AAMC PS Self-Assessment Official Q&A

[gettheleadout]

This thread shall serve as the official site for discussion of the AAMC Self-Assessment in MCAT Physical Sciences.

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[justadream]

@pr2med4lf

#27. The material that is making up the round thing has a higher index of refraction than does the air. Thus, when light enters, it bends TOWARD the normal. When the light leaves the ball and enters the air again, it bends AWAY from the normal. Only choice C shows that.


#28 I forgot the context of this question (is this a discrete?). In any case, just think about this intuitively. Changing the angle of refraction changes the angle at which light bends. For example, if light is going from air to transparent glass, you know light will bend toward the normal. If you have light going from air to opaque glass, the light will still bend toward the normal but do so to an even greater degree. Thus, the light coming out the other end will be shifted to different degrees (depending on how much the incoming light is bent).

 

[pr2med4lf]

@justadream, first of all thank you SO much for helping! I understand #28 now using your way of thinking intuitively (And FYI it was not a discrete--below is the passage)

As for #27, I'm still a little confused. I understand the concept of light bending based on the new index of refraction, but I'm unclear about how to identify the normal. Also, how come the light ray converges at a point on the far end of the bead?

Thanks again for your kindness!

 

[pr2med4lf]

@justadream, first of all thank you SO much for helping! I understand #28 now using your way of thinking intuitively (And FYI it was not a discrete--below is the passage)

As for #27, I'm still a little confused. I understand the concept of light bending based on the new index of refraction, but I'm unclear about how to identify the normal. Also, how come the light ray converges at a point on the far end of the bead?

Thanks again for your kindness!

@justadream : I think after taking another look, I understand it. The normal is vertically down the center of the diagram (AKA perpendicular to the horizontal line). Correct?

 

[pr2med4lf]

Eureka! I think I might have figured out #39.

The wording of the question stem says, ¨control,¨ the reaction. I think that might imply slowing down this sort of fast reaction or some chain reaction. And maybe, the wrong answer choices would either result in speeding up the reaction or make no difference. I'm not sure this is right. Still working on it...

I think "absorbing neutrons" (C) is the correct answer because it is the only choice that is listed in the passage as something that can affect a reactor's reaction. The exact sentence I'm getting this from is the last sentence of paragraph 1: "Other important conditions for the operation of a natural reactor are concentration of uranium in the soil, size, and shape of the deposit; presence of low atomic mass elements (e.g. those in water) with which higher-energy fission neutrons may collide; and absence of elements that strongly absorb neutrons."

 

[justadream]

@pr2med4lf

To identify the normal, draw a line that goes straight out the surface (perpendicular to the surface). If the surface is round, I'm sure you know how to draw a tangent line to the curve (think of calculus!). Now draw a line PERPENDICULAR to that tangent line. That is the normal.

I'm not sure what you mean by light "converging at a point"? The incident light is a beam.

 

[pr2med4lf]

@pr2med4lf

To identify the normal, draw a line that goes straight out the surface (perpendicular to the surface). If the surface is round, I'm sure you know how to draw a tangent line to the curve (think of calculus!). Now draw a line PERPENDICULAR to that tangent line. That is the normal.

I'm not sure what you mean by light "converging at a point"? The incident light is a beam.

Okay, makes sense! My earlier confusion was because I was trying to draw 2 normal lines (one perpendicular to the tangent line at the ENTRY beam and one perpendicular to the tangent line at the EXIT beam). After this, it was hard to tell which beam was bending towards the normal. (Sorry if that didn't make sense. My train of thought was a bit messed up initially).

As for the "converging point", I meant why does the incident beam meet at a point inside the bead. But now I see that I missed a sentence in the passage that said "Light is refracted as it enters a bead, then reflected off the back of the bead."

Thanks again!!

 

[pr2med4lf]

Chemistry #28:

The boiling point of glycerine in comparison with that of isopropyl alcohol (CH3)2CHOH, is:
A. more than 10 °C higher
B. less than 10 °C higher
C. less than 10 °C lower
D. more than 10 °C lower

I understand that the BP would be higher but after that I only happened to guess that it was more than 10°C higher. Does anyone know how we were supposed to know that? Note: the passage shows the striation of glycerine, which has 3 times the OH bonds as isopropyl alcohol. Is there some quantitative relationship between number of hydrogen bonds and BP? Thanks for any insight!

 

[justadream]

@pr2med4lf

OH groups raise the BP by more than 10 degrees. This is just something for which you need to have an "intuition". 10 degrees is not that much of a difference. Hydroxyl groups should raise it by more than that.

 

[pr2med4lf]

Physics #81: A square-wave voltage signal is sent into a resistor-capacitor circuit as shown [in the attachment]. Which plot gives the typical voltage response between points A and B? (Answer choices are also in the attachment).

 

[pr2med4lf]

Physics #73:

Two cars, each of mass 1000 kg traveling at 20 m/s in opposite directions, have a head-on inelastic collision. How much heat and deformation energy is produced?
A. 2 x 10^5 J
B. 4 x 10^5 J

Why is B the right answer if the passage states "During such accidents, a (potentially large) fraction of the kinetic energy is rapidly and irreversibly converted to thermal energy and deformation of the car structure? Choice B is the total amount of kinetic energy so shouldn't the only answer less than that number be correct?

 

[justadream]

Physics #81: A square-wave voltage signal is sent into a resistor-capacitor circuit as shown [in the attachment]. Which plot gives the typical voltage response between points A and B? (Answer choices are also in the attachment).

I think the answer is A.

This is a graph of the capacitor being charged and then discharged. This process is gradual. Just knowing that will allow you to get the answer.

 

[justadream]

Physics #73:

Two cars, each of mass 1000 kg traveling at 20 m/s in opposite directions, have a head-on inelastic collision. How much heat and deformation energy is produced?
A. 2 x 10^5 J
B. 4 x 10^5 J

Why is B the right answer if the passage states "During such accidents, a (potentially large) fraction of the kinetic energy is rapidly and irreversibly converted to thermal energy and deformation of the car structure? Choice B is the total amount of kinetic energy so shouldn't the only answer less than that number be correct?

Because the two cars initially have the same speed and same mass, at the end of the collision, they are both stationary.

Thus, final KE = 0.

So ALL of the KE must have been converted into heat/deformation energy.

 

[pr2med4lf]

Physics #19

If you use the equation given for black body radiation, you'd get that the wavelength of light is 483.

I'm just curious why it doesn't emit red/orange light?

That's a good point but I think the key is that red light is emitted, BUT it's not the only color emitted, hence the white blackbody.

 

[tshank]

Physics #106

What is the magnitude of the horizontal component of air resistance on the projectile at any point during the flight? (Note: Vx= horizontal speed.)

Can someone explain the answer to this one? Their explanation is that ultimately PVx^2 is the same as (Vx/V)PV^2 Wouldn't that get you (P)(V)(Vx) not PVx^2? I thought those were different numbers since V and Vx are different values.

 

[tshank]

PS #56:

Is there a simpler way to come to the same conclusion for this? They seem to complicate bernoulli's principle and give a convoluted answer when I think it is probably really simple.

 

[tshank]

PS #61:

Can anyone explain which equation we were supposed to use, I think I used to wrong one. It doesn't make sense to me that their explanation has why its (Pi/4) x (1.0 x 10^-2)^2 and not (1/2) x (1.0 x 10^-2/2)^2 I just don't see where the 5Pi comes from.

 

[BBVET123]

I have a question from the AAMC general chem assessment (#75). I looked at previous posts and some addressed this question but I'm still confused.

I'm aware of the rule of 'like dissolves like' and was trying to apply that rule for this question by thinking that 'acid would dissolve in acid' would be a correct notion.

The question asks us to pick the substance that is more likely to be soluble in 1.0M HCL than in 1.0M NaOH. The options are HI, NaCl, Pb(OH)2, and CaF2.

I thought the answer would be HI because it's an acid. The answer is actually Pb(OH)2, the base.

Thanks for your help!

 

[ZMA]

Consider any reactions that would occur in addition to applying the 'like dissolves like' rule, as well as Le Chatelier. I'm also pretty sure everything mentioned is (aq).

 

[kraskadva]

^ditto

All the compounds listed are soluble, but...
HI is a strong acid, so yes it will dissolve, but the ions will just sit there, so nothing else is happening.
NaCl would give you a buffer, again, nothing much happening.
CaF2 would do a double substitution (to CaCl2, which breaks apart as ions, and HF, which will just sit there)

Pb(OH)2 is more fun though. That will also do a double substitution to PBCl2 and 2 H2O. Lead (II) chloride is insoluble in water and will precipitate out, driving the reaction (Le Chatlier's) forward, which would pull more Pb(OH)2 into solution, and then on to product. So that would be the most soluble, since you can keep adding more until you use up the Cl in solution.

 

[Teleologist]

Sodium chloride (ionic) can be solvated by (dissolved in) water because the ionic bonds between its component ions are comparable in strength to the ion-dipole bonds formed between water (solvent) and ion (solute) particles. In addition there is an entropic advantage to losing the ionic intermolecular association. Nonetheless, the dissolution is slightly endothermic. The driving force though behind the dissolution is entropy.

 

[gettheleadout]

I have a question from the AAMC general chem assessment (#75). I looked at previous posts and some addressed this question but I'm still confused.

I'm aware of the rule of 'like dissolves like' and was trying to apply that rule for this question by thinking that 'acid would dissolve in acid' would be a correct notion.

The question asks us to pick the substance that is more likely to be soluble in 1.0M HCL than in 1.0M NaOH. The options are HI, NaCl, Pb(OH)2, and CaF2.

I thought the answer would be HI because it's an acid. The answer is actually Pb(OH)2, the base.

Thanks for your help!

Acid-base reagents are a case where the opposite is true. As a general rule, acids are more soluble in bases and vice versa.

^ditto

All the compounds listed are soluble, but...
HI is a strong acid, so yes it will dissolve, but the ions will just sit there, so nothing else is happening.
NaCl would give you a buffer, again, nothing much happening.
CaF2 would do a double substitution (to CaCl2, which breaks apart as ions, and HF, which will just sit there)

Pb(OH)2 is more fun though. That will also do a double substitution to PBCl2 and 2 H2O. Lead (II) chloride is insoluble in water and will precipitate out, driving the reaction (Le Chatlier's) forward, which would pull more Pb(OH)2 into solution, and then on to product. So that would be the most soluble, since you can keep adding more until you use up the Cl in solution.

Just want to add and correct a few things here.

First of all, the question is asking which of the choices will be more soluble in 1M HCl than in 1M NaOH, so saying "[all] are soluble, but..." isn't relevant, since we're not considering plain water. In fact, it's not even true in this case because Pb(OH)2 will be insoluble in 1M NaOH.

(Point 1.5, neutral salts will be equally soluble in solutions of any pH given no common ions, so NaCl cannot be correct. Acids will be less soluble in acid than base, not more, so HI cannot be correct either. The answer is either Pb(OH)2 or CaF2.)

Second, adding NaCl to either solution will not produce a buffer. NaCl is a neutral salt and cannot form a buffer with any strong reagent.

Third, the important feature of CaF2 is that because the fluoride ion (F–) is weak base, upon solvation it will react with water to form HF and OH–. This reaction will be more favorable in a solution lower in OH– (ergo more acidic) than in one higher in OH– (more basic). Thus, CaF2 is a potential correct answer choice, but in this case Pb(OH)2 is better, since as you said, it has the added benefit of the Pb2+ ions crashing out with the chloride and further driving the dissolution forward.

Edit: Also, merging with official AAMC PS Self-Assessment thread.

 

[texan2414]

PS #61:

Can anyone explain which equation we were supposed to use, I think I used to wrong one. It doesn't make sense to me that their explanation has why its (Pi/4) x (1.0 x 10^-2)^2 and not (1/2) x (1.0 x 10^-2/2)^2 I just don't see where the 5Pi comes from.

would like to know this as well..
http://puu.sh/cgf7t.png

 

[simmac]

#70 typo???

The question asks "When a nucleus emits 2.5 MeV gamma ray, by how much does the nuclear mass decrease?"

I thought gamma ray emision doesn't affect nuclear mass!!

The answer is to convert MeV into amu (understandable) and then use mass of alpha particle to convert to kg...WHY would they use alpha particle mass for gamma ray emission???? Assuming the question meant to ask about alpha ray emission???



Please any help would be appreciated, this is really confusing me. thank you!!

 

[tshank]

The alpha particle converts to kg because E=mc^2. m is mass. energy can be converted to mass.

The rest of the explanation requires provided information from the passage. AAMC's explanation works pretty well in this case. Its simply plugging in equations using the information given.

I hope that helps.

 

[waynestate313]

Physics Self Assessment #14

Q: By what factor would a string's tension need to be changed to raise its fundamental frequency by a perfect fifth?

A: The relation between string frequency and tension is given in the passage as f is proportional to T^(1/2). To raise the string frequency by a perfect fifth (a factor of 3/2) the tension must be increased by a factor of (3/2)^2 = 9/4.



In the passage it says that a perfect fifth is a difference in frequency of 2/3. Can someone please explain to me how they came up with 3/2 and squaring it? Thanks in advance.

 

[tshank]

The first paragraph of the passage says that each frequency is separate by a perfect 5th, which is equivalent to 3/2. The question then essentially asks: "If F = 3/2 (as 3/2 = 5th harmony separation) then T = ??

3/2 = T^1/2. To find T you square both sides to cancel out the ^1/2.

--> T = (3/2) ^ 2 = 9/4. T = 9/4

Does that make sense to you?
The first part in identifying that 3/2 = F, if F is a harmonic frequency, is key.

 

[waynestate313]

Thank you for your quick reply. I understand all your explanation expect when it comes to the fraction 3/2. The passage says "The fundamental tones of these strings are separated by a perfect fifth, which means the fundamental frequency of each string is 2/3 that of the next higher frequency string." What I would have done to solve it is (2/3)^2. I didn't know how they came up with the fraction 3/2?

 

[tshank]

Thank you for your quick reply. I understand all your explanation expect when it comes to the fraction 3/2. The passage says "The fundamental tones of these strings are separated by a perfect fifth, which means the fundamental frequency of each string is 2/3 that of the next higher frequency string." What I would have done to solve it is (2/3)^2. I didn't know how they came up with the fraction 3/2?

Ohhh, right I get you. I don't know, it looks like they are wrong to me.

 

[popopopop]

Aren't you going from lower frequency to higher frequency? That's why you increase the tension of the string (T) by 9/4, not lowering the tension. If you use 2/3, you're decreasing the frequency and will find a lower tension. We want to see what T would be if we increased frequency by 2/3, so

(2/3)(higher f that we are finding) = 1(lower f)
1/(2/3) = higher f = T^.5

Someone please correct me if I'm wrong!

 

[tshank]

Aren't you going from lower frequency to higher frequency? That's why you increase the tension of the string (T) by 9/4, not lowering the tension. If you use 2/3, you're decreasing the frequency and will find a lower tension. We want to see what T would be if we increased frequency by 2/3, so

(2/3)(higher f that we are finding) = 1(lower f)
1/(2/3) = higher f = T^.5

Someone please correct me if I'm wrong!

I don't think you are correct. The passage says that each perfect fifth harmonic is 2/3 away from the others. So, I don't think the fraction would be inverted.

 

[popopopop]

I don't think you are correct. The passage says that each perfect fifth harmonic is 2/3 away from the others. So, I don't think the fraction would be inverted.

We start with a lower frequency string and we are trying to see how much T to increase to get a higher frequency. The passage states "each string is 2/3 that of the next higher frequency string." If I set up the lower frequency as simply 1(lower frequency) = 2/3(higher frequency), then the higher f = 3/2(lower frequency).
Also #10 also confirms this as the increment from one octave to another increases by 3/2, not 2/3 (which doesn't make sense because your tension will be lower if you squared 2/3) or 9/4.


Had you use (n<1) = T^.5, you are stating T would be lower going to a higher octave. Intuitively, lowering tension on a string gives a lower frequency sound. Pluck a higher tension string and it'll be more pingy. The answer has to be either 3/2 or 9/4 off the bat.

 

[tshank]

We start with a lower frequency string and we are trying to see how much T to increase to get a higher frequency. The passage states "each string is 2/3 that of the next higher frequency string." If I set up the lower frequency as simply 1(lower frequency) = 2/3(higher frequency), then the higher f = 3/2(lower frequency).
Also #10 also confirms this as the increment from one octave to another increases by 3/2, not 2/3 (which doesn't make sense because your tension will be lower if you squared 2/3) or 9/4.


Had you use (n<1) = T^.5, you are stating T would be lower going to a higher octave. Intuitively, lowering tension on a string gives a lower frequency sound. Pluck a higher tension string and it'll be more pingy. The answer has to be either 3/2 or 9/4 off the bat.

Ah yes I see now. Thank you very much. My intuition was this, but I couldn't prove it mathematically like you did, therefore assumed it was wrong. It seems so simple now. Thank you! This problem is a good applicable example of MCAT ratio questions.

 

[Divji Medved]

Physics #47

I put A cause the system initially has translational & rotational KE. Seems like a cheap shot.

Thoughts?

 

[TheRhymenocerous]

Physics #47

I put A cause the system initially has translational & rotational KE. Seems like a cheap shot.

Thoughts?

Once it's skidding the axle is locked so it's no longer rotating.

 

[Silence in ER]

Physics #26, I somehow still don't really understand the answer they gave I don't understand their turn of 90 degrees, at all.

 

[Silence in ER]

So you mean that retroflected angle is kind of like a reflected angle which reflects at 90 degrees, that means that both the other incident and reflected angle must add up to 90 and since the incident angle must equal the reflected angle the answer has to be 45.. if I'm getting you right?

 

[Silence in ER]

Ok thanks man

 

[OrangeMed]

I need help on #47. I thought that the initial translational kinetic energy (KE) would be greater than the work done to stop the car from skidding because initially we have Total energy= KE of the car (say 30J) +E lost due to static friction (say -3J)=27J. So when you start applying breaks you have Kinetic Friction acting on the tires that provide the entire 27J to stop the car from skidding. So isn't the initial KE greater than the energy required to stop the tire from skidding because of opposing static friction?

Thank you!

 

[OrangeMed]

I'm lost on #52, the one with organ pipe. Any guidance will be much appreciated.

 

[OrangeMed]

PS #56:

Is there a simpler way to come to the same conclusion for this? They seem to complicate bernoulli's principle and give a convoluted answer when I think it is probably really simple.

Yeah, just use the third Newtonian equation. You will end up with velocity=sqrt(2gH). Plug in g=10m/s2 and H=5m and you'll get the answer.

 

[ayocaptain628]

q91:
Because comets shine predominantly by reflected sunlight,
what one sees when viewing a comet is:
a. coma gas
b. coma dust
c. tail gas
d. ices

why is the answer coma dust?

 

[ayocaptain628]

q86. assume that the major axis (the length) and the eccentricity (the ratio of the length to the width) of mercury's orbital ellipse are both constant over time. as mercury's perihelion precesses, the figure traced by the perihelion point is a:
a. circle.
b. hyperbola
c. parabola
d. sphere

can someone reword this question to be less verbose and more layman friendly lol? what was the basic concept being tested here

 

[erythrocyte666]

q91:
Because comets shine predominantly by reflected sunlight,
what one sees when viewing a comet is:
a. coma gas
b. coma dust
c. tail gas
d. ices

why is the answer coma dust?

I got that wrong too. I feel like it was a trick question where you're just supposed to recognize that the solid dust particles can reflect better than the gas. I tried to rationalize it by stating that the gas particles are far enough apart to cause light scattering and hence unable to reflect light while dust particles are close enough together to be able to reflect light. Also, the passage states something about the ice being within the nucleus so not visible.

 

[OrangeMed]

#77 The question asks to find how far the object will fall in 0.01 sec if a rail gun shoots it at a speed of 2km/s horizontally. I thought this would be simple Distance=Speed * Time as the question gives the horizontal speed and gives the time, but in the solutions it is finding the distance the object fell from the height it was shot from the rail gun. Can anyone please explain the question to me? I keep reading the question but I always think of finding the horizontal distance and not the height.
Thank you!

 

[ayocaptain628]

#77 The question asks to find how far the object will fall in 0.01 sec if a rail gun shoots it at a speed of 2km/s horizontally. I thought this would be simple Distance=Speed * Time as the question gives the horizontal speed and gives the time, but in the solutions it is finding the distance the object fell from the height it was shot from the rail gun. Can anyone please explain the question to me? I keep reading the question but I always think of finding the horizontal distance and not the height.
Thank you!

this is a free fall problem in disguise. this is similar to those problems where something like a bullet is ejected off the top of a cliff with a certain horizontal velocity. that information is irrelevant since it's asking how far the object falls; you want vertical displacement. if the wording was like "how far from the origin" or "what range", those'd be the buzz words for horizontal distance.

 

[ayocaptain628]

q105. what is the approximate total kinetic energy of the projectile at the highest point in its path? assume that the effects of air resistance are negligible.
a. 92 j
b. 133 j
c. 184 j
d. 225 j

how do you solve this?

 

[ayocaptain628]

for anyone who has done passage 22, is there a video that i can watch of this? i could not even visualize what this electric motor was supposed to look like or do, esp. the question with the lifting.

 

[ayocaptain628]

q42. a nuclear reactor operates most efficiently if its shape contains the minimum surface area for neutrons to escape for any specific volume. the most efficient reactor would be in the shape of a:
a. cube
b. cylinder
c. sphere
d. slab

i don't understand what physics concept we are being tested on here?

 

[ayocaptain628]

the glass that is used as a beam splitter is replaced with glass that is identical except that it has a 10% higher index of refraction. which of the following changes will occur to the pinhole image?

a. it will move
b. it will become larger
c. it will become smaller
d. it will get more clearer

i dont understand? what relationship was i supposed to see with higher index?

 

[OrangeMed]

q105. what is the approximate total kinetic energy of the projectile at the highest point in its path? assume that the effects of air resistance are negligible.
a. 92 j
b. 133 j
c. 184 j
d. 225 j

how do you solve this?

Use Energy= 1/2 mv^2. At the highest point vertical velocity is 0, but there is still horizontal velocity which is V cos (theta). Use that in the energy equation and m=0.5 kg and you'll get the answer.