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Official GRE Physics Exam 2 Questions with Correct Answers

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Official GRE Physics Exam 2 Questions with Correct Answers

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GRE Psychology
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GRE Psychology

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Uploaded on
June 6, 2025
Number of pages
44
Written in
2024/2025
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Official GRE Physics Exam #2 Questions
with Correct Answers
Which of the following best illustrates the

acceleration of a pendulum bob at points

a through e ? - CORRECT ANSWER -The pendulum sways back and forth with its acceleration
pointing infinitely upwards toward the center of motion. Choice (C).



The coefficient of static friction between a small

coin and the surface of a turntable is 0.30. The

turntable rotates at 33.3 revolutions per minute.

What is the maximum distance from the center

of the turntable at which the coin will not slide?

(A) 0.024 m

(B) 0.048 m

(C) 0.121 m

(D) 0.242 m

(E) 0.484 m - CORRECT ANSWER -33.3 rev/min = 33.3*2*pi/60 rad/s = 1.11*pi rad/s. u(v^2)/r = g
(masses cancel out). r = u*(v^2)/g = [(0.30)(1.11*pi)^2]/(9.8 m/s^2) = 0.30*12.3/9.8 = 0.30
*1.25 = 0.3 m, which is closest to choice (D).



A satellite of mass m orbits a planet of mass M

in a circular orbit of radius R. The time required

for one revolution is

(A) independent of M

(B) proportional to m

(C) linear in R

,(D) proportional to R3/2

(E) proportional to R2 - CORRECT ANSWER -Kepler's Third Law of Planetary Motion ==> R^3 =
T^2, which implies the time for one revolution is proportional to R^(3/2). Choice (D).



In a nonrelativistic, one-dimensional collision,

a particle of mass 2m collides with a particle of

mass m at rest. If the particles stick together after

the collision, what fraction of the initial kinetic

energy is lost in the collision?

(A) 0

(B) 1/4

(C) 1/3

(D) 1/2

(E) 2/3 - CORRECT ANSWER -Initial Momentum: 2m(v) + m(0) = 2mv

Final Momentum is also 2mv but different mass of 3m. The kinetic energy is KE = (p^2)/(2m), so
the ratio is (p,f^2)/(p,i^2) = m,i/m,f = 2m/3m = 2/3. So 1 - 2/3 = 1/3 is the fraction of kinetic
energy lost. Choice (C).



A three-dimensional harmonic oscillator is in

thermal equilibrium with a temperature reservoir

at temperature T. The average total energy of the

oscillator is

(A) (1/2)kT

(B) kT

(C) (3/2)kT

(D) 3kT

,(E) 6kT - CORRECT ANSWER -A three-dimensional oscillator at thermal equilibrium has kT
average total energy for each degree of freedom, which is represented by choice (D).



An ideal monatomic gas expands quasi-statically

to twice its volume. If the process is isothermal,

the work done by the gas is Wi

. If the process

is adiabatic, the work done by the gas is Wa .

Which of the following is true?

(A) Wi

= Wa

(B) 0 = Wi

< Wa

(C) 0 < Wi

< Wa

(D) 0 = Wa < Wi

(E) 0 < Wa < Wi - CORRECT ANSWER -A adiabatic expansion is more ideal in nature and allows
for no heat to be taken or given to the environment, whereas an isothermal process does the
opposite. Choice (E).



Two long, identical bar magnets are placed under

a horizontal piece of paper, as shown in the figure

above. The paper is covered with iron filings.

When the two north poles are a small distance

apart and touching the paper, the iron filings

move into a pattern that shows the magnetic field

lines. Which of the following best illustrates the

, pattern that results? - CORRECT ANSWER -The two bar magnets will create a field which is
similar to that of two positive charges in proximity of each other with their electric field. Choice
(B).



A positive charge Q is located at a distance L

above an infinite grounded conducting plane,

as shown in the figure above. What is the total

charge induced on the plane?

(A) 2Q

(B) Q

(C) 0

(D) -Q

(E) -2Q - CORRECT ANSWER -The positive charge induces a negative coulomb charge on the
plate which is -Q. Choice (D).



Five positive charges of magnitude q are

arranged symmetrically around the circumference

of a circle of radius r. What is the magnitude of

the electric field at the center of the circle?

( ) k = 1 4p 0

(A) 0

(B) kq r2

(C) 5 2 kq r

(D) ( / ) cos / kq r2 a f 2 5 p

(E) ( / ) cos / 5 25 2 kq r a f - CORRECT ANSWER -Anytime all similar charges are arranged in
circle and are symmetrically distributed, the net electric field becomes zero due to total
cancellation. Choice (A).

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