200 K = 927°C.
The following MCQ is meant for testing your understanding of the work done in thermodynamic processes:

Starting from the same initial conditions an ideal gas expands from volume V₁ to volume V₂ in three different ways: (i) Adiabatically, doing work W₁. (ii) Isothermally, doing work W₂. (iii) Isobarically, doing work W₃. Then,

(a) W₁ = W₂ = W₃ (b) W₁ > W₂ > W₃ (c) W₃ > W₂ > W₁ (d) W₂ > W₁ > W₃ (e) W₁ > W₃ > W₂

Since the work done is the area under the corresponding curve in a PV diagram, you can easily verify that the work done is the largest in the isobaric case since it is a straight line parallel to the volume axis. The adiabatic curve is the steepest one so that the area under it is the smallest. The correct option therefore is (c).
You can easily show that the slope of the adiabatic curve is γ times the slope ofthe isothermal curve as follows:
In the case of an adiabatic change, the pressure and volume are related as PV^γ = constant.
Differentiating, P γV^γ–1 dV + V^γdP = 0
Slope of adiabatic curve = dP/dV = (– γPV^γ–1)/V^γ = – γP/V
In the case of an isothermal change, the pressure and volume are related as PV= constant.
Differentiating, PdV + VdP = 0.
Slope of isothermal curve = dP/dV = – P/V.
This show that the slope of the adiabatic curve is γ times the slope ofthe isothermal curve.
The following MCQ also pertains to adiabatic compression:
A gas having a volume of 800 cm³ is suddenly compressed to 100cm³. If the initial pressure is P, the final pressure is (γ = 5/3)
(a) P/32 (b) 24P (c) 32P (d) 8P (e) 16P
We have PV^γ = constant so that P×800^5/3 = P'×100^5/3 from which P' = P×8^5/3 = P×2⁵ = 32P.

Now, see this simple question:
An ideal gas expands isothermally from volume V₁ to volume V₂. It is then compressed to the original volume V₁ adiabatically. The initial pressure is P₁, final pressure is P₂ and the net work done by the gas during the entire process is W. Then
(a) P₁ = P₂, W>0 (b) P₁> P₂, W>0 (c) P₂ > P₁, W>0 (d) P₂ > P₁, W=0 (e) P₂ > P₁, W<0

The adiabatic compression will increase the temperature of the gas so that the final pressure (P₂) when the volume is restored to the value V₁ is greater than the initial pressure P₁. Since the pressure is greater during the adiabatic compression, more work has to be done on the gas. The work done on the gas is thus greater than the work done by the gas. In other words, the net work done by the gas during the entire process is negative. So, the correct option is (e).

Questions on Optical Instruments

The essential points you have to remember to work out questions on optical instruments are the following:
(1) Myopia (short sightedness) is corrected by a concave lens of focal length ‘d’ where ‘d’ is the distance of the far point in the case of the defective eye. [Note that for a normal eye the far point will be at infinity where as for the defective eye, it will be at a finite distance‘d’].

Therefore, f = –d
(2) Hypermetropia (long sightedness) is corrected by a convex lens of focal length ‘f’ given by f = dD/(d – D) where ‘d’ is the distance of the near point for the defective eye and D is the least distance of distinct vision (25cm), [Note that for the normal eye, the near point will be at D and for the defective eye it will be at a greater distance ‘d’].
(3) Presbyopia is corrected by bifocal lens with the upper portion concave and the lower portion convex. [Note that for the defective eye in this case, the far point is nearer (and not at infinity) while the near point is farther away (and not at D)].
(4) Astigmatism is corrected by cylindrical lens.
(5) Magnifying power of a simple microscope M = 1+ D/f if the image is formed at the least distance of distinct vision ‘D’.
If the image is formed at infinity (normal setting or setting for relaxed eye), M= D/f .
(6) Magnifying power of compound microscope (M):
(i) M = (v_o/u_o)(1 + D/f_e) if the final image produced by the eye piece is at the least distance of distinct vision ‘D’. Here u_o is the distance of object from the objective, v_o is the distance of the image produced by the objective and f_e is the focal length of the eye piece.
(ii) If the final image is at infinity (normal setting or setting for relaxed eye),
M = (v_o/u_o)(1 + D/f_e)
Approximate expressions for the magnifying power of a compound microscope in the two cases are M = (L/f_o)(1 + D/f_e) and M = (L/f_o)(D/f_e) in the two cases respectively. Here L is the tube length of the microscope, which is the distance between the objective and the eye piece.
(7) (a)Limit of resolution (minimum separation between two point objects which can be resolved) of a microscope, dmin = λ /2n sinθ where λ is the wave length (in vacuum) of the light used for illuminating the object, ‘n’ is the refractive index of the medium between the object and the objective and θ is the semi angle of the cone of light proceeding from the object to the objective. This is Abbe’s expression for the resolving power when the object is not luminous and is therefore illuminated, as is usually the case. [Note that nsinθ is the numerical aperture]
(b) Resolving power of microscope = 1/d_min
(8) Magnifying power of telescope(M)
(i) M = β/α = fo/fe for normal adjustment (image at infinity) where β is the angle subtended at the eye by the image, α is the angle subtended at the eye by the object, fo is the focal length of the objective and fe is the focal length of the eye piece.
(ii) M = (f_o/f_e)(1 + f_e/D) if the final image is formed at the least distance of distinct vision ‘D’.
(9) Length of a telescope (in normal setting) = f_o + f_e
(10) Limit of resolution of telescope (minimum angular separation between two point objects that can be resolved) dθ = 1.22λ/a where λ is the wave length of light proceeding from the object and ‘a’ is the aperture (diameter) of the objective.
(11) Resolving power of a telescope = 1/dθ = a/1.22λ
Now let us now consider the following MCQ:
The length of a microscope is 12 cm and its magnifying power is 25 for relaxed eye.The focal length of the eye piece is 4 cm. The distance of the object from the objective is
(a) 2 cm (b) 2.5 cm (c) 3 cm (d) 3.5 cm (e) 4 cm
The image produced by the objective is formed at the focus of the eye piece since the final image is formed at infinity (for relaxed eye). If v_o is the distance of the image formed by the objective, the tube length of the microscope (distance between the objective and eye piece) is v_o+ f_e = v_o+ 4 = 12 cm from which v_o = 8 cm.
The magnifying power (for relaxed eye) is given by M = (v_o/u_o)(D/f_e). Substituting the known values, 25 = (8/u_o)(25/4) from which u_o = 2 cm. [Option (a)].
The following question appeared in MPPMT 2000 question paper:
The length of the tube of a microscope is 10 cm. The focal lengths of the objective and eye lenses are 0.5 cm and 1 cm. The magnifying power of the microscope is about
(a) 5 (b) 166 (c) 23 (d) 500
The magnifying power, M = (L/f_o)( D/f_e) = (10/0.5)( 25/1) = 500.
Consider now the following MCQ which appeared in EAMCET 2000 question paper:
In a compound microscope, the focal lengths of two lenses are 1.5 cm and 6.25 cm. If an object is placed at 2 cm from the objective and the final image is formed at 25 cm from the eye lens, the distance between the two lenses is
(a) 6 cm (b) 7.75 cm (c) 9.25 cm (d) 11 cm
You should note that the focal length of the objective in a microscope is less than that of the eye piece. (In a telescope it is the other way round). Therefore, f_o = 1.5 cm and f_e = 6.25 cm.
The distance between the objective and the eye piece is the sum of the image distance(vo) for the objective and the object distance (ue) for the eye piece.
From the equation, 1/f = 1/v – 1/u as applied to the objective, we have
1/1.5 = 1/v_o – 1/(–2).
Note that we have substituted the object distance as –2 in accordance with the Cartesian sign convention discussed in the post dated 22-11-06 (Questions (MCQ) on Refraction at Spherical Surfaces). This yields the image distance in the case of the objective as v_o = 6 cm.The image distance for the eye piece is similarly given by
1/6.25 = 1/(–25) – 1/u_e.
Note that the sign of the object distance is negative in accordance with the Cartesian sign convention. This equation yields u_e = –5 cm. The negative sign just shows that the object distance for the eye piece is measured opposite to the direction of the incoming rays.In fact, we know that the real image (of the object) formed by the objective serves as the object for the eye piece and therefore its distance is negative. However, since u_e is an unknown quantity, we did not bother about its sign. If we had substituted its sign as negative in the law of distances, we would have obtained the value as +5 cm.
The distance between the objective and the eye piece is 6+5 =11 cm.
Here is a simple question involving myopia:
A man who cannot see clearly beyond 10m wants to see stars clearly. He should use a lens of power (in dioptre)
(a) 10 (b) – 10 (c) –1 (d) 0.1 (e) – 0.1
His defect is myopia and hence he should use a concave lens of focal length equal to the distance of the far point, which is given as 10m. The power of the lens is 1/(–10) = – 0.1 dioptre.

Now consider the following question:
The near point of a person with defective eye is at 65 cm. To correct his defect, he should use spectacle lenses of focal length
(a) 65 cm (b) 55.6 cm (c) 50.6 cm (d) 45,5 cm (e) 40.6 cm
As his defect is hypermetropia (long sightedness), he should use convex lenses of focal length ‘f’ given by f = dD/(d – D) where ‘d’ is the distance of his far point and ‘D’ is the least distance of distinct vision. Therefore, f = 65×25/(65 – 25) = 40.6 cm.
Here is a typical simple question on telescopes:
The magnifying power of a small telescope is 25 and the separation between its objective and eye piece is 52 cm in normal setting. The focal lengths of its objective and eye piece are respectively
(a) 50 cm and 2 cm (b) 27 cm and 25 cm (c) 45 cm and 7 cm (d) 50.5 cm and 1.5 cm (e) 47 cm and 5 cm
We have magnifying power, M = f_o/f_e so that 25 = f_o/f_e, from which f_o = 25 fe.
Sincethe length of the telescope = f_o + f_e, we have 52 = 25f_e + f_e = 26 f_e so that f_e = 2 cm and hence fo = 50 cm.
The following question pertains to the resolving power of a telescope:
The distance between the earth and the moon is nearly 3.8×10⁵ km. What is the separation of two points on the moon that can be just resolved using a 400 cm telescope, using light of wave length 6000 Ǻ?
(a) 46.5 m (b) 56 m (c) 69.5 m (d) 78.6 m (e) 85.5 m
The limit of resolution dθ = 1.22λ/a. This is the minimum angular separation between objects that can be just resolved.
The linear separation between the objects that can be just resolved is rdθ = 3.8×10⁸×1.22×6000×10^–10/4 = 69.5 m. [Note that we have converted all distances in to metre].
The following m.c.q. appeared in the MPPMT 2000 question paper:
The focal lengths of the eye piece and objective of a telescope are respectively 100 cm and 2 cm. The moon subtends an angle of 0.5º at the eye. If it is looked through the telescope, the angle subtended by the moon’s image will be
(a) 100º (b) 25º (c) 50º (d) 10º
the magnifying power of a telescope is given by M = β/α = f_o/f_e with usual notations. Therefore, β = α f_o/f_e = 0.5×100/2 = 25º [Option (b)].

Now Consider the following MCQ:
A telescope has an objective lens of focal length 1.5m and an eye piece of focal length 5 cm. If this telescope is used in normal setting to view a tower of height 100m located 3km away, what will be the height of the image of the tower?
(a) 10 cm (b) 15 cm (c) 20 cm (d) 25 cm (e) 50 cm
The angle subtended at the objective by the tower will be the same as the angle subtended (at the objective) by the image produced by the objective so that we have
100/3000 = h/1.5 where ‘h’ is the height of the image produced by the objective. [Note that this image is at the focus of the objective].
From the above, h = 0.05m = 5 cm.
The magnifying power of the eye piece in normal adjustment is D/f_e = 25cm/5cm = 5.
The height of the final image = 5h = 25 cm.

Kerala Medical and Engineering Entrance Examinations- 2007

The Commissioner of Entrance Examinations (Govt. of Kerala) has invited applications for the Entrance Examinations for admission to the following Professional Degree courses in Kerala for 2007-08:
(a) Medical (i) MBBS (ii) BDS (iii) BPharm (iv) BSc (Nursing) (v) BSc (MLT) (vi) BAMS (vii) BHMS (viii) BSMS (Siddha) (ix) BSc–Nursing (Ayurveda) and (x) BPharm (Ayurveda) .
(b) Agriculture (i) BSc (Agriculture) (ii) BFSc (Fisheries) (iii) BSc (Forestry)
(c) Veterinary BVSc & AH
(d) Engineering B.Tech [including BTech (Agricultural Engg. / BTech (Dairy Sc. & Tech) courses under the Kerala Agr iculture University]
(e) Architecture B.Arch
Time Table for the Entrance Examinations:
The Entrance Examinations will be held in all the District Centres in Kerala, New Delhi and Dubai (UAE), on the dates mentioned below as per Indian Standard Time.
Engineering Entrance Examination (For Engineering courses except Architecture):
23-04-2007 Monday 10.00 A.M. to 12.30 PM Paper-I: Physics & Chemistry.
24-04-2007 Tuesday 10.00 A.M. to 12.30 PM Paper-II: Mathematics.
Medical Entrance Examination [For Medical (including B.Pharm), Agriculture and Veterinary Courses]:
25-04-2007 Wednesday 10.00 A.M. to 12.30 PM Paper-I: Chemistry & Physics.
26-04-2007 Thursday 10.00 A.M. to 12.30 PM Paper-II: Biology.
Appearance in the two papers of the concerned Entrance Examination is compulsory for being considered for inclusion in the Engineering/Medical rank lists (except Architecture)
Admission to the BPharm course will be based on a separate rank list prepared on the basis of the performance of candidates in the ‘Chemistry & Physics’ Paper of the Medical Entrance Examination. Those who wish to be considered for admission to the BPharm course should therefore appear for the Chemistry & Physics Paper of the Medical Entrance Examination.
Application Forms:
The application form and Prospectus will be sold from 07-02-2007 to 05-03-2007 through selected Canara Bank branches in Kerala and outside the State.
Cost of Application form: General candidates: Rs. 700/- ; SC/ST candidates : Rs. 350/- ( Candidates opting Dubai centre should enclose a bank draft for Rs 7000/- along with the application)
Last date and time for receipt of filled in Application Forms: The filled in Application Form along with the OMR DATA SHEET and other relevant documents to be submitted with the Application form is to be sent in the printed envelope bearing the address of the Commissioner for Entrance Examinations supplied along with the application form so as to reach him before 5 p.m. on 07.03 200 7, by Hand Delivery / Registered Post/ Speed Post.
Visit the site www.cee-kerala.org for complete details about the eligibility for applying for the examinations, sale centres of application forms, provision for downloading the application form in the case of certain categories etc. Visit the site to be informed of the latest developments in this regard.

New Template, New Look

After waiting for some time, I have switched over to the new version of Blogger today. The switch over was quite easy. I will have to spend some time on effecting a few changes in the lay out as well as the look of my blog. The priority will be in posting new questions and solution since examinations are fast approaching.
Thank you Blogger Staff for effecting the transition out of beta in a relatively short time span!

Multiple Choice Questions on Waves

The section on waves may appear to be somewhat boring and difficult to some of you. But you should not ignore this section because you will usually get a couple of questions from this. Remember the following important relations:
(1) Speed of transverse waves in a stretched string, v = √(T/m) where T is the tension and ‘m’ is the linear density (mass per unit length) of the string
(2) Frequency of vibration of a string, n = (1/2l)√(T/m) where ‘l’ is the length of the string.
Note that this is the frequency in the fundamental mode. Generally, the frequency is given by n = (s/2l)√(T/m) where s = 1,2,3,….etc. In the fundamental mode, s = 1.
(3) Speed of sound (v) in a medium is generally given by v = √(E/ρ) where E is the modulus of elasticity and ρ is the density of the medium.
(a) Newton-Laplace equation for the velocity (v) of sound in a gas:
v = √(γP/ρ) where γ is the ratio of specific heats and P is the pressure of the gas.
(b) Velocity of sound in a solid rod, v =√(Y/ρ) where Y is the Young’s modulus.
(4) Equation of a plane harmonic wave:
You will encounter the wave equation in various forms. For a progressive wave proceeding along the positive X-direction, the wave equation is
y = A sin [(2π/λ)(vt–x) +φ]
where A is the amplitude, λ is the wave length, v is the velocity (of the wave) and φ is the initial phase of the particle of the medium at the origin.
If the initial phase of the particle at the origin (φ) is taken as zero, the above equation has the following forms:
(i) y = A sin [(2π/λ)(vt–x)]
(ii) Since λ = vT and 2π/T = ω, where T is the period and ω is the angular frequency of the wave motion, y = A sin [(2π/T)(t – x/v)] and
(iii) y = A sin ω(t–x/v)
(iv) A common form of the wave equation (obtained from the above) is
y = A sin [2π(t/T – x/ λ)]
(v) Another form of the wave equation is y = A sin (ωt – kx), which is evident from the form shown at (iii), where k = ω/v = 2π/λ.
Note that unlike in the case of the equation of a simple harmonic motion, the wave equation contains ‘x’ in addition to ‘t’ since the equation basically shows the variation of the displacement ‘y’ of any particle of the medium with space and time.
It will be useful to remember that the velocity of the wave,

v = Coefficient of t /Coefficient of x
(6) Equation of a plane wave proceeding in the negative X-direction is
y = A sin [2π(t/T + x/ λ)] or
y = A sin [(2π/λ)(vt + x)] or
y = A sin ω(t + x/v) or
y = A sin (ωt + kx).
Note that the negative sign in the case of the equation for a wave proceeding along the positive X-direction is replaced with positive sign.
(7) Equation of a stationary wave is y = 2A cos(2πx/λ) sin(2πvt/λ) if the stationary wave is formed by the superposition of a wave with the same wave reflected at a free boundary of the medium (such as the free end of a string or the open end of a pipe).
If the reflection is at a rigid boundary (such as the fixed end of a string or the closed end of a pipe), the equation for the stationary wave formed is
y = – 2A sin(2πx/λ) cos(2πvt/λ).
Don’t worry about the negative sign and the inter change of the sine term and the cosine term. These occurred because of the phase change of π suffered due to the reflection at the rigid boundary.The important thing to note is that the amplitude has a space variation between the zero value (at nodes) and a maximum vlue 2A (at the anti nodes). Further, the distance between consecutive nodes or consecutive anti nodes is λ/2.
Now consider the following MCQ:
The equation, y = A sin [2π/λ(vt – x)] represent a plane progressive harmonic wave proceeding along the positive X-direction. The equation, y = A sin[2π/λ(x – vt)] represents
(a) a plane progressive harmonic wave proceeding along the negative X-direction (b) a plane progressive harmonic wave with a phase difference of π proceeding along the negative X-direction
(c) a plane progressive harmonic wave with a phase difference of π proceeding along the positive X-direction
(d) a periodic motion which is not necessarily a wave motion
(e) a similar wave generated by reflection at a rigid boundary.
The equation, y = A sin[2π/λ(x – vt)] can be written as y = – A sin [2π/λ(vt – x)] and hence it represents a plane progressive harmonic wave proceeding along the positive X-direction itself, but with a phase difference of π (indicated by the negative sign).
The following question appeared in IIT 1997 entrance test paper:
A traveling wave in a stretched string is described by the equation, y = A sin(kx– ωt). The maximum particle velocity is
(a) Aω (b) ω/k (c) dω/dk (d) x/t
This is a very simple question. The particle velocity is v = dy/dt = –Aω cos(kx– ωt) and its maximum value is Aω.
Consider the following MCQ:
The displacement y of a wave traveling in the X-direction is given by y = 10^–4 sin (600t–2x + π/3) metres. where x is expressed in metres and t in seconds. The speed of the wave motion in ms^–1 is
(a)200 (b) 300 (c) 600 (d) 1200
This MCQ appeared in AIEEE 2003 question paper. The wave equation given here contains an initial phase π/3 but that does not matter at all. You can compare this equation to one of the standard forms given at the beginning of this post and find out the speed ‘v’ of the wave. If you remember that velocity of the wave, v = Coefficient of t /Coefficient of x, you get the answer in notime: v = 600/2 =300 ms^–1.
Here is a question of the type often found in Medical and Engineering entrance test papers:
Velocity of sound in a diatomic gas is 330m/s. What is the r.m.s. speed of the molecules of the gas?
(a) 330m/s (b) 420m/s (c) 483m/s (d) 526m/s (e) 765m/s
At a temperature T, the velocity of sound in a gas is given by v = √(γP/ρ) where γ is the ratio of specific heats and P is the pressure of the gas. The r.m.s. velocity of the molecules of the gas is given by c = √(3P/ρ). Therefore, c/v = √(3/γ). For a diatomic gas, γ = 1.4 so that c/v = √(3/1.4). On substituting v=330m/s, c = 483m/s.
Now consider the following MCQ which appeared in EAMCET 1990 question paper:
If two waves of length 50 cm and 51 cm produced 12 beats per second, the velocity of sound is
(a) 360 m/s (b) 306 m/s (c) 331 m/s (d) 340 m/s
If n₁ and n₂ are the frequencies of the sound waves, n₁– n₂ =12 or, v/λ₁– v/λ₂ =12. Substituting the given wave lengths (0.5m and 0.51m), the velocity v works out to 306 m/s.
Here is a typical simple question on stationary waves:
A stationary wave is represented by the equation, y = 3 cos(πx/8) sin(15πt) where x and y are in cm and t is in seconds. The distance between the consecutive nodes is (in cm)
(a) 8 (b) 12 (c) 14 (d) 16 (e) 20
This stationary wave is in the form y = 2A cos(2πx/λ) sin(2πvt/λ) so that 2π/λ = π/8 from which λ = 16cm. The distance between consecutive nodes is λ/2 = 8cm.
You can find all posts on waves in this site by clicking on the label 'waves' below this post or on the side of this page.

Questions on Centre of Mass

The following question involving the motion of centre of mass is worth noting:

Two spheres of masses m₁ and m₂ (m₁>m₂) respectively are tied to the ends of a light, inextensible string which passes over a light frictionless pulley. When the masses are released from their initial state of rest, the acceleration of their centre of mass is
(a) [(m₁–m₂)/(m₁+m₂)]g
(b) [(m₁–m₂)/2(m₁+m₂)]g
(c) [(m₁–m₂)/4(m₁+m₂)]g
(d) [(m₁–m₂)/(m₁+m₂)]²g
(e) [4(m₁–m₂)/(m₁+m₂)]g
If r₁ and r₂ are the position vectors of the centres of the spheres, the position vector of their centre of mass is given by R = (m₁r₁ + m₂r₂)/(m₁+m₂). The acceleration of the centre of mass is therefore given by
a = d²R/dt² = (m₁d²r₁/dt² + m₂ d²r₂/dt²)/(m₁+m₂).
But d²r₁/dt² and d²r₂/dt², which are the accelerations of the masses m1 and m2 have the same magnitude (m₁–m₂)g/(m₁+m₂). If we take the acceleration of m₁ (which is downwards) as positive, that of m₂ is negative.
Therefore, a = [m₁(m₁–m₂)g/(m₁+m₂) – m₂(m₁–m₂)g/(m₁+m₂)] /(m₁+m₂).
On simplifying, this yields a = [(m₁–m₂)/(m₁+m₂)]²g. [Option (d)].
The following MCQ has been popular among question setters:
In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Ǻ. The approximate distance of the centre of mass (from the hydrogen atom) of the HCl molecule, assuming the chlorine atom to be about 35.5 times as massive as the hydrogen atom is
(a) 0.27 Ǻ
(b) 0.56 Ǻ
(c) 1 Ǻ
(d) 1.24 Ǻ
(e) 2.26 Ǻ
The centre of mass of this two particle system is on the line joining the two atoms and in between these atoms. If ‘x’ is the distance of the centre of mass from the hydrogen atom, we have x = (1×0 + 35.5×1.27)/ (1+35.5) = 1.24 Ǻ, approximately.
Note that the equation giving the value of ‘x’ is the usual equation for the position vector of the centre of mass: R = (m₁r₁ + m₂r₂)/(m₁+m₂). We have taken the origin to be at the centre of the hydrogen atom so that r₁ = 0 and r₂ = 1.27 Ǻ.
[You can easily find the centre of mass of two particle systems by equating the ‘moments’ of the masses about the centre of mass. In the present case, 1×x = 35.5(1.27–x) from which x = 1.24 Ǻ].
Now consider the following simple question:
A proton and an electron, initially at rest, are allowed to move under their mutal attractive force. Their centre of mass will
(a) move towards the proton
(b) move towards the electron
(c) remain stationary
(d) move in an unpredictable manner.

The correct option is (c) because you require an external force to move the centre of mass of a system of particles. The mutual force of electrical attraction is an internal force which can not affect the position of the centre of mass of the system.