physics_auth wrote:blackcat007 wrote:Really nice pool of questions.. i really appreciate it.. can you please post the answers too.. i am planning to take them as a short test..
by the way.. i might be a bit nosy but you don't seem to be a student.. who are you exactly? .
No, I am a graduate student who has finished with all tests (not from US), but due to the fact that I missed the November deadlines for PGRE (I sat it in spring and) I have to wait some more time in order to be able to apply for the fall 2010 or the spring 2009, though I have limited opportunities in the latter case. Anyway, I plan to ask advise when is better to do so... .
P.S.: If I knew how to post images taken from pdfs i would have posted even more questions. Some questions require a figure or else they will tend to be very lengthy and probably appalling for many of you. The good thing about all this is that when I constructed some questions, I "bumped into" them in the real test, but the irony is that I didn't pay much attention to them before! One such question was about Coriolis effect.
Physics_auth
physics_auth wrote:Question 27:
This is a hard PGRE question. At first, I would like to say that I omitted to say that the soap film is surrounded by air. Anyway. I corrected it. The reflected beam includes the ray that is reflected by the upper surface of the film and that that is reflected by the lower surface of it. The phase difference is in general:
Δφ = phase difference due to different optical path + phase shift due to reflection
In our case it is:
phase difference due to different optical path = (2π/λ) *(2nd) = 4πnd/λ
phase shift due to the reflection off the upper surface of the film = π
thus Δφ = 4πnd/λ + π and for constructive interference it should be Δφ = 2mπ, where m = integer. For constructive interference it is therefore
4πnd/λ + π = 2mπ => {constant quantity} = 4nd = (2m - 1)λ (1)
From (1) it is that as λ increases, the order of interference should decrease so that their product is constant (since d and n are given, constant quantities).
Let's assume that the order of interference that corresponds to λ1 = 480 nm is m1, then since no other wavelength between 480 nm and 800 nm is intensified in the reflected beam, it should be that
m2 = m1 - 1 (2),
where m2 the order of interference of wv λ2 = 800 nm, since according to (1) for λ2 > λ1 it should be m2 < m1 (since the left hand product of (1) is constant quantity). This explains why in (2) we got minus 1 instead of plus 1.
Applying (1) for m1 and m2 and equating the righ-hand members of the equations (i.e eliminating the constant product) we find a relationship between m1 and m2. If in this last equation we substitute m2 from (2), then we have an equation of only one variable -> m1. After some algebra it is found that m1 = 3. By substituting m1 = 3 into (1) and solving for d we end up to answer (D).
blackcat007 wrote:physics_auth wrote:blackcat007 wrote:Really nice pool of questions.. i really appreciate it.. can you please post the answers too.. i am planning to take them as a short test..
by the way.. i might be a bit nosy but you don't seem to be a student.. who are you exactly? .
No, I am a graduate student who has finished with all tests (not from US), but due to the fact that I missed the November deadlines for PGRE (I sat it in spring and) I have to wait some more time in order to be able to apply for the fall 2010 or the spring 2009, though I have limited opportunities in the latter case. Anyway, I plan to ask for advice when is better to do so... .
P.S.: If I knew how to post images taken from pdfs i would have posted even more questions. Some questions require a figure or else they will tend to be very lengthy and probably appalling for many of you. The good thing about all this is that when I constructed some questions, I "bumped into" them in the real test, but the irony is that I didn't pay much attention to them before! One such question was about Coriolis effect.
Physics_auth
AFAIK spring sems are poorly funded, and since you are an intl student, may be it will be exorbitant for you. i think fall 2010 will be better
well if you have your homepage you can load them there and give the link here. or you may even mail them to the interested members.
hmm coriolis effect?? i thought they never ask such questions.. what else did you encounter new?
oh and yes.. please post the answers for these questions
physics_auth wrote:physics_auth wrote:Question 27:
This is a hard PGRE question. At first, I would like to say that I omitted to say that the soap film is surrounded by air. Anyway. I corrected it. The reflected beam includes the ray that is reflected by the upper surface of the film and that that is reflected by the lower surface of it. The phase difference is in general:
Δφ = phase difference due to different optical path + phase shift due to reflection
In our case it is:
phase difference due to different optical path = (2π/λ) *(2nd) = 4πnd/λ
phase shift due to the reflection off the upper surface of the film = π
thus Δφ = 4πnd/λ + π and for constructive interference it should be Δφ = 2mπ, where m = integer. For constructive interference it is therefore
4πnd/λ + π = 2mπ => {constant quantity} = 4nd = (2m - 1)λ (1)
From (1) it is that as λ increases, the order of interference should decrease so that their product is constant (since d and n are given, constant quantities).
Let's assume that the order of interference that corresponds to λ1 = 480 nm is m1, then since no other wavelength between 480 nm and 800 nm is intensified in the reflected beam, it should be that
m2 = m1 - 1 (2),
where m2 the order of interference of wv λ2 = 800 nm, since according to (1) for λ2 > λ1 it should be m2 < m1 (since the left hand product of (1) is constant quantity). This explains why in (2) we got minus 1 instead of plus 1.
Applying (1) for m1 and m2 and equating the righ-hand members of the equations (i.e eliminating the constant product) we find a relationship between m1 and m2. If in this last equation we substitute m2 from (2), then we have an equation of only one variable -> m1. After some algebra it is found that m1 = 3. By substituting m1 = 3 into (1) and solving for d we end up to answer (D).
physics_auth wrote:
EXTRA QUESTION 3:
Infrared electromagnetic radiation of broad range (it covers near and far infrared) is sent to pass through a gaseous mixture which consists amongst others of the following diatomic molecules
H_2, O_2, CO, HCl, N_2, HF, HI
Which of the above molecules will absorb the incident radiation?
(A) All of them
(B) HCl, HF, HI and O_2 only
(C) H_2, O_2 and N_2 only
(D) CO, HCl, HF and HI only
(E) None of the above
sravanskarri wrote:Can you pls explain the solution for extra question 5. I think we need to apply rocket eq but I could not get to the answer. Was this on the sample GRE Qs ?
Thanks for the previous posts
sravanskarri wrote: when the string comes off the cylinder it is a_tangential = v^2/r and a_radial = 0 leaving the magnitude constant.
sravanskarri wrote:Thanks for the reply. But isn't the change in magnitude of acceleration of the "string" zero; when it is in contact with the cylinder it has a_radial = v^2/r and a_tangential =0 => total magnitude is a_radial and when the string comes off the cylinder it is a_tangential = v^2/r and a_radial = 0 leaving the magnitude constant.
I thought there will be some change in velocity due to some mass leaving the cylinder + string system...giving it some acceleration.May be I overdid it.
sravanskarri wrote:My bad. Agree with the solution. As I mentioned earlier, I kind of mis-understood the Q, at this part
>>"As each small segment of string leaves the cylinder, by what amount does its acceleration change"
I never thought "its acceleration" implied "string acceleration" but "cylinder acceleration" which is a foolish mistake on my part. I did not mean to say the questions you had need to be from ETS. Please keep posting ,the questions are really good and helping me identify my shortcomings.
betelgeuse1 wrote:small problem with problem 16: The Idea is to integrate 2/L * Sin^2(n pi x/L) from 5L/6 to L/2
I obtain Integral(1/2-Cos(2 n pi x/L)/2)dx and that gives (1/2)x from L/2 to 5L/6 minus a term depending on Sin(2 n pi x/L) from L/2 to 5L/6.
Now if the last term (the one depending on Sin) is zero the I obtain the probability for the particle to be there as 1/3 and the prob of not being there as 2/3 i.e. answer A but the problem is that Sin (2 n pi x/L) for n=2 and x=5L/6 is not zero as it is for x=L/2 so the Sine term is not zero so you won't get 2/3.
Do I make a mistake here?
betelgeuse1 wrote:small problem with problem 16: The Idea is to integrate 2/L * Sin^2(n pi x/L) from 5L/6 to L/2
I obtain Integral(1/2-Cos(2 n pi x/L)/2)dx and that gives (1/2)x from L/2 to 5L/6 minus a term depending on Sin(2 n pi x/L) from L/2 to 5L/6.
Now if the last term (the one depending on Sin) is zero the I obtain the probability for the particle to be there as 1/3 and the prob of not being there as 2/3 i.e. answer A but the problem is that Sin (2 n pi x/L) for n=2 and x=5L/6 is not zero as it is for x=L/2 so the Sine term is not zero so you won't get 2/3.
Do I make a mistake here?
betelgeuse1 wrote:what if the interaction is weak? As I know you should specify something about the life time of the stable initial state, otherwise you could choose D too if strangeness is not conserved... I agree with the conservation of barion number though
betelgeuse1 wrote:me again : problem 34: I don't have any idea about it now... some hlp! I guess quantum tunneling is not the solution...
blackcat007 wrote:betelgeuse1 wrote:small problem with problem 16: The Idea is to integrate 2/L * Sin^2(n pi x/L) from 5L/6 to L/2
I obtain Integral(1/2-Cos(2 n pi x/L)/2)dx and that gives (1/2)x from L/2 to 5L/6 minus a term depending on Sin(2 n pi x/L) from L/2 to 5L/6.
Now if the last term (the one depending on Sin) is zero the I obtain the probability for the particle to be there as 1/3 and the prob of not being there as 2/3 i.e. answer A but the problem is that Sin (2 n pi x/L) for n=2 and x=5L/6 is not zero as it is for x=L/2 so the Sine term is not zero so you won't get 2/3.
Do I make a mistake here?
physics_auth wrote:blackcat007 wrote:betelgeuse1 wrote:small problem with problem 16: The Idea is to integrate 2/L * Sin^2(n pi x/L) from 5L/6 to L/2
I obtain Integral(1/2-Cos(2 n pi x/L)/2)dx and that gives (1/2)x from L/2 to 5L/6 minus a term depending on Sin(2 n pi x/L) from L/2 to 5L/6.
Now if the last term (the one depending on Sin) is zero the I obtain the probability for the particle to be there as 1/3 and the prob of not being there as 2/3 i.e. answer A but the problem is that Sin (2 n pi x/L) for n=2 and x=5L/6 is not zero as it is for x=L/2 so the Sine term is not zero so you won't get 2/3.
Do I make a mistake here?
apart from the method mentioned by Physics_auth here is another one: 2nd excited means n=3, (n=1 is ground state) and the probability of not finding the particle in between L/2 and 5L/6 is (1- Probability of finding the particle between L/2 and 5L/6 )[/quote
Yeah man. It is impossible that ETS will ask you to use Bragg's formula for a non-simple cubic lattice, since in this case one needs to know things about Miller indices, and the edge of the cube also, in order to use a formula for the distance between lattice planes. At least me I have never met such question in PGRE.
Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
blackcat007 wrote:Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
well i think it should be monochromatic, because if its a continuous spectrum, then for a given lattice spacing, the final pattern of diffraction will be very cumbersome.. there will resolution problem (Rayleigh's Criterion), ie difficult to resolve, since dispersion d(theta)/d(lambda) will be very large and thus overlapping will prevent proper identification of the angle diffreacted.
The wavelength is continuous, so in 2*d*sin(theta)=n*(lambda), since lambda is continuous we will get a continuous pattern and thus difficult to differentiate.
physics_auth wrote:blackcat007 wrote:Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
well i think it should be monochromatic, because if its a continuous spectrum, then for a given lattice spacing, the final pattern of diffraction will be very cumbersome.. there will resolution problem (Rayleigh's Criterion), ie difficult to resolve, since dispersion d(theta)/d(lambda) will be very large and thus overlapping will prevent proper identification of the angle diffreacted.
The wavelength is continuous, so in 2*d*sin(theta)=n*(lambda), since lambda is continuous we will get a continuous pattern and thus difficult to differentiate.
Good attempt, but we use radiation of continuous spectrum. In fact, it is difficult to find the correct direction of the incident ray -assuming it is monochromatic- for which it can lead to Bragg diffraction. Besides the detector assumes only a small solid angle in space - detectors of solid angle of 4π are only very rarely used due their high cost (and probably other factors that I miss... for the time being). By using a continuous radiation, Bragg diffraction picks automatically out all these wavelengths that satisfy Bragg's law and in the pattern in the output we see well resolved peaks sitting upon an almost flat background. The peaks do correspond to Bragg reflections (or diffractions) and then we try to find out which possible groups of lattice planes could give a specific peak (this is done by the formula that connects d -the distance between consecutive planes in a group- with the Miller indices and the dimensions of the cell.) The point is that we don't know in advance which group of planes (i.e. the d) has given a specific Bragg reflection ... .
blackcat007 wrote:physics_auth wrote:Blackcat -> By the way what type of radiation do you think is used in practice to illustrate Bragg's law? Monochromatic or of continuous spectrum and why (give a possible explanation)?
well i think it should be monochromatic, because if its a continuous spectrum, then for a given lattice spacing, the final pattern of diffraction will be very cumbersome.. there will resolution problem (Rayleigh's Criterion), ie difficult to resolve, since dispersion d(theta)/d(lambda) will be very large and thus overlapping will prevent proper identification of the angle diffreacted.
The wavelength is continuous, so in 2*d*sin(theta)=n*(lambda), since lambda is continuous we will get a continuous pattern and thus difficult to differentiate.
Good attempt, but we use radiation of continuous spectrum. In fact, it is difficult to find the correct direction of the incident ray -assuming it is monochromatic- for which it can lead to Bragg diffraction. Besides the detector assumes only a small solid angle in space - detectors of solid angle of 4π are only very rarely used due their high cost (and probably other factors that I miss... for the time being). By using a continuous radiation, Bragg diffraction picks automatically out all these wavelengths that satisfy Bragg's law and in the pattern in the output we see well resolved peaks sitting upon an almost flat background. The peaks do correspond to Bragg reflections (or diffractions) and then we try to find out which possible groups of lattice planes could give a specific peak (this is done by the formula that connects d -the distance between consecutive planes in a group- with the Miller indices and the dimensions of the cell.) The point is that we don't know in advance which group of planes (i.e. the d) has given a specific Bragg reflection ... .
betelgeuse1 wrote:Now that I jumped into the pool and came out in a poor condition but still alive I would like to see the other pool... you said something about harder problems. Where are they?
betelgeuse1 wrote:First about problem 34: I don't understand how is it possible to solve this problem using R=(k1-k2)^2/(k1+k2)^2, k1~sqrt(E) and k2~sqrt(E-V) in 2 minutes. I am sure there may be a quicker solution but I don't get it.
colonblow wrote:hi, shouldn't the answer to #1 problem be letter D, which is 3g/L? If 3g/2L is the correct answer, can someone please tell me how it was solved? thanks thanks!
QUESTION 35: (Magnetism - CR question)
The magnetic dipole moment of a current-carrying loop of wire is in the positive z direction. The magnetic dipole is placed in space where there is a magnetic field B = B0*i + B0*j, where B0 = const. and i , j the unit vectors along directions x and y respectively. What is the direction of the magnetic torque on the loop?
(A) It is along the negative z direction.
(B) It is along the line y = -x in the forth quadrant.
(C) It is along the line y = x, in the third quadrant.
(D) It is along the line y = -x, in the second quadrant.
(E) It is along the line x = y = z towards positive x,y and z.
QUESTION 40: (Cosmology - CR question)
Which of the following facts could be a remnant of the Big Bang theory for the evolution of the universe from a space-time singularity?
I. Uniform distribution of microwave background radiation.
II. Uniform distribution of background electrons.
III. Uniform distribution of background neutrinos.
IV. Uniform distribution of gluons and quarks.
(A) I and II only
(B) I, II and III only
(C) I and III only
(D) I, III and IV only
(E) II and III only
blackcat007 wrote:QUESTION 35: (Magnetism - CR question)
The magnetic dipole moment of a current-carrying loop of wire is in the positive z direction. The magnetic dipole is placed in space where there is a magnetic field B = B0*i + B0*j, where B0 = const. and i , j the unit vectors along directions x and y respectively. What is the direction of the magnetic torque on the loop?
(A) It is along the negative z direction.
(B) It is along the line y = -x in the forth quadrant.
(C) It is along the line y = x, in the third quadrant.
(D) It is along the line y = -x, in the second quadrant.
(E) It is along the line x = y = z towards positive x,y and z.
I think B0>0 should be mentioned for avoiding any ambiguity between B and D option.
B0 is supposed to represent the magnitude of magnetic field component which is always a non-zero quantity. To avoid confusion with the direction in space I used the unit vectors i and j. If I said B0 *(-i) this means that the i-th component of the magnetic induction is towards negative x-direction. Anyway, I corrected it to avoid ambiguity as you said ... . Thanks!QUESTION 40: (Cosmology - CR question)
Which of the following facts could be a remnant of the Big Bang theory for the evolution of the universe from a space-time singularity?
I. Uniform distribution of microwave background radiation.
II. Uniform distribution of background electrons.
III. Uniform distribution of background neutrinos.
IV. Uniform distribution of gluons and quarks.
(A) I and II only
(B) I, II and III only
(C) I and III only
(D) I, III and IV only
(E) II and III only
but the recent WMAP found anisotropy in the cosmic background of the order of micro kelvins and astrophysicists (esp Smoot and Fire) have made some prediction of structure formations from these anisotropy.
noospace wrote:Where did there questions come from? Are they really official ETS? If so how come they're not available on the website?
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