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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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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 GR0177; Problem 61; Page 0 Posted by Grandpsykick, on 2010-03-11 15:20:18.
	Even more simply put: 1) E fields don't propagate through conductors (if they are normal to the surface). 2) The energy in the field MUST be transmitted through the surface. 3) E is zero (0), thus B cannot be zero (0); thus, the only answer is C. NOTE: you didn't need any math, or "wave dynamics", or anything but simple knowledge of EM and conservation of energy.           jump to this problem
2	Exam GR9677; Problem 98; Page 0 Posted by cilginfizikci, on 2010-03-09 02:36:30.
	why the energy can not be "0" ??? dont tell me that this is harmonic oscillator... coz u already change the system... ????????????           jump to this problem
3	Exam GR9677; Problem 87; Page 0 Posted by cilginfizikci, on 2010-03-09 01:06:55.
	For the solution at the top: The induced magnetic field direction can not be opposite to what it was before... on the contrary it s the same direction, so that the magnetic flux will be kept constant... Len'z law says that the closed loops doesnt like the change in magnetic flux... there was a magnetic field at the beginning and hen suddenly it s gone so is the magnetic flux... the two cgarhe system will begin rotating to acquire the former magnetic flux value... This approach looks the most formal one, and i tried to solve in this way, however the angular momentum i found is ( mdB(pi^2)(R^2) )/q u0 .... so i dunno what was the guy who prepared this question was thinking...           jump to this problem
4	Exam GR9677; Problem 83; Page 0 Posted by cilginfizikci, on 2010-03-08 23:42:31.
	you have solved the equation wrongly... V= (gd)^0.5... and according to ur solution the answer is A... check ur calc. pls           jump to this problem
5	Exam GR0177; Problem 1; Page 0 Posted by chrisiskey, on 2010-03-08 22:14:25.
	No, this makes sense. At the endpoints the velocity of the bob is zero, so there is indeed a need for a nonzero acceleration for the bob to return to the center position; however, the bob has a nonzero velocity at the center position so it is not necessary for a left- or right-pointing acceleration to be present for the bob to continue in the left/right direction. Consider circular motion - a centrally-pointing force is present yet at any instant the bob is moving perpendicular to the direction of the force vector.           jump to this problem
6	Exam GR9277; Problem 6; Page 0 Posted by fredsp7, on 2010-03-08 14:18:38.
	Sorry about the messages, first time. Real Message: I tryied to calculate like you said with any angle but my answer came as being: and it also provides the correct answer with           jump to this problem
7	Exam GR9277; Problem 6; Page 0 Posted by fredsp7, on 2010-03-08 09:27:42.
	I tryied to calculate like you said with any angle \theta but my answer came as being: M\leq \frac{2\mu m}{(2\cos \theta \sen \theta - \mu)} and it also provides the correct answer with \theta=45           jump to this problem
8	Exam GR9277; Problem 6; Page 0 Posted by fredsp7, on 2010-03-08 09:09:31.
	I tryied to calculate like you said with any angle \theta but my answer came as being: M\leq \frac{2\mu m}{(2\cos \theta \sen \theta - \mu)} and it also provides the correct answer with \theta=45           jump to this problem
9	Exam GR8677; Problem 97; Page 0 Posted by Tatyana, on 2010-03-08 04:41:12.
	Actually rotational angular momentum is determined as only about rotational axis. We cannot calculate total angular momentum about the point P as . rnThe line of reasoning leading us to the right answer is folowing. The collision is absolutely nonelastic as it is indicated in the formulation of the problem: "two discs stick together". In nonelastic collisions colliding bodies move as a united body after the collision. It means that relative motion of bodies about the center of mass of the system ceased after the collision. So discs will move translationally as a united body. The rotational part will be lost. Consequently the angular moment about the point P - the center of mass of the system in the moment of the collision - will be equal to zero. A is the right answer.           jump to this problem
10	Exam GR8677; Problem 97; Page 0 Posted by Tatyana, on 2010-03-08 04:40:28.
	Actually rotational angular momentum is determined as only about rotational axis. We cannot calculate total angular momentum about the point P as . rnThe line of reasoning leading us to the right answer is folowing. The collision is absolutely nonelastic as it is indicated in the formulation of the problem: "two discs stick together". In nonelastic collisions colliding bodies move as a united body after the collision. It means that relative motion of bodies about the center of mass of the system ceased after the collision. So discs will move translationally as a united body. The rotational part will be lost. Consequently the angular moment about the point P - the center of mass of the system in the moment of the collision - will be equal to zero. A is the right answer.           jump to this problem