11Physics discretes test w solutions

Physics Discretes
Test
Time: 30 Minutes
Number of Questions: 30

This
test
consists
of
30
discrete
questions—questions that are NOT based on a
descriptive passage. These discretes comprise
15 of the 77 questions on the Physical
Sciences and Biological Sciences sections of
the MCAT.


MCAT

PHYSICS DISCRETES TEST
DIRECTIONS: The following questions are not
based on a descriptive passage; you must select
the best answer to these questions. If you are
unsure of the best answer, eliminate the choices
that you know are incorrect, then select an
answer from the choices that remain. Indicate
your selection by blackening the corresponding
circle on your answer sheet. A periodic table is
provided below for your use with the questions.

PERIODIC TABLE OF THE ELEMENTS
1
H
1.0

2
He
4.0

3
Li
6.9

4
Be
9.0

5
B
10.8

6
C
12.0

7
N
14.0

8

O
16.0

9
F
19.0

10
Ne
20.2

11
Na
23.0

12
Mg
24.3

13
Al
27.0

14
Si
28.1

15
P
31.0

16
S
32.1

17
Cl
35.5

18
Ar
39.9

19
K
39.1

20
Ca
40.1

21
Sc
45.0

22
Ti
47.9

23

V
50.9

24
Cr
52.0

25
Mn
54.9

26
Fe
55.8

27
Co
58.9

28
Ni
58.7

29
Cu
63.5

30
Zn
65.4

31
Ga
69.7

32
Ge
72.6

33
As
74.9

34
Se
79.0

35
Br
79.9

36
Kr
83.8

37
Rb
85.5

38

Sr
87.6

39
Y
88.9

40
Zr
91.2

41
Nb
92.9

42
Mo
95.9

43
Tc
(98)

44
Ru
101.1

45
Rh
102.9

46
Pd
106.4

47
Ag
107.9

48
Cd
112.4

49
In
114.8

50
Sn
118.7

51
Sb
121.8

52
Te
127.6

53

I
126.9

54
Xe
131.3

55
Cs
132.9

56
Ba
137.3

57
La *
138.9

72
Hf
178.5

73
Ta
180.9

74
W
183.9

75
Re
186.2

76
Os
190.2

77
Ir
192.2

78
Pt
195.1

79
Au
197.0

80
Hg
200.6

81
Tl
204.4

82

Pb
207.2

83
Bi
209.0

84
Po
(209)

85
At
(210)

86
Rn
(222)

87
Fr
(223)

88
Ra
226.0

89
Ac †
227.0

104
Unq
(261)

105
Unp
(262)

106
Unh
(263)

107
Uns
(262)

108
Uno
(265)

109
Une
(267)

*

58
Ce
140.1

59
Pr
140.9

60
Nd
144.2

61
Pm
(145)

62
Sm
150.4

63
Eu
152.0

64
Gd
157.3

65
Tb
158.9

66

Dy
162.5

67
Ho
164.9

68
Er
167.3

69
Tm
168.9

70
Yb
173.0

71
Lu
175.0

†

90
Th
232.0

91

Pa
(231)

92
U
238.0

93
Np
(237)

94
Pu
(244)

95
Am
(243)

96
Cm
(247)

97
Bk
(247)

98
Cf
(251)

99
Es
(252)

100
Fm
(257)

101
Md
(258)

102
No
(259)

103
Lr
(260)

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2

as developed by


Physics Discretes Test

1 . An ideal
experiences:

gas

compressed

adiabatically

A . a decrease in temperature and an increase
internal energy.
B . a decrease in temperature and a decrease
internal energy.
C . an increase in temperature and an increase
internal energy.
D . no change in either the temperature
internal energy.

in
in

4 . An object is moving counterclockwise in a circle
at constant speed. Which of the following
diagrams correctly indicates both its velocity
vector and its acceleration vector?
A.

in
or

at
v

2 . Two identical rods having temperatures 100°C
and 50°C, respectively, are brought into contact.
If the system is isolated from the environment,
its entropy:

3

a
v

a

B.

A.
B.
C.
D.

C.

v

increases.
decreases.
remains the same.
cannot be determined.

ar

D.

v
a

5 . An ambulance originally at rest has its siren
going. The ambulance starts accelerating away
from a boy standing nearby. He will hear:
A.
B.
C.
D.

a decrease in the pitch of the siren.
no change in the pitch of the siren.
an increase in the pitch of the siren.
an increase, and then a decrease in the pitch
of the siren.

A 1.0 kg object moving east with a velocity of
10 m/s has a head-on collision with a 0.5 kg
object which is at rest. Neglecting friction, what
is the momentum of the system after collision?
A.
B.
C.
D.

15 kg • m/s; east
15 kg • m/s; west
10 kg • m/s; east
10 kg • m/s; west

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3


6 . The diagram shows energy levels E1, E 2, E 3, and
E4 in an atom. If the transition from energy E3
to E2 gives rise to a photon of radiation of a
particular wavelength, which of the following
transitions could give rise to a photon of
radiation of longer wavelength?

9 . An object (O) is placed within the focal length f
of a converging lens. The image (I) is located at
which of the following positions?
O

A.
f

E4

f'

Energy (eV)

E3

I

I

B.

O

E2

f
C.

f'
O
f'

E1

A.
B.
C.
D.

E4 to E1
E3 to E1
E2 to El
E4 to E3

f
D.

I
O

f

f'

7 . The activity of a radioactive source falls to onesixteenth of its original value in 32 minutes.
What is the half-life, in minutes, of this decay
process?
A.
B.
C.
D.

2
4
8
16

min
min

min
min

I

1 0 . When the 2/3 Ω resistor is replaced by an 8/3 Ω
resistor, the voltage across the resistor changes
from:
9V

8 . A parallel plate air capacitor is connected across a
battery of constant emf. If the separation of the
plates is decreased, which of the following
increases?
I. Charge on the plates of the capacitor
II. Potential difference between the plates
of the capacitor
III. Capacitance of the capacitor
A.
B.
C.
D.

III only
I and II only
I and III only
II and III only

2/3 Ω
1/3 Ω

A.
B.
C.
D.

4
4
6
6

V
V
V
V

to
to
to
to

6
8
4
8

V.
V.
V.
V.

Physics Discretes Test

1 1 . From the data shown below, estimate the halflife of element X.

Number of Undecayed Nucleii

10 x 105

1 3 . In the diagram below a stream of electrons leaves
the electron gun at G and strikes the fluorescent
screen at P. When the current is switched on at
S, it flows through the wire coils in an
anticlockwise direction as seen by the observer at
O; the observer sees the spot of light at P:

Element X
7.5 x 105

G

5 x 105

P

O

2.5 x 105

0

1.5

3.0

4.5

6.0

S

Years elapsed

A.
B.
C.
D.

1.5 years
3.0 years
4.5 years
6.0 years

1 2 . A planet has a diameter one half of the Earth’s
diameter and a mass that is one half of the
Earth’s mass. What would be the acceleration due
to gravity on the planet?
A.
B.
C.

D.

2.45 m/s2
4.90 m/s2
9.80 m/s2
19.6 m/s2

A.
B.
C.
D.

move downwards.
move to the left.
move to the right.
remain still.

1 4 . Object A of mass M is released from height H
while object B of mass 0.5 M is released from
height 2H. What is the ratio of the velocity of
object A to the velocity of object B immediately
before they hit the ground? (Note: Assume that
air resistance is negligible.)
A.
B.
C.
D.

1:1
1: 2

1:2
1:4

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5


1 5 . In the hydraulic lift shown below, piston A has a
cross-sectional diameter of 2 m and piston B a
cross-sectional diameter of 4 m. If the force
exerted by piston A on the liquid is doubled, the
force lifting up piston B is:

A

A.
B.
C.
D.

B

1 7 . A fluid flows through a pipe such that the
velocity at point A is 1/4 that at point B. What
is the ratio of the radius of the pipe at point A to
that at point B?
A.

B.
C.
D.

1:4
2:1
4:1
1:2

1 8 . A cube of wood whose sides are each 10 cm
weighs 16 N in air. When half submerged in an
unknown liquid it weighs only 10 N. What is the
density of the liquid? (Note: Assume g = 10
m/s2.)

decreased by a factor of 4.
increased by a factor of 2.
increased by a factor of 4.
increased by a factor of 8.

A . 600 kg/m3
B.
800 kg/m3
C . 1,000 kg/m3
D . 1,200 kg/m3

1 6 . Which of the following is true of the adiabatic
expansion from point Q to R?

P

T

Q
S

R

1 9 . Two fixed charges of +Q and +3Q repel each
other with a force, F. If an additional charge of
–2Q is now added to each of the two charges,
what is the new force between them?
A.
B.
C.
D.

Zero
F/6 attraction
F/3 repulsion
F/3 attraction

V
A . The internal energy at Q is greater than at R.
B . The internal energy at R is greater than at Q.
C . Heat is released by the gas in moving from
Q to R.
D . Heat is absorbed by the gas in moving from
Q to R.

2 0 . If the binding energy of a nucleus is 186.2 MeV,
what is its mass defect? (Note: 1 amu yields 931
MeV of energy, 1 amu is equivalent to 1.66 ×
10–27 kg.)
A.
B.
C.
D.

3.3
6.6
8.3
3.1

×
×
×
×

10–28
10–28
10–27
10–25

kg
kg
kg
kg

Physics Discretes Test

2 1 . A block starts from rest and slides down along
one quadrant of a frictionless circular track whose
radius is one meter. Which of the following is
true?

2 4 . In a frictionless horizontal mass spring system,
as the mass moves from the point of maximum
expansion through the equilibrium position to
the point of maximum compression, the
potential energy of the system:
A.
B.
C.
D.

1m
A . The speed of the block at the bottom will be
the same as if it had fallen vertically.
B . Mechanical energy will not be conserved.
C . The acceleration of the block is constant
throughout the descent.
D . The total work done by gravity is greater
than if the block had fallen vertically.

decreases continually .
increases then decreases.
decreases then increases.
remains the same.

2 5 . Which of the following diagrams accurately
depicts the electric field lines between 2 charges
of equal magnitude but opposite sign?
A.

C.

B.

D.

2 2 . Which one of the following is a scalar quantity?
A.
B.
C.
D.

Velocity
Force
Impulse
Kinetic energy

2 3 . A fireboat is fighting a fire to its right but also
has its hoses going full blast into the water to its
left. Which law of physics best explains the
reason for this?
A.
B.
C.

D.

Newton’s law of gravity
Newton’s second law
Newton’s third hew
Conservation of energy

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KAPLAN

7


MCAT

2 6 . Four positively charged spheres (+1 C each) and
four negatively charged spheres (–1 C each) are
distributed at equally spaced intervals around a
circular hoop whose radius is 1 m as shown
below. What is the electrostatic potential at the
center of the hoop? (Note: k is the electrostatic
constant.)

2 8 . In an attempt to restore the natural rhythm of a
heart attack victim, electric shock plates are
sometimes used. The shock plates are run off a
10 V battery and draw current at a constant 30 A.
If each shock draws 300 J of electrical energy,
what is the duration of the shock?

A.
B.
C.
D.

+
+

+

+

–

–

–
–

2 9 . A bat locates obstacles in its flight path by
emitting sound in the same direction as it travels.
When the sound encounters an obstacle, it is
reflected back to the bat. Assuming that the bat’s
velocity and the frequency of the emitted sound
are constant, and assuming the obstacle is
stationary, which of the following describes the
frequency observed by the bat?
A . It is greater than the frequency emitted by the
bat.
B . It is less than the frequency emitted by the

bat.
C . It is the same as the frequency emitted by the
bat.
D . Its character cannot be determined.

A. 0V
B . 4k V
C . 8k V
D . 16k V

2 7 . Which of the following quantities can be
expressed dimensionally as M • L2/T2, where M
is mass, L is length, and T is time?

0.1 sec
0.3 sec
1.0 sec
3.0 sec

3 0 . An object is moved from 2f towards f, where f is
the focal length of this converging lens. The
image formed on the other side of the lens will
move from:

I. Torque
II. Work
III. Energy
A.
B.
C.

D.

I only
I and II only
II and III only
I, II, and III

2f

f

A.
B.
C.
D.

f

f to 2f.
2f to f/2.
2f to f.
2f to infinity.
END OF TEST

2f


Physics Discretes Test

THE ANSWER KEY IS ON THE NEXT PAGE

KAPLAN

9


MCAT

ANSWER KEY:
1. C
11. B
2. A
12. D
3. C
13. D
4. A
14. B
5. A
15. B
6.
7.
8.
9.
10.

10

D
C
C

B
C

16.
17.
18.
19.
20.

A
B
D
D
A

21.
22.
23.
24.
25.

A
D
C
C
A

26.
27.
28.

29.
30.

A
D
C
A
D


Physics Discretes Test

EXPLANATIONS
1.

C
This is a first law of thermodynamics question, which states that the change in internal energy, U, equals
the heat, Q, minus the work done, W. We are asked to find what happens to a gas that is compressed adiabatically.
To solvethis problem one must know that adiabatic means that no heat enters or leaves the system. This means that
Q is zero, and the first law simplifies to U = –W.
Because the system is compressed, work is done on the system, and not by the system. Work is therefore
negative. Another way we could have reached this conclusion from the formula W = PV for an ideal gas. Since in
this case the final volume is less than its initial volume, V is negative and so is W.
Since W is negative, –W is positive and so the change in internal energy is positive. Choices B and D can
therefore be eliminated. Now all that is left is to decide whether an increase in internal energy would increase or
decrease the temperature. Since temperature increases with the average kinetic energy of the gas molecules, as the
internal energy increases, so would the temperature.
2.

A

In an isolated system, the entropy never decreases. The only possibilities are entropy remains the same (if
the process is reversible) or it increases (if the process is irreversible). Under what category does this process fall?
When the two rods are put in thermal contact, heat will flow from the hot rod to the cold one until both
rods are at the same temperature (in thermal equilibrium). Is it possible for the original situation (hot rod/cold rod) to
redevelop spontaneously? No: the process is irreversible. The answer is thus choice A.
Note that the rod that is hot originally will experience a decrease in entropy as it cools. This rod by itself is
however not an isolated system (energy is flowing out of it into the cold rod) and therefore it doesn’t violate what we
have just said. The increase in entropy of the cold hot upon warming is larger in magnitude than the decrease in
entropy of the hot rod as it cools, thus leading to a net increase in entropy.
3.

C
The two concepts necessary to keep in mnd to solve this problem are (1) momentum is conserved in a
collision and (2) momentum is the product of mass and velocity. Conservation of momentum states that the
momentum immediately before a collision equals the momentum immediately after the collision. Before the
collision, the 1.0-kg object has a momentum of 1.0 kg × 10 m/s = 10 kg•m/s. This is the total initial momentum
since the other object is at rest (v = 0, hence p = mv = 0). This initial momentum of the system points in the same
direction as the velocity of the moving object, i.e. towards the east. This must then also be the momentum of the
system after collision.
4.

A
The object is undergoing uniform circular motion as its trajectory is circular and it is moving at a constant
speed. The velocity of the object is always tangential to the circle. On this criterion alone, choices B and C can be
eliminated. For uniform circular motion, there is no tangential component to the acceleration. The acceleration is
therefore entirely radial and is called the centripetal acceleration, with magnitude equal to v2/r. It points towards the
center of the circle.
5.

A

This is a Doppler effect question where we have a source moving away from a stationary observer. The
exact form of the equation that describes this situation is:
f’ =

fv
(v + v s )

where f’ is the observed frequency, f is the actual frequency (from the point of view of the source), v the velocity of
sound in the medium, and vs the velocity of the source. We do not need to remember this equation specifically
though to answer the question. (In fact it is very unlikely that the MCAT will expect you to have memorized this.)
We should, however, be able to reason out that if the source is constantly moving away from the observer,
successive wave crests would take longer and longer to reach the observer, leading to a perception of a lower
frequency or pitch. The faster the source is moving away, the lower the perceived pitch would be. Here the
ambulance is accelerating from the boy; its speed is therefore increasing. The pitch would then be decreasing.
6.

D

KAPLAN

11


MCAT

We are asked for a transition that would give rise to a photon of a longer wavelength. When a particle (for
example an electron) relaxes from a higher energy state to a lower energy state, the excess energy is released in the
form of a photon. The energy of a photon is directly proportional to its frequency via E = hf, but is inversely
proportional to the wavelength: E = hc/λ. We are therefore looking for the transition that corresponds to a smaller
energy difference between the initial and final states than between E3 and E2. The energy difference between levels is

easily read off from the figure as the vertical distance between the energy levels involved. Out of all the choices, only
choice D, a transition from level 4 to level 3, represents a transition that releases less energy than the transition from
level 3 to level 2.
7.

C
When one half-life has passed there is one-half of the original sample left. When two half-lives have passed,
one quarter of the original sample is left. Continuing in the same way, we find that when four half-lives have passed
there is one sixteenth of the sample left. This means that in 32 minutes 4 half-lives have passed. This corresponds to
8 minutes per half-life.
8.

C
The capacitance of a parallel-plate capacitor is approximately given by:
C = κε0

A
d

where κ is the dielectric constant of the medium between the plates, ε0 is the permittivity of free space, A is the area
of overlap between the two plates, and d is the separation of the two plates. From this equation we can see that if the
separation of the plate decreases, the capacitance will increase.
A more general relationship involving capacitance, which applies to capacitors that are not necessarily of
the parallel-plate variety, is C = Q/V, where Q is the charge stored on one of the pair of conductors making up the
capacitor, and V the voltage across the two. We are told in the question stem that the capacitor is connected across a
battery of constant emf. Applying Kirchhoff’s law, this means that the voltage across the plates will have to
constant as well—in fact, its magnitude will be the same as the emf of the battery. Since C increases while V has to
remain constant, Q, the charge, will have to increase as well.
9.

B
This is one of those questions where you may already know the answer, but if you don't you can reason it
out. For any object that is placed inside the focal point of a converging lens, the image is on the same side of the
lens, erect and enlarged. The converging lens is then a magnifying lens. If you didn't know this, you could have
1
1
1
selected some sensible numbers and plugged them into the equation
=
+
. If we assume that the focal
f
i
o
length is 3 centimeters and the object distance is 2 centimeters, we can calculate an approximate value for the image
distance. Rearranging the equation and putting in the numbers, we get:
1
1
1
1
=
–
=–
i
3
2
6
The image distance is then –6 centimeters. Since the value is negative, the image is virtual. For a lens this
means that the image is located on the same side of the lens as the object. There is only one answer choice where
this is the case, and that is answer choice B.

10.

D
In both cases the two resistors are in series. Case 1 has a 2/3-Ω resistor and a 1/3-Ω resistor in series,
while case 2 has an 8/3-Ω resistor and a 1/3-Ω resistor in series. Let's first look at case 1. The total resistance,
since they are in series, is 1 Ω, i.e. 2/3 + 1/3 and, since the potential difference in this circuit is 9 volts, the current
from I = V/R is 9 volts/1 Ω or 9 amps. It follows then that the potential drop across the 2/3-Ω resistor is 9 A x 2/3
Ω or 6 volts from V = IR. For case 2 the current in the circuit is 9 volts/3 Ω or 3 amps. The potential drop across
the 8/3-Ω resistor is IR, or 3 amps x 8/3 Ω , which is 8 volts. The change in voltage drop across that resistor is
from 6 volts to 8 volts, and that is choice D.
11.

12

B


Physics Discretes Test

This question requires you to interpret a graph. On the x-axis you are given the time in years, and on the yaxis you are given the number of undecayed nuclei remaining. You are asked to determine the half-life.
The half-life is the time it takes for one-half of a sample to decay. We start off at time zero with 10 x 105
nuclei. When one half-life elapses, one-half of the sample decays, and that implies that there is one-half of the
sample remaining. One-half of 10 x 105 is 5 x 105. At what time do we have 5 x 105 nuclei remaining? From the
graph, you see that it is 3 years, and therefore 3 years is the half-life.
If you wanted to make sure, you could check to see when the next half-life occurs. After 3 years we have 5
x 105 nuclei left. In another half-life, only half of 5 x 105 or 2.5 x 105 are left. When does that occur? Looking at the
graph, that occurs in 6 years. So another 3 years have passed. Therefore, the half-life is 3 years, answer choice B.
12.

D

Any two objects with mass attract each other with a particular force given by Newton's law of gravity
which says F=Gm1m2/R2, where G is the universal gravitational constant, m1 and m 2 are the two masses, and R is
the distance between their centers. Let's look at the case when an object on the Earth's surface is attracted to the
Earth. Under these conditions, we'll let m1 be the Earth's mass, m2 be the mass of the object on the Earth's surface,
and R be the distance between the centers of the two masses, which is equal to the Earth's radius. However, we also
know that this gravitational force on m 2 is equal to m 2g, where g is the acceleration due to gravity at the Earth's
surface.If we equate these two expressions for the force of attraction on m2 and cancel m 2 which appears in both, we
see that the acceleration due to gravity g is equal to Gm 1/R 2. Moving to another planet, the analysis would be the
same but m1 and R would be different for the new planet. In this problem the new m1 and R are one-half of the Earth
values. So the mass of the new planet equals m1/2, and the radius of the new planet equals R/2. This gives us that
the new acceleration a is
a=G

m 1/2
m 1/2
m1
=G 2
= 2G 2
2
(R/2)
R /4
R

Now Gm 1/R2 is the acceleration due to gravity on Earth so substituting in gives us that a = 2g. The
gravitational acceleration due to gravity on the Earth is 9.8 m/s 2 which would make the acceleration on the other
planet two times that or 19.6 m/s2.
13.

D
This question asks you to determine how an electron beam is deflected when current flows in the wire coils.

When current flows in the wire coils, it creates a magnetic field. This magnetic field can exert a force on the electron
beam, causing the beam to be deflected. In order to find the direction of the magnetic force, we first need to know the
direction of the magnetic field. To find the direction of the field, we put the thumb of our right hand in the direction
of the current. Our remaining fingers curl around the wire until we get to the point in question. The direction of the
field is the direction in which our fingers point.
We are told that the current travels counter-clockwise with respect to the observer. So if you imagine
yourself at O looking at the tube, the current travels from your right to your left at the top of the tube. Therefore as
seen from O, the thumb of your right hand should point towards the left across the top of the tube. Now curl your
fingers around. Inside the tube, where the electron beam is, your fingers go along the dashed line from G to P. In
terms of the electron beam, your fingers and therefore the magnetic field are going in the same direction as the
electron beam.
Now whenever the magnetic field is in the same direction or in the opposite direction to the motion of the
charged particle, the magnetic force on the particle is zero. This can be seen from the equation F=qvBsinθ. If the
magnetic field is in the same, or opposite direction, then θ is either 0° or 180°. Therefore the sine of θ is 0, so the
magnetic force is also zero. When no magnetic force is present, there is no deflection of the beam.
14.

B
We are asked to find the ratio of the velocity between two objects just before they hit the ground. Object A
has a mass of M and is released from a height of H. Object B has a mass of M/2 and dropped from a height of 2H.
We want to find the ratio of the velocity of object A to object B just before they hit the ground.
How can we relate final velocity to height? Let's use the kinematic equation: v2 = v o 2 + 2as where v is the
final velocity, vo is the initial velocity, as is the acceleration -- in this case, it is the acceleration due to gravity, g, -and s is the distance traveled. In our problem, the two objects are released from a given height. This implies that
they had no initial velocity, so v o = 0. This simplifies our equation to v2 = 2gs, or v = (2gs)1/2. Notice that this

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13

MCAT

equation says that velocity is independent of mass. That is true. Specifying the mass in the question is superfluous,
because all objects free-fall at the same rate regardless of their mass. So the velocity of object A is given by va =
. For object B, vb =

2g2h =

2gh

4gh . Now taking the ratio we find that
va
=
vb

2gh
=
4gh

1/2 = 1/ 2

15.

B
The hydraulic lever is based on Pascal's principle, which states that an applied pressure is transmitted
undiminished to all portions of a fluid. Therefore, the pressure under piston A equals the pressure under piston B.
Since pressure equals force divided by area, we can say F1/A1=F2/A2. In the question stem we are told that piston A
has a diameter of 2 meters, and that the diameter of piston B is 4 meters. But notice that the areas of the pistons do
not change. The information that the question stem gives you about the diameters of the two pistons is not needed to
answer the problem. The point is that if the areas remain constant, the force on piston A is directly proportional to

the force on piston B, so if the force on A is doubled, then the force on B must also double.
16.

A
An adiabatic expansion means that there is no heat flow in or out of the system. That fact alone
immediately eliminates choices C and D. When a gas expands, it does work on the environment. The magnitude of
the work done by the gas is expressed as PV. P is the pressure that it expands against and V is the increase in
volume. Now since under adiabatic conditions no heat flows into the system, the energy needed to do this work of
expansion must come from the internal energy of the gas itself. So the internal energy of the gas after adiabatic
expansion will have to be less than the internal energy before the expansion. From points Q to R the volume of the
gas has expanded and work has been done by the gas so that the internal energy of the gas must be greater at point Q
than it is at R.
17.

B
To answer this question we must use the continuity equation, which states that the product of the crosssectional area and the velocity is a constant. This means that we equate the product of the cross-sectional area and the
velocity at point A with that at point B, so we have that AAvA=ABvB, where A is the area, and v is the velocity.
Now, in the question stem we are told that the velocity at point A is one quarter that at point B, so vA = v B /4, or vB
= 4vA. If we let the radius at point A be rA, and the radius at point B be rB , we can write the cross-sectional area at
point A as being r A 2, and the cross sectional-area at point B as being r B 2 . Substituting into the continuity
equation, we get that rA2vA equals rB2vA. Canceling out the 's and the vA's and rearranging
the terms, we are left
with rA2/rB2 = 4. Taking the square root of this we get that rA/rB = 2. Therefore, the ratio of the radius at point A to
that at point B is 2 to 1, which is answer choice B.
18.

D
To solve this question properly you need a good understanding of Archimedes' Principle, which states that a
body wholly or partially immersed in a fluid will be buoyed up by a force equal to the weight of the fluid that it
displaces. In other words, the buoyant force, Fb, equals the weight of the displaced fluid, which is the mass of the

displaced fluid times g, the acceleration due to gravity. But we're looking for the density of the fluid, so we need to
define the buoyant force in terms of density. The equation for density is ρ = m/V, where ρ is the density, m is the
mass, and V is the volume. We can rewrite this equation to get an equation for m, and we get that m = ρV. So now
we have the buoyant force Fb = ρVg. Solving for the density of the liquid, we find ρ1 = Fb/V1g, where the subscript
1 denotes liquid.
The problem tells you to assume that g equals 10 m/s2. Now we have to calculate the buoyant force and the
volume of liquid displaced. The buoyant force is the difference between an object's weight in air and its weight in the
fluid. We're told that the cube weighs 16 newtons in air and only 10 newtons in the liquid. So, the buoyant force is
equal to the difference, or 16 – 10 = 6 N. We're told that the block is only half submerged, so, the volume of liquid
displaced V1 equals half of the volume of the cube. Since the volume of any cube is its side cubed and each side is 10
cm or 0.1 m, we find that the volume of the block is 10–3 m3. And so the volume of the water displaced is one-half
the volume of the block, or 10–3/2, or 5 x 10–4 m3.

14


Physics Discretes Test
In sum, we found the buoyant force, Fb = 6 newtons, the volume of the liquid displaced, V1 = 5 x 10 –5 m 3 ,
and the acceleration due to gravity, g = 10 m/s 2 . Now we just have to substitute into the equation, ρ1 = F b/V 1g.
Doing the math we find that the density of the liquid equals 1,200 kg/m3, which is answer choice D.
19.

D
This is an electrostatics question that requires you to know Coulomb's law, which gives the magnitude of
the electrostatic force. The force F = kQ1Q2/r2, where k is the electrostatic constant, Q1 and Q2 are the charges of the
two particles and r is the distance separating them.
Now, in this question we originally have two particles with charges of +Q and +3Q. From Coulomb's law,
the force, which we are given as F, = k(Q)(3Q)/r2. So we can say that F = 3kQ2/r2.
Now the situation changes. A charge of –2Q is added to each of the two charges. So the +Q becomes –Q,
and the +3Q becomes +Q. So now we have two charged particles at +Q and –Q. Since the question specifies that the

charges are fixed, we can assume that r has not changed. Let's plug these values into Coulomb's law. The force
equals kQ(–Q)/r2 which equals –kQ2/r2. Remember that the original force F was equal to 3kQ2/r2. So this new force
is actually one third the original force, and this eliminates answer choices A and B.
Now we have to determine if the force is an attractive one or a repulsive one. Oppositely charged particles
attract, and similarly charged particles repel. Initially we have charges of +Q and +3Q. Since they are both positive,
they repel each other. However, when we add the –2Q to each charge, one of the charges becomes +Q and the other
becomes –Q. Since they have opposite signs, the force is attractive. Therefore, answer choice D, F/3 attraction is
correct.
20.

A
The binding energy, E, equals the mass defect, m, times c2. c is the speed of light in vacuum, and has the
value 3 x 108 m/s. However, you do not need to know this, and in fact, if you try to plug this value in and do
calculations, you are more likely to make mistakes. We are given the fact in the question stem that 1 amu yields 931
MeV of energy, and so we have:
E = mc2
931 MeV = (1 amu)c2
c2 term is therefore equal to 931 MeV/amu. This may be a weird unit to report the square of a velocity, but
all we are after is some proportionality constant between E and m and as long as we keep our units consistent we
are fine.
In the problem we are given the binding energy as 186.2 MeV. hence, the mass defect, expressed in amu, is
m=

E
1 amu
= 186.2 MeV x
= 0.2 amu
c2
931 MeV

But we want the answer =in kilograms. We are also told that in 1 amu there are 1.66 x 10 –27 kilograms.
Multiplying 0.2 amu by 1.66 x 10-27 kg/amu, we find the mass defect is 0/33 x 10–27 or 3.3 x 10–28 kilograms.
21.

A
Here we have a block sliding down one quadrant of a frictionless circular track whose radius is one meter,
and we are asked which of the statements A, B, C, or D is true. Let's go through each of the answer choices in turn.
Answer choice A states that the speed of the block at the bottom will be the same as if it had fallen vertically.
Initially the block has a potential energy of mgh, where m is the mass, g is the acceleration due to gravity, and h is
the height. Now we are told that the radius of the track is 1 meter. So the initial potential energy of the block is mg.
Now this potential energy converts to kinetic energy as the block moves down the track, and since the track
is frictionless, when the block reaches the bottom all the potential energy will have been converted to kinetic energy.
Kinetic energy is given by the equation K = (1/2)mv2, and this is equal to the potential energy which we found to be
mg. So we get that (1/2)mv2 = mg, or v2 = 2g.
For an object in free fall the equationn that relates the final velocity v, to the initial velocity vo, and the
height h, is v2 = vo2 + 2ah, where a is the acceleration, which in this case is g, the acceleration due to gravity. Now
we know that the initial velocity is zero since it starts from rest, and we also know that the height is 1 meter since it
starts from the same height as the block that slides down the track. So our equation becomes v2 = 2g. This is exactly
the same as the result that we obtained from the block sliding down the track, which tells us that the final velocity
in both cases is equal. Therefore answer choice A is correct.

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MCAT

Let's go through the other answer choices and see why they are incorrect. Answer choice B states that
mechanical energy will not be conserved. This statement is false since we have already said that because the track is

frictionless there are no non-conservative forces acting on the block, and so the sum of the potential and kinetic
energy at any point is always constant. In other words, mechanical energy is conserved.
Answer choice C states that the acceleration of the block is constant throughout the descent. You may have
thought that this answer was correct since the acceleration due to gravity remains constant. However, you must
remember that the block is not in free fall but is following a circular path, and so the direction of the acceleration
vector is continually changing. In order for a vector to be constant, both the magnitude and the direction must remain
constant, so the fact that the direction changes along the circular track is sufficient to eliminate choice C.
Answer choice D states that the total work done by gravity is greater than if the block had fallen vertically.
This is untrue in this case since, as we said before, when the block moves down the track there are no nonconservative forces acting on the block, so no extra work has to be done by gravity. Thus answer choice D is
incorrect, and once again the correct answer is choice A.
22.

D
A scalar is a quantity that has a magnitude but no dierection as opposed to a vector which has both
magnitude and direction. Velocity, choice A, has both a magnitude and direction. An example of a velocity vector is
3 meters per second, south. The magnitude in this case is 3 meters per second and the direction is south.
Answer choice B suggests force. We draw force diagrams all the time. For instance, a girl pushes a box to
the left with a force of 5 newtons. Now, the magnitude of the force is 5 newtons and the direction is to the left. So
force is also a vector.
Impulse is defined as the force vector times the time. Time is a scalar, but when you multiply a scalar by a
vector you get a vector, so impulse is a a vector. Impulse is also equal to the change in momentum and changes in
momentum are vectors. So again, impulse must be a vector.
Answer choice D suggests kinetic energy.Energy is a scalar quantity. "But wait a minute," you say,
"Doesn't kinetic energy also depend on velocity?" No, kinetic energy depends on the square of the speed, and the
speed, which is the magnitude of the velocity vector, is a scalar. Kinetic energy is a scalar, and answer choice D is
correct.
23.

C
Here we have a fireboat fighting a fire to its right, but it also has its hoses throwing water to the left. We

are asked to find which of the given physical laws best explains why it does this. The answer is choice C -Newton's third law, which states that to every action there must be an equal and opposite reaction. When the fireboat
is fighting the fire, it has its hoses going to the right, and it is exerting a force to the right on that water. Now, by
Newton's third law, the water provides a reactive force that is equal in magnitude but opposite in direction towards
the left, and this causes the fireboat to be pushed towards the left. In order to prevent the boat from moving towards
the left, the fireboat must also have its hoses pushing water to the left, creating a second reactive force on it which is
to the right.
The other answer choices don't imply that the boat would have to blast water in the opposite direction to
stay in place. Newton's law of gravitation, choice A, states that there exists a force of mutual attraction between all
matter which is proportiomal to the product of the masses and inversely proportiomal to the square of the distance
separating them. The constant of proportionality is G, the universal gravitational constant. Newton's second law,
choice B, states that a force exerted on an object equals the mass of the object times the acceleration. And finally
conservation of energy, choice D, states that in the absence of non-conservative forces like friction or air resistance,
the total energy of a system must remain constant. None of the other answer choices directly affects the situation.
24.

C
This question tests how well you understand energy exchanges in a mass-spring system. There are two
types of energy: energy of motion, or kinetic energy, and energy of position, or potential energy. Kinetic energy in
any system, including a mass-spring system, is defined as mv2/2. Potential energy does not have such a universal
formula that applies for every case; the potential energy function for gravity is different from the potential energy for
the mass-spring system which is different from the electrostatic potential energy, etc. For a mass-spring system,
potential energy is one-half times the spring constant times the distance from the equilibrium point squared:
(1/2)kx2. The other important piece of information you should know about energy in a mass-spring system is that,
provided there are no frictional forces acting, energy is conserved. This means that the total mechanical energy,
which is the sum of the kinetic and potential energies, stays constant.
Here we have a spring traveling from the point of maximum extension to the point of maximum
compression . Let's consider the energy at maximum extension. The words maximum expansion should tell you that

16

Physics Discretes Test

the spring will not extend any more. At this point the speed should be zero. If it were not, then the spring would
continue to expand. If the speed is zero at this point, the kinetic energy must also be zero. Now we know from
conservation of energy that the total mechanical energy is a constant. So when the kinetic energy equals, zero, the
potential energy must be a maximum. If potential energy is a maximum at this point, it cannot increase. This
eliminates answer choice B.
Now we are told that the spring moves from this position through the equilibrium point. At the
equilibrium point there is no expansion or compression. Here, the speed of the mass is a maximum, and therefore
the kinetic energy is a maximum. Now from conservation of energy if kinetic energy is a maximum, potential
energy must be a minimum. So we can eliminate answer choice D since the potential energy does change.
Now the mass moves to the point of maximum compression. The story at this point is much the same as
at the point of maximum expansion. At the point of maximum compression the speed is zero, otherwise the spring
would continue to compress. If the speed is zero, the kinetic energy must be also. By conservation of energy when
the kinetic energy is a minimum, potential energy must be a maximum. So the potential energy has gone from its
maximum value at maximum extension, to its minimum value at the equilibrium point, to its maximum value at
maximum compression. In other words, the potential energy has decreased and then increased. Therefore, answer
choice C is the correct answer.
If you know that the potential energy was given by the formula U = kx2/2, where U is the potential energy,
k is the spring constant, and x is the distance from the equilibrium point, you could have also reached the same
conclusion. At maximum compression or expansion, x is a maximum, and therefore the potential energy is a
maximum. At the equilibrium point, x equals zero so the potential energy is zero. Therefore, the potential energy
goes from maximum, to zero to maximum. This means that it decreases and then increases.
25.

A
An electric field line shows how a positively charged test particle would be accelerated in an electric field.
Keep in mind that a positive test charge will be attracted to a negative particle and will be repelled from a positive
particle. We know that because opposite charges attract and like charges repel in electrostatics. So we are looking for

an answer choice in which the lines point away from the positive charge and toward the negative charge.
Now in answer choice A, we see field lines directed away from the positive charge and toward the negative
charge. Answer choice B shows the opposite situation. So it is wrong. Answer choices C and D are the kind of field
patterns you would expect from two similarly charged particles. They repel each other. Therefore, the lines don't
meet and, in fact, they diverge from each other. If you didn't recognize the pattern, you could have eliminated answer
choice C because it depicts field lines going away from, not toward, a negative particle. Similar reasoning could be
used for answer choice D -- for a positive charge the field lines point away from, not toward, the charge.
26.

A
We are asked to find the electrostatic potential at a specific point given a distribution of charged spheres. We
have four spheres charged to +1 coulomb and another four spheres charged to –1 coulomb. These spheres are arranged
in a circle of radius 1 meter.
What is the electrostatic potential? The electrostatic potential is defined as the work per unit charge required
to move a test charge from infinity to a particular point. Mathematically, we can say V = W/q, where V is the
potential, W is the work, and q is the charge of the test charge. Now, work is defined as the force times the distance.
So we can also say that V = Fr/q. Now the force is the electrostatic force of Coulomb's law: F = kq1q2/r2. Plugging
that into the formula V = Fr/q, we find that one of the r's cancel and one of the q's cancel. We are then left with V =
kq/r.
Let's calculate the potential for one negatively charged sphere. The charge q equals –1 coulomb, and the
distance from the charge to the point in question, which is the center of the circle, is 1 meter. From the equation V =
kq/r, the potential due to each negatively charged sphere is –k. Now to calculate the potential caused by each
positively charged sphere, we use q = +1 coulomb, and r = 1 meter. We find that each positively charged sphere
produces a potential of +k.
Now since electrostatic potential is scalar, we can add the effect from each charge directly. In other words,
there is no directional component to worry about. So, Vtotal = 4(+k) + 4(–k) which equals 4k – 4k = 0.
27.

D
In this Roman numeral question we are being asked which of the given quantities -- torque, work, and

energy -- can be expressed dimensionally as ML2/T2. The first thing to note is that work and energy have the same
units. So you can immediately eliminate any answer choice that includes II but not III or that includes III but not II.
This enables us to eliminate choice B. You can also consider the remaining choices strategically. If the formula is

KAPLAN

17


MCAT

applicable to work and energy, then the answer is either choice C or choice D; and if it's wrong for work and energy,
then the answer is choice A.
So let's look at the units of work. Work is force times distance; let's see how those units compare to the
formula ML2/T2. Distance has units of length, L. What about force? From Newton's second law, force is mass
which is M and acceleration, which has units of meters over seconds-squared or L/T2. So force is ML/T2, and force
times distance is ML2/T2. So work and therefore energy can be expressed dimensionally as ML2/T2.
Now, torque is also force times distance. Even though torque has a physical meaning different from work or
energy, it has the same units of ML 2/T2. Therefore all three Roman numeral statements are correct, which means
that the correct answer is choice D.
28.

C
In this medical application of physics we want to find the duration of a shock given to restore natural
rhythm to a heart attack victim. We are told that the shock plates are run off a 10 volt battery with a current of 30
amps, and the shock draws 300 joules of energy.
We want to relate work and time here, so we can use the equation P = W/t, where P is the power, W is the
work or energy, and t is the time. We don't directly have a value for power, but we can figure it out. In any electrical
system the power is the product of the current, i, and the voltage, V. In other words, P = iV. We have values for
both of those quantities, so we can equate our two expressions for power, and find that W/t = iV. We want to solve

this for the time, t, so we find that t = W/iV. Now it's just a case of plugging in the numbers. We are told that the
work or energy drawn is 300 joules, the current is 30 amps, and the voltage is 10 volts, so the time that the shock
lasts is 300/(10 x 30), or 300/300 -- and that's just 1 second, answer choice C.
29.

A
The bat emits sound that is reflected off a stationary obstacle then travels back to the bat. We want to know
about the frequency heard by the bat relative to the frequency that it emitted.
First the bat emits the sound, then the sound bounces off a stationary object, and then the sound returns to
the bat. The Doppler effect says that when an observer and a source are moving with respect to each other, the
frequency perceived by the detector will be different than the frequency emitted, and when they are approaching each
other the perceived frequency will be higher. There is a formula associated with this that allows us to calculate the
actual frequency perceived by the detector, but we can get the answer without having to use the formula. Since the
bat and the stationary object are approaching each other, the bat intercepts more wavelengths per second than it
would intercept if it were stationary. Since frequency is the number of wavelengths per second, more wavelengths
mean a higher frequency, answer choice A.
30.

D
To find the image distance we use the formula 1/o + 1/i = 1/f, where o is the object distance, i is the image
distance, and f is the focal length. We're interested in the image distance, so we can rewrite the equation to get 1/i =
1/f – 1/o. Since it's a converging lens, f is positive; and since it's a single-lens system, o will also be positive.
Now looking at the answer choices you might notice that the answers for the image distance when the
object is at f are all different. So we can just find that distance; that will be enough to locate the correct answer. So
when the object distance is f, our equation for i becomes 1/i = 1/f – 1/f = 0. To find i we must take the inverse. 1/0
is infinity, so answer choice D is correct.
For the sake of completeness, let's figure out what the image distance is when the object is at a distance of
2f. Our equation becomes 1/i = 1/f – 1/(2f), which equals 1/(2f). And taking the reciprocal we find that i = 2f. So, as
an object is moved from 2f to f, the image formed will move from 2f to infinity, answer choice D.

18