NOW: WITH FREE PRINTABLE SOLUTIONS. See Gold Linkbar above.

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.

A Brief Guide to Basic Site Navigation:

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.)

In the Meta-Search below, you can search for problem(s) by type and subtype---or search * in solutions or user-comments.

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!

Meta Search (Beta):

Top 10 latest comments:

1	Exam GR8677; Problem 18; Page 0 Posted by Kabuto Yakushi, on 2010-09-02 09:49:11.
	I just plugged the wave function into the schrodingers equation. is obviously zero so V(x) equals the second time derivative of psi time the . Solving one gets B). Thanks for the great site Yosun.           jump to this problem
2	Exam GR8677; Problem 8; Page 0 Posted by Kabuto Yakushi, on 2010-09-02 09:25:45.
	Just use the Impulse-Momentum theorem. F t = M V the answer falls out.           jump to this problem
3	Exam GR9277; Problem 19; Page 0 Posted by ashestofeonix, on 2010-08-31 01:54:18.
	uhh... I hate to be a stickler about something as silly as correct units but.... the density of water is 1 g/cm, that is 1kg/dm^3... And most of the earth is not water... most of the SURFACE of the earth is covered by water, but most of the volume is iron & nickel. There really is a big difference.           jump to this problem
4	Exam GR9277; Problem 29; Page 0 Posted by shak, on 2010-08-29 15:55:08.
	right           jump to this problem
5	Exam GR9677; Problem 91; Page 0 Posted by shak, on 2010-08-20 17:19:23.
	good trap:)))           jump to this problem
6	Exam GR8677; Problem 93; Page 0 Posted by his dudeness, on 2010-08-19 18:37:17.
	Yeah I definitely fell for it too... Here's Wikipedia to the rescue: "Occasionally it is maintained that µ is always < 1, but this is not true. While in most relevant applications µ < 1, a value above 1 merely implies that the force required to slide an object along the surface is greater than the normal force of the surface on the object. "           jump to this problem
7	Exam GR9677; Problem 81; Page 0 Posted by shak, on 2010-08-19 07:02:42.
	Can u please explain me, why are u dividing frequency of A4 to the frequency of D2 to find harmonics? even , if then answer is close to 6 , but it is smaller than 6, and therefore answer should be 5,i.e lowest? thank you           jump to this problem
8	Exam GR8677; Problem 31; Page 0 Posted by his dudeness, on 2010-08-18 13:12:00.
	kroner, the situation is the exact opposite of what you said. If point B has a negative voltage with respect to point A, then the electric field points from A-->B. This means that positive charges want to drop in voltage and electrons want to go up in voltage. So we need the detector to be at a negative value that's sufficiently high to stop the electron current which arises naturally (even at zero voltage). Tricky stuff!           jump to this problem
9	Exam GR8677; Problem 1; Page 0 Posted by atp_ch, on 2010-08-17 05:43:03.
	test           jump to this problem
10	Exam GR9677; Problem 66; Page 0 Posted by shak, on 2010-08-16 21:09:25.
	good analysis!!!!!thank u           jump to this problem