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GRE Physics Solutions
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Welcome to GRE Physics Solutions. This is where you'll find complete in-depth solutions to all problems from the ETS released GRE Advanced Physics Examinations.

Free solutions, always and forever. The site aims to provide the most comprehensive and easy-to-access source of solutions to released GRE Physics exams.

 

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1-liner: Jump directly to a particular problem via the QuickLoad bar. (On the grey bar on top of the page, you can quickly jump to a problem by selecting GRxxxx, Problem y.)

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Alternate solutions are contributed by users like you in the comments section appending each problem. Your questions pertaining to a particular problem are, in turn, answered by other users. If you find any errors or have an alternate solution you'd like to share---feel free to post a comment!

 

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Top 10 latest comments:

1 Exam GR9277; Problem 93; Page 0
Posted by his dudeness, on 2010-09-05 15:55:11.
 
     
Yosun's method is clearly by far the best for this particular problem. However, if anyone is curious about how the relation in (E) is actually derived, here you go: The acceleration can be split into radial and transverse components. Since gravity is the only force contributing to transverse acceleration, We have a_t=gcos(\theta). To calculate a_r, we note that a_r=\frac{v^2}{r}. From conservation of energy, we have \frac{1}{2}mv^2=mgsin(\theta), so we get a_r=2gsin(\theta). All together now: a_{tot} = \sqrt{a_r^2+a_t^2} = \sqrt{gcos(\theta)^2+4gsin(\theta)^2}. A simple trig identity gets us the (E).    
     
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2 Exam GR9277; Problem 89; Page 0
Posted by his dudeness, on 2010-09-05 14:52:10.
 
     
also bear in mind that ETS usually begins indexing at n=0, as opposed to n=1, which most books tend to use. this convention interchanges even and odd functions -- I find that a quick sketch usually helps me out a lot.    
     
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3 Exam GR9677; Problem 10; Page 0
Posted by Kabuto Yakushi, on 2010-09-05 09:56:05.
 
     
One can get the answer without any knowledge of internal conversion. (C), (D), and (E) are all products of radioactive decay which are results of unstable nuclei. The problem tells us that after internal conversion the atom (not the nucleus) is in an exited state leaving it up to the electrons to return the atom to ground state. This leaves just (B). (A) is just plain silly.    
     
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4 Exam GR9677; Problem 8; Page 0
Posted by Kabuto Yakushi, on 2010-09-05 09:39:38.
 
     
One can see at once that the answer must be either (B) or (A). We know that when a linearly resistive force is introduced to harmonic oscillations the period: \frac{1}{\omega}=\frac{1}{(\omega_o^2-\frac{b^2}{4m^2})^{\frac{1}{2}}} will not be same as the natural period. Since (B) and (A) cannot both be false it must be one or the other, this takes out (C), (D), and (E). Common sense tells us that the oscillations won't speed up when a resistive force is applied! Therefore the answer is (A). QED. Not very rigorous, but saves time on the GRE.    
     
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5 Exam GR9677; Problem 7; Page 0
Posted by Kabuto Yakushi, on 2010-09-05 09:14:40.
 
     
It probably would be a good idea to memorize the velocity formulas for elastic collisions: v_{1f}=\frac{m_{1}-m_{2}}{m_{1}+m{2}}v_{1i}+\frac{2m_{2}}{m_{2}+m_{1}}v_{2i} and v_{2f}=\frac{m_{2}-m_{1}}{m_{2}+m_{1}}v_{2i}+\frac{2m_{1}}{m_{1}+m_{2}}v_{1i} once you get the pattern down the formulas aren't as difficult as they look.    
     
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6 Exam GR9277; Problem 77; Page 0
Posted by his dudeness, on 2010-09-05 07:02:29.
 
     
I think this kind of reasoning could potentially get you into trouble. For example, I could say "magnetic moment, in classsical terms, is generally current times area. since a nucleus has a much larger "area" than an electron, it should have a larger magnetic moment!" which is clearly wrong. A better way to think about this problem is to realize that particle magnetic moments are generally proportional to angular momentum, with the constant of proportionality being the "gyromagnetic ratio" (charge over mass). Since nuclei and electrons have the same charge and spin angular momentum, the only difference will be in the masses, so an electron's magnetic moment will be ~2000 times bigger.    
     
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7 Exam GR9277; Problem 70; Page 0
Posted by his dudeness, on 2010-09-04 19:26:17.
 
     
I meant to write p=100mc there at the end... This site is awesome in every way, but I wish there was a way to edit posts...    
     
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8 Exam GR9277; Problem 70; Page 0
Posted by his dudeness, on 2010-09-04 19:24:04.
 
     
In a similar vein, we can just use the two equations E=\gamma mc^2 and p=\gamma mv. From the first equation, we get \gamma =100. This crazy value of \gamma tells us that v is essentially equal to c, so the second equation becomes E=100mv.    
     
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9 Exam GR9277; Problem 69; Page 0
Posted by his dudeness, on 2010-09-04 19:17:48.
 
     
my personal favorite is "poop loops"    
     
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10 Exam GR9277; Problem 44; Page 0
Posted by his dudeness, on 2010-09-04 13:42:09.
 
     
well done, brah    
     
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