Tuesday, December 26, 2006

Additional MCQ on Bohr Model of Hydrogen Atom

The following questions are in continuation of the post dated August 26, 2006 (Questions on Bohr Atom Model):
(1) Suppose the energy required to remove all the three electrons from a lithium atom in the ground state is ‘E’ electron volt. What will be the energy required (in electron volt) to remove two electrons from the lithium atom in the ground state?
(a) 2E/3 (b) E – 13.6 (c) E – 27.2 (d) E – 40.8 (e) E – 122.4
The energy of the electron in a hydrogen like atom in the ground state is – 13.6Z2 electron volt. Therefore, after removing two electrons from the lithium (Z=3) atom, the third electron has energy equal to – 13.6×32 eV = 122.4 eV. The energy needed to remove two electrons from the lithium atom in the ground state is therefore equal to (E –122.4) eV.
(2) How many revolutions does the electron in the hydrogen atom in the ground state make per second? (h = 6.63×10-34 Js, mass of electron = 9.11×10-31 kg, Bohr radius = 0.53 A.U.)
(a) 6.55×10-15 (b) 3.28×10-15 (c) 3.28×10-16 (d) 1.64×10-15 (e) 9.11×10-15
The angular momentum (Iω) of the electron is an integral multiple of h/2π. Therefore, Iω = nh/2π, from which, for the first orbit (n=1), ω = h/2πI = h/2πme r2 . The orbital frequency of the electron is given by f = ω/2π = h/4π2m r2 = 6.55×10-15 per second, on substituting for h, m and r.
(3) The ionisation energy of the hydrogen atom is 13.6 eV. If hydrogen atoms in the ground state absorb quanta of energy 12.75 eV, how many discrete spectral lines will be emitted as per Bohr’s theory?
(a) 1 (b) 2 (c) 4 (d) 6 (e) zero
On absorbing 12.75 eV, the energy of the electron in the hydrogen atom will become (–13.6 + 12.75) eV which is – 0.85 eV. This is an allowed state (with n=4) for the electron, since the energy in the 4th orbit is – 13.6/n2 = – 13.6/42 = 0.85 eV. From the 4th orbit, the electron can undrego three transitions to the lower orbits (4→3, 4→2, 4→1). From the third orbit, the electron can undergo two transitions (3→2 and 2→1). The electron in the second orbit can undergo one transition (2→1). So, altogether 6 transitions are possible, giving rise to 6 discrete spectral lines [Option (d)].
You can work it out as n(n–1)/2 = 4(4–1)/2 =6.
(4) The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following siatements is true?
(a) Its kinetic energy increases and its potential and total energies decrease.
(b)Its kinetic energy decreases, potential energy increases and its total energy remains the same.
(c) Its kinetic and total energies decrease and its potential energy increases.
(d) Its kinetic, potential and total energies decrease.
This question appeared in IIT 2000 entrance test paper. The correct option is (a). You should note that the kinetic energy is positive while the potential energy and total energy are negative. Further, the kinetic energy and total energy are numerically equal and the numerical value is equal to half the potential energy.
The total energy is –13.6 Z2/n2. In lower orbits (with smaller n), the potential energy is smaller since it has a larger negative value. The total energy also is therefore smaller. But, the kinetic energy is greater since it has a larger positive value.
(5) If the binding energy of the electron in a hgydrogen atom is 13.6 eV, the energy required to remove the electron from the first excited state of Li++ is
(a) 122.4 eV (b) 30.6 eV (c) 13.6 eV (d) 3.4 eV
Questions of this type often appear in entrance test papers.
Since Li++ is a hydrogen like system with a single electron revolving round a nucleus of proton number Z = 3, the energy of the electron in orbit of quantum number n is
E = –13.6 Z2/n2 eV.
The energy in the first excited state (second orbit) is – 13.6×9/4 eV = – 30.6 eV. The energy to be supplied to the electron to remove it from the first excited state is therefore + 30.6 eV [Option (b)].
(6) The wave lengths involved in the spectrum of deuterium (1D2) are slightly different from that of hydrogen spectrum, because
(a) the attraction between the electron and the nucleus is different in the two cases
(b) the size of the two nuclei are different
(c) the nuclear forces are different in the two cases
(d) the masses of the two nuclei are different.
This MCQ appeared in AIEEE 2003 questionn paper. The answer to this will not be easy if you stick on to the elementary theory of the Bohr model in which the energy (En) of the electron of quantum number n (nth orbit) is given by
En = – me4/8ε0n2h2 where m is the mass of the electron, e is its charge, ε0 is the permittivity of free space and h is Planck’s constant.
In the elementary theory we take ‘m’ as the mass of the electron on the assumption that the nucleus has a very large mass compared to the mass of the electron and hence the electron is moving round with the nucleus at the centre. The real situation is that both the electron and the nucleus are moving along circular paths with the centre of mass as the common centre. Instead of the actual mss of the electron, the reduced mass of the electron and the nucleus is to be substituted in the expression for energy. The modified form of the expression is
En = – μ e4/8ε0 n2h2 where μ is the reduced mass of electron and the nucleus, given by
μ = Mme/(M+me), M and me being the masses of the nucleus and the electron respectively. [Generally, for a hydrogen like system with proton number Z, the expression for energy is En = – μ Z2e4/8ε0n2h2 ].
The nucleus of deuterium contains a proton and a neutron and has very nearly twice the mass of the hydrogen nucleus (proton). So, the reduced mass and the energy levels of deuterium are slightly greater than those of hydrogen and this is the reason for the difference in wave length. [The wave lengths are slightly shorter].The correct option is (d).
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An important point you should remember in the light of the above discussion is the drastic change in the energy levels and the spectrum of positronium compared to hydrogen. Positronium is a highly unstable neutral atom with an electron revolving round a positron. [ You can as well say, a positron revolving round an electron!]. The concept of reduced mass is absolutely necessary in this case since the positron has the same mass as that of the electron so that the reduced mass of positronium is mm/(m+m) = m/2 where ‘m’ is the mass of positron as well as the electron.
Now consider the following MCQ:
A positronium atom undergoes a transition from the state n = 4 to n = 2. The energy of the photon emitted in this process is
(a) 1.275 eV (b) 2.55 eV (c) 3.4 eV (d) 5.1 eV (e) 13.6 eV
The expression for energy of positronium is En = – μ e4/8ε0n2h2 where μ is the reduced mass of positron and electron, given by μ = mm/(m+m) = m/2. Therefore, the mass of the electron (m) used in the expression for the energy of a hydrogen atom (Bohr’s theory) is to be replaced by m/2. All energy levels are therefore reduced to half of the hydrogen levels. Since the energies for states n=4 and n=2 for hydrogen are –13.6/16 eV(=
0.85 eV) and –13.6/4 eV (= –3.4 eV) respectively, the energy of the photon emitted in the case of hydrogen is [(–0.85) – (3.4)] eV = 2.55 eV. In the case of positronium, the energy will be half of this. So, the answer is 2.55/2 eV = 1.275 eV.

Sunday, December 24, 2006

Graduate Record Examinations (GRE) -- Physics Test

Many of you might have already noted that the posts here are useful for preparing for the GRE Physics Test which consists of about 100 multiple choice questions with 5 options. The needs of the GRE Physics Test takers will be considered while discussing questions here. You may make use of the facility for comments for communications in this context.

Wednesday, December 20, 2006

Birla Institute of Technology & Science (BITS)-- BITSAT-2007 Online Tests

The BITSAT-2007 Online tests (for admission to the academic year 2007-08) will be conducted during 7th May - 10th June 2007. These tests are for admitting students to the Integrated First Degree Programmes of BITS, Pilani, Rajasthan, at Pilani Campus and Goa Campus. You will find details here

Tuesday, December 19, 2006

Multiple Choice Questions from Nuclear Physics

Questions in Nuclear Physics at the level expected of you are simple and interesting. You should be careful not to omit questions from this section. Consider the following MCQ which appeared in Kerala Medical Entrance 2000 test paper:
A radioactive isotope has a half life T years. The time after which its activity is reduced to 6.25% of its original activity is
(a) 2T years (b) 4T years (c) 6T years (d) 8T years (e) 16T years
Since the activity of a sample is directly proportional to the number of nuclei present at the instant, we can express the activity ‘A’ after ‘n’ half lives in terms of the initial activity ‘A0’ as,
A = A0/2n.
[The well known radioactive decay law is, N = N0e-λt where N0 is the initial number of nuclei, N is the number remaining undecayed after time ‘t’ and λ is the decay constant. This equation, modified in terms of half life can be written as N = N0/2n where N is number of nuclei remaining undecayed after ‘n’ half life periods. Since the activity, A = dN/dt = -λN, it follows that the activity also can be expressed in the same manner as we express N. Therfore, A = A0/2n].
Therefore, we have, 6.25 = 100/2n, taking initial activity as 100. This yields n = 4, which means that 4 half lives have been elapsed to reduce the activity to 6.25% of the initial activity. The time required is therefore 4T, given in option (b).
Let us discuss another MCQ:
Out of 10 mg of a radio active sample, 1.25 mg remains undecayed after 6 hours. The mean life of the sample in hours is
(a) 0.693 (b) 2/0.693 (c) 4/0.693 (d) 0.693/4 (e) 0.693/2
We have, N = N0/2n so that 1.25 = 10/2n, from which 2n = 8 and n = 3. So, 3 half life periods is 6 hours so that the half life of the sample is 2 hours.
Mean life, T = 1/λ = Thalf /0.693 = 2/0.693 [Option (b)].
Let us consider one more question involving radio activity:
The half life of a radioactive sample is 3.02 days. After how many days 10% of the sample will remain undecayed?
(a) 10 (b) 12.5 (c) 15 (d) 20 (e) 25
Using the relation, N = N0/2n where ‘n’ is the number of half life periods in which the sample decays from N0 to N, we have, 10 = 100/2n. From this, 2n = 10. Taking logarithms, n log 2 = log 10.
You can work this out even if you don’t have a calculator or logarithm tables since you definitely remember that log 2 is 0.3010. Therfore, n = 1/0.3010 = 3.32.
So you require 3.32 half lives or, 3.32×3.02 days = 10 days for the sample to decay to 10%.
Now consider the following question:
The density ‘d’ of nuclear matter varies with nucleon number ‘A’ as
(a) d α A-1 (b) d α A-2 (c) d α A (d) da A2 (e) d α A0
The correct option is (e).
The mass of a nucleus is directly proportional to the number (A) of the nucleons. The volume of the nucleus is (4/3)πR3 where R is the nuclear radius. But, R = 1.1×10-15A. So, the volume of the nucleus also is directly proportional to the nucleon number A. Since density is the ratio of mass to volume, it follows that the density of nuclear matter is independent of the nucleon number A.
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
You can easily show that the density of nuclear matter is of the order of 1017 kg/m3 as follows:
Mass of nucleus = 1.67×10-27×A since the mass of a nucleon is approximately 1.67×10-27 kg
Volume of nucleus = (4/3)πR3 where R is the nuclear radius.
But, R = 1.1×10-15A metre so that density of nuclear matter,
d = (1.67×10-27×A) /[(4/3)π×(1.1×10-15 × A)3. This works out to approximately 3×1017 kg/m3.
The following simple MCQ appeared in AIEEE 2004 question paper:
A nucleus disintegrates into two nuclear parts which have their velocities in the ratio 2:1. The ratio of their nuclear sizes will be
(a) 2:1 (b) 1:2 (c) 3½:1 (d) 1:3½
Since the momenta of the parts have to be equal in magnitude in accordance with the law of consevation of momentum, we have, m1v1 = m2v2 so that m1 /m2 = v2/v1 =1/2. The masses being directly proportional to the volumes, and the volumes being directly proportional to the cube of the radii, the radii (sizes) are directly proportional to the cube root of the masses. So, the answer is 1:2.
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Suppose a radioactve mother nucleus emits a β-particle. Are the mother and daughter nuclei isotones or isobars?
When a nucleus emits a β-particle, a neutron in the nucleus becomes a proton and hence the neutron number is changed. So the mother and daughter are not isotones. The mass number is not changed but the proton number (atomic number) is changed. So they are isobars.
* * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * *
Take note of the following question on calculation of the activity of a sample:
Half life of radium (88Ra 226 ) is 1620 years. What is the activity of 2gram of radium? (a) 3.6×1010 Bq (b) 7.2×1010 Bq (c) 1.44×1011 Bq (d) 2.88×1011 Bq (e) 5.76×1011 Bq
Note that becquerel (Bq) is the unit of radioactivity in SI and is equal to one disintegration per second.
Number of nuclei (or, atoms) in 2 gram of radium is (2/226)×NA = (2/226)×6.025×1023 = 5.33×1021. [NA is the Avogadro number, 6.025×1023].
Activity = λN = (0.693/Thalf)×N = [0.693/(1620×3.16×107)] × (5.33×1021).
Note that we have converted the half life in years into seconds. [ 1 year = 325.25×24×60×60 seconds = 3.16×107 s].
The activity works out to 7.2×1010 becquerel.
Now consider the following MCQ which involves Einstein’s mass- energy relation:
How much mass has to be converted into energy to produce electric power of 100 MW for one hour? (Assume that the conversion efficiency is 100%).
(a) 1mg (b) 2 mg (c) 4 mg (d) 20 mg (e) 40mg
At the rate of 100 MW, total energy produced for 1 hour is, P×t = (100×106) ×(60×60) joule = 3.6×1011 J
[We have converted megawatt into watts and hour into seconds].
Since this is equal to mc2 in accordance with Einstein’s mass energy relation, we have,
m = (3.6×1011)/(3×108)2 = 4×10-6 kg = 4 mg.
Here is a question involving relativistic increase of mass:
The rest mass of a proton is ‘m0’. Its linear momentum when it moves with half the speed of light ‘c’ in free space is
(a) 3m0c/4 (b) m0c/2 (c) m0c (d) 2m0c/√3 (e) m0c/√3
The mass of the proton while moving with velocity ‘v’ is given by
m = m0/√[1- v2/c2] so that when v = c/2, m = m0/√[1- ¼] = 2m0/√3.
The momentum of the proton is mc/2 = m0c/√3.

Tuesday, December 12, 2006

Multiple Choice Questions from Electronics

Questions in Electronics are simple at the Higher Secondary/Plus two level. Many of you might be interested in Electronics and your attitude towards this subject will make it seem to be simpler!
Consider the following MCQ which appeared in AIEEE 2003 question paper:
The difference in the variation of resistance with temperature in a metal and a semiconductor arises essentially due to the difference in the
(a) variation in the scattering mechanism with temperature (b) crystal structure (c) variation in the number of charge carriers with temperature (d) type of bonding
The correct option is (c). On raising the temperature of a semiconductor, more charge carriers are produced by the breaking of the covalent bonds, unlike in the case of a metal. Therefore, as you might have noted, semiconductors have negative temperature coefficient of resistance.
Now, consider the following question:
In an npn power transistor, the collector current is 20 mA. If 98% of the electrons injected in to the base region reach the collector, the base current in mA is nearly
(a) 2 mA (b) 1 mA (c) 0.5 mA (d) 0.4 mA (e) 0.2 mA
Since 98% of the electrons reach the collector, the collector current is 98% of the emitter current and the base current is 2% of the emitter current. As the collector current is nearly equal to the emitter current, the base current is 2% of the collector current. Therefore, base current = 20×2/100 = 0.4 mA [Option (d)].
The following question involving the common base current transfer ratio (current gain) α also is simple:
The current gain α of a transistor is 0.995. If the change in emitter current is 10 mA, the change in base current is
(a) 50 μA (b) 100 μA (c) 500 μA (d) 25 μA (e) 5 μA
We have ∆IB = ∆IE-∆IC where ∆IB, ∆IE and ∆IC represent the changes in the base current, emitter current and the collector current respectively.
Since α = ∆IC/∆IE, we have, ∆IC = α∆IE = 0.995 ×10 = 9.95 mA.
Therefore, change in base current, ∆IB = 10 – 9.95 = 0.05 mA = 50 μA
The following question is designed to check whether you have grasped the method of transistor biasing in the simplest possible practical circuit:
In the common emitter amplifier circuit shown, the transistor has a current gain β equal to 400. The collector load resistor RL = RC =3kΩ. If the collector-to-emitter voltage under no-signal (quiescent) condition is to be equal to half of the supply voltage VCC, what should be the value of the base biasing resistor RB? (Neglect the base to emitter voltage drop)

(a) 2.4 MΩ (b) 1.2 MΩ (c) 120 kΩ (d) 6 MΩ (e) 600 kΩ
In the circuit, the supply voltage used is 12 volts and half of it (6V) should appear between the collector and emitter. The remaining half (6V) should appear across the 3 kΩ collector load resistor, on account of the collector current flowing through it.
Therefore, collector current, IC = 6V/3kΩ = 2 mA. [Note that we have substituted the resistance in kilo ohm itself to obtain the collector current in milliampere].
We have IB = IC/β = 2 mA/400 = 0.005 mA = 5 μA.
Therefore, 5 μA should flow through the base biasing resistor RB. Since the base to emitter voltage drop is allowed to be neglected in the question, the voltage across RB is the full supply voltage, 12V. [If you cannot neglect the base to emitter voltage, you will have 11.3 volts instead, in the case of a silicon transistor, since you will have to subtract the forward voltage of 0.7 volts].Therefore, RB = 12V/5μA = 2.4 MΩ [Option (a)].
[Note that we substituted the current in microampere itself to obtain the resistance in megohm (mega ohm).
Now,
what is the function of the capacitors on the input side and the output side of the amplifier circuit shown above?
Note that they are meant for preventing the direct biasing voltages from reaching the signal source and the load and for allowing the varying signal voltages to pass through.
Let us consider another question:
In a common emitter amplifier circuit drawing a quiescent collector current of 1 mA, the input resistance of the transistor is 1.2 kΩ and the collector load resistance is 3 kΩ. If the common emitter current gain of the transistor is 400, what is the voltage gain of the amplifier?
(a) 500 (b) 600 (c) 700 (d) 800 (e) 1000
The quiescent collector current of 1 mA given in this question is just a distraction.
The voltage gain (or, voltage amplification, Av) of a common emitter amplifier is given by Av = βRL/Ri where RL is the load resistance and Ri is the input resistance of the amplifier. Therefore, voltage gain = 400×3/1.2 = 1000. [Note that we have substituted the resistance value in kilohm (kilo ohm) itself since we have a ratio of resistances in the expression].
Consider now the following three questions which appeared in Kerala Engineering Entrance 2006 test paper:
(1) If α and β are the current gains in the CB and CE configurations respectively of the transistor circuit, then (β – α)/αβ =
(a) ∞ (b) 1 (c) 2 (d) 0.5 (e) zero
We have, β = α/(1- α). Therefore, the given ratio, (β – α)/αβ = [α/(1- α) – α]/αβ = [1/(1- α) -1]/β = [1-(1- α)]/(1-α)β = α/(1- α)β = β/β =1. Therefore, the correct option is (b).
(2) The number densities of electrons and holes in pure germanium at room temperature are equal and its value is 3×1016 per m3. On doping with aluminium the hole density increases to 4.5×1022 per m3. Then the electron density in doped germanium is
(a) 2×1010m-3 (b) 5×109m-3 (c) 4.5×109m-3 (d) 3×109m-3 (e) 4×1010m-3
This is a very simple question based on the law of mass action, NeNh = Ni2 , where Ne and Nh are the number densities (number per unit volume) of the electrons and holes respectively in the doped semiconductor and Ni is the number density of electrons as well as holes in the intrinsic (pure) semiconductor.
Therefore, Ne = Ni2/Nh = (9×1032)/(4.5×1022) = 2×1010 m-3.
(3) The input resistance of a CE amplifier is 333 Ω and the load resistance is 5 kΩ. A change of base current by 15 μA results in the change of collector current by 1 mA. The voltage gain of the amplifier is
(a) 550 (b) 51 (c) 101 (d) 501 (e) 1001
The current gain of the transistor used, in the common emitter configuration is β = ∆IC/∆IB = (1000 μA)/(15 μA) = 1000/15. [Note that we have converted the collector current into micro ampere since the base current is in micro ampere].
The voltage gain of the amplifier is βRL /Ri = (1000/15) ×(5000/333) = 1001.
Here is a question which appeared in IIT 1998 entrance test paper:
In a p-n junction diode not connected to any circuit,
(a) the potential is the same everywhere
(b) the p-type side is at higher potential than the n-type side
(c) there is an electric field at the junction directed from the n-type side to the p-type side
(d) there is an electric field at the junction directed from the p-type side to the n-type side.
Even if no voltages are applied to a junction diode, the n-side is at a higher potential (of a few hundred millivolts, the exact value depending on the type of the semiconductor) compared to the p-side. This results due to the diffusion of electrons from the n-side to the p-side and similarly, the diffusion of holes from the p-side to the n-side. The n-side is therefore left with a net positive charge and the p-side is left with a net negative charge. This inherent reverse bias across the junction therefore produces an inherent electric field at the junction, directed from the n-type side to the p-type side[Option (c)].

Friday, December 08, 2006

More Multiple Choice Questions on Rotational Motion & Moment of Inertia

Here is an interesting question which appeared in
Kerala Engineering Entrance - 2006 test paper
( and also in IIT screening 2000 question paper):
A thin wire of length ‘L’ and uniform linear mass density ρ is bent into a circular loop with centre at O and radius ‘r’ as shown. The moment of inertia of the loop about the axis XX’ is
(a) 3ρL3/8π2 (b) ρL3/16π2 (c) 3ρL3/8π2r (d) ρL3/8π2r (e) 3ρL3/16π2
The moment of inertia of a circular ring about a diameter is ½ mr2, with usual notations. The axis of rotation in the question is a tangent to the ring. The moment of inertia of the ring about the tangent is ½ mr2 + mr2 = (3/2)mr2, on applying the parallel axis theorem. Now, m= ρL and the radius ‘r’ is given by 2πr = L, from which r = L/2π
On substituting for ‘m’ and ‘r’, moment of inertia = (3/2) Lρ×L2/ 4π2 = 3ρL3/8π2, given in option(a).
The following simple question appeared in Kerala Medical Entrance 2006 test paper:
Moment of inertia of a body does not depend upon its
(a) mass (b) axis of rotation (c) shape (d) distribution of mass (e) angular velocity
The correct option, as you might be knowing, is (e). Occasional simple questions like this will help you in saving your time for spending on other difficult questions. Good question setters will usually include a few simple questions for boosting your morale!
The following MCQ which also appeared in Kerala Medical Entrance 2006 test paper is worth noting:
A solid cylinder rolls down an inclined plane of height 3m and reaches the bottom of the plane with angular velocity of 2√2 rad.s-1. The radius of the cylinder must be
(a) 5 cm (b) 0.5 m (c) √10 m (d) √5 m (e) 10 cm
The initial gravitational potential energy (Mgh) of the cylinder is converted to rotational and translational kinetic energy when the cylinder reaches the bottom of the plane so that we can write,
Mgh = ½ Iω2 + ½ Mv2 where M is the mass, ‘v’ is the linear velocity, I is the moment of inertia and ‘ω’ is the angular velocity of the cylinder.
Since v = ωR and I = ½ MR2, the above equation becomes
Mgh = ½ ×(½ MR2) ω2 + ½ Mω2R2, which yields R = √[4gh/3ω2] = √5, on substituting for g,h and ω.
Now see whether you can solve the following problem in a minute:
A solid sphere rolls (without slipping) down a plane inclined at 30˚ to the horizontal. The distance traveled in √7 seconds after starting from rest is (g = 10 ms-2)
(a) 5.5 m (b) 6.25 m (c) 12.5 m (d) 15 m (e) 17.5 m
If you remember that the acceleration of a body rolling down an incline of angle θ is (gsinθ)/[1+ (k2/R2), where ‘k’ is the radius of gyration and R is the radius of the rolling body, you will be able to solve this in one minute. The radius of gyration of a solid sphere about its diameter is (2/5)R2 since its moment of inertia, I = (2/5)MR2, which can be equated to Mk2. Therefore, k2/R2 = 2/5 for a solid sphere. Its acceleration = (g sin30˚)/[1+(2/5)] = (10×½)/(7/5) = 25/ 7.
We have s = ut + ½ at2 = 0 + ½ ×(25/7) ×7 = 12.5 m [Option (c)].
You will find more multiple choice questions (with solution) on rotational motion here.

Wednesday, December 06, 2006

Additional Questions (MCQ) on Properties of Fluids

Even though you have been studying basic properties of fluids from lower classes, questions based on them are repeatedly found in Medical and Engineering Entrance Examinations. Here are some questions which may be interesting and useful to you:
(1) A cubical block of wood with a glass bead placed on it, is floating in water contained in a beaker. The height of water column in the beaker in this condition is ‘h’ and the extent to which the wooden block is within water is ‘d’. If the glass bead is gently transferred to the water in the beaker,
(a) ‘h’ will increase and ‘d’ will decrease (b) ‘h’ will decrease and ‘d’ will increase (c) ‘h’ and ‘d’ will be unchanged (d) ‘h’ and ‘d’ will increase (e) ‘h’ and ‘d’ will decrease
The glass bead can displace more water when it is resting on the wooden block since a floating body will displace a volume of liquid having the weight of the floating body. Inside the water, it can displace water having its own volume only. This volume is less since the density of glass is greater than that of water. Therefore, ‘h’ is decreased on transferring the glass bead to the water in the beaker.
The extent (d) to which the wooden block is within water also is decreased since the weight of the glass bead is relieved from the block. So, the correct option is (e).
(2) A tank is filled with water up to height ‘H’. A hole is made on the side of the tank at a distance ‘h’ below the level of water. What will be the horizontal range of the water jet?
(a) 2√[h(H-h)] (b) 4√[h(H+h)] (c)2√[h(H-h)]
(d) 2√[h(H+h)] (e) √[h(H+h)]
The water jet at the hole on the side of the tank will be directed horizontally with a velocity (of efflux) equal to √(2gh), as given by Torricelli’s theorem. The vertical fall of the jet is through the distance (H-h) and the time of fall (t) is √[2(H-h)/g], given by the equation of linear motion, H-h = 0 + ½ gt2.
The horizontal range R = Horizontal velocity × Time = √(2gh) ×√[2(H-h)/g] = 2√[h(H-h)].
So, the correct option is (c). [ It is interesting to note that this is independent of g so that the range on the moon will be the same as that on the earth for given values of H and h].
By putting dR/dh equal to zero, you can show that the horizontal range ‘R’ is maximum when h = H/2, which means that the hole should be at the middle of the water column.
(3) Two identical cylindrical vessels with base area ‘A’ containing a liquid of density ‘ρ’ are placed on a horizontal table. The heights of liquid column in them are h1 and h2. If they are connected by a narrow pipe to make the liquid levels in them same, the work done by gravity is
(a) ½ Aρg(h1+h2) (b) ½ Aρg(h1+h2)2 (c) ½ Aρg(h1 - h2)2
(d) ¼ Aρg(h1 + h2)2 (e) ¼ Aρg(h1 - h2)2
The work done in obtaining a cylindrical liquid column of height ‘h’ and cross section area ‘A’ is ½Aρgh2 and this is stored as potential energy in the liquid column. [ You can easily obtain this by integrating Aρgxdx between limits 0 and h. Note that (Adxρg) is the weight of liquid column (slice) of small height dx and hence (Adxρg)x is the work done against gravity to lift this liquid slice through a height x. The total work done is the integral between the limits 0 and h].
[Alternatively, you can get it by arguing that the weight of the entire liquid column is Ahρg and its centre of gravity has been raised through a height h/2 so that the work done against gravity is Ahρg(h/2), which is equal to ½Aρgh2].
The work done by gravity on equalizing the levels in the cylinders is the difference between the initial and final potential energies of the two liquid columns.
Therefore, work done = (½Aρgh12 + ½Aρgh22) - 2×½Aρg [(h1+ h2)/2]2. Note that finally the two liquid columns have the same height [(h1+ h2)/2].
The above expression for work done simplifies to (¼)Aρg(h1 - h2)2 given in option (e).
Let us now discuss the following MCQ which appeared in Har PMT 2000 question paper:
When a bubble rises from the bottom of a lake to the surface, its radius doubles. The atmospheric pressure is equal to that of a column of water of height H. The depth of the lake is
(a) 8H (b) 2H (c) 7H (d) H
This simple question involves the application of Boyle’s law: P1V1 = P2V2 with usual notations. The initial pressure is the total pressure due to the atmosphere and the water column of height ‘h’ in the lake. So, the initial pressure of the bubble is equivalent to that due to water column of height H+h. If the initial volume is V, the final volume is 8V since the radius of the bubble is doubled. The final pressure is the atmospheric pressure ‘H’.
Expressing the pressures in terms of height of water column itself, we have,
(H+h)V = H×8V, from which h = 7H [Option (c)].
You will find more questions (with solution) on properties of fluids here as well as here.

Friday, December 01, 2006

All India Engineering/Architecture Entrance Examination (AIEEE) 2007

Application Form and the Information Bulletin in respect of the All India Engineering/Architecture Entrance Examination 2007 to be conducted by Central Board of Secondary Education (CBSE) on 29th April 2007 will be distributed from 1-12-2006 to 5-1-2007. The Information Bulletin and the Application Form can be obtained personally from designated branches of Syndicate Bank/Institutions and Regional Offices of CBSE. Designated branches of Syndicate Bank in Kerala are:
(1) Thiruvananthspuram- Syndicate Bank, Co-operative Agriculture and Rural Development Bank Building, P .B.No.314, Thiruvananthapuram-695001
(2) Ernakulam- Syndicate Bank, Pioneer Towers, First Floor, P.B.No.2616, Shanmugham Road, Ernakulam-682031
(3) Kozhikode- Syndicate Bank, P.B.No.61, Cherootty road, Kozhikode-673001
(4) Kottayam- Syndicate Bank,, P.B.No.121, Baker Junction, Kottayam-686001
(5) Trichur- Syndicate Bank, Palace Road, Trichur-680020
The Information Bulletin and the Application form can be obtained by post from the Joint Secretary (AIEEE), Central Board of secondary Education, 17, Rouse Avenue, Institutioal Area (Near Bal Bhavan), New Delhi-110002.
The cost of Information Bulletin containing the Application Form, inclusive of the examination fee for BE/B.Tech. only or B.Arch/B.Planning only is Rs.300/- and Rs.150/- for General and SC/ST candidates respectively. Candidates appearing for both BE/B.Tech and B.Arch/B.Planning together should send their application form along with additional fee in the form of Demand Draft of Rs.200/- for General and Rs.100/- for SC/ST candidates in favour of Secretary, CBSE, payable at Delhi/New Delhi.
To obtain Information Bulletin containing Application Form by post, candidates should send their request to The Joint Secretary (AIEEE), Central Board of Secondary Education, 17, Rouse Avenue, Institutional Area (Near Bal Bhavan), New Delhi-110002, along with a bank draft of Rs.350/- for General Category and Rs.200/- for SC/ST candidates in favour of The Secretary, CBSE, payable at Delhi/New Delhi and a self addressed envelope of 12”×10”.
Important information at a glance in this context:
(1)
a. Sale of AIEEE Information Bulletin containing Application Form - 01.12.2006 to 05.01. 2007
b. Online submission of application on website
www.aieee.nic.in/ - 01.12.2006 to 05.01.2007
(2) Last date for
a. Receipt of request for Information Bulletin and Application Form by Post at CBSE Office, Shiksha Sadan, 17, Rouse Avenue, Institutional Area, (Near Bal Bhawan), New Delhi-110002 - 25.12.2006
b. Sale of Information Bulletin at designated branches of Syndicate Bank, Regional Offices of the CBSE and designated institutions - 05.01.2007
c. Online submission of applications - 05.01.2007
d. Receipt of complete applications “by post” including Registration Forms with Bank Draft at AIEEE Unit, CBSE, Shiksha Sadan, 17, Rouse Avenue, Institutional Area, (Near Bal Bhawan), New Delhi – 110002 -
10.01.2007
(3)
Date of dispatch of Admit Card - 09.03.2007 to 26.03.2007
(4) Issue/dispatch of duplicate admit card(on request only with fee of Rs. 50/- + postal charges of Rs. 30/- extra for out station candidate). - 10.04.2007 to 29.04.2007 (By Hand)
- 10.04.2007 to 20.04.2007 (By Post)
(5) Dates of Examination
PAPER–1: 29.04.2007 (0930-1230 hrs) PAPER–2: 29.04.2007 (1400-1700 hrs)
By visiting the site www.aieee.nic.in complete information in this regard can be obtained. Make it a point to visit the site to be informed of information updates.
Details of ‘Online’ application and extra information for candidates opting for examination centres in foreign countries also can be obtained from the site.