Kaplan MCAT Physics - Discretes Test

Physics DiscretesTest

Time: 30 MinutesNumber of Questions: 30

This test consists of 30 discretequestionsquestions that are NOT based on adescriptive passage. These discretes comprise15 of the 77 questions on the PhysicalSciences and Biological Sciences sections ofthe MCAT.

MCAT

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PHYSICS DISCRETES TEST

DIRECTIONS: The following questions are notbased on a descriptive passage; you must selectthe best answer to these questions. If you areunsure of the best answer, eliminate the choicesthat you know are incorrect, then select ananswer from the choices that remain. Indicateyour selection by blackening the correspondingcircle on your answer sheet. A periodic table isprovided below for your use with the questions.

PERIODIC TABLE OF THE ELEMENTS

1H1.0

2He4.0

3Li6.9

4Be9.0

5B

10.8

6C

12.0

7N

14.0

8O

16.0

9F

19.0

10Ne20.2

11Na23.0

12Mg24.3

13Al

27.0

14Si

28.1

15P

31.0

16S

32.1

17Cl

35.5

18Ar39.9

19K

39.1

20Ca40.1

21Sc

45.0

22Ti

47.9

23V

50.9

24Cr52.0

25Mn54.9

26Fe

55.8

27Co58.9

28Ni

58.7

29Cu63.5

30Zn

65.4

31Ga69.7

32Ge72.6

33As

74.9

34Se

79.0

35Br

79.9

36Kr83.8

37Rb85.5

38Sr

87.6

39Y

88.9

40Zr

91.2

41Nb92.9

42Mo95.9

43Tc

(98)

44Ru

101.1

45Rh

102.9

46Pd

106.4

47Ag

107.9

48Cd

112.4

49In

114.8

50Sn

118.7

51Sb

121.8

52Te

127.6

53I

126.9

54Xe

131.355Cs

132.9

56Ba

137.3

57La *138.9

72Hf

178.5

73Ta

180.9

74W

183.9

75Re

186.2

76Os

190.2

77Ir

192.2

78Pt

195.1

79Au

197.0

80Hg

200.6

81Tl

204.4

82Pb

207.2

83Bi

209.0

84Po

(209)

85At

(210)

86Rn

(222)87Fr

(223)

88Ra

226.0

89Ac 227.0

104Unq(261)

105Unp(262)

106Unh(263)

107Uns

(262)

108Uno(265)

109Une(267)

*

58Ce

140.1

59Pr

140.9

60Nd

144.2

61Pm

(145)

62Sm

150.4

63Eu

152.0

64Gd

157.3

65Tb

158.9

66Dy

162.5

67Ho

164.9

68Er

167.3

69Tm

168.9

70Yb

173.0

71Lu

175.0

90Th

232.0

91Pa

(231)

92U

238.0

93Np

(237)

94Pu

(244)

95Am

(243)

96Cm

(247)

97Bk

(247)

98Cf

(251)

99Es

(252)

100Fm

(257)

101Md

(258)

102No

(259)

103Lr

(260)

Physics Discretes Test


KAPLAN 3

1 . An ideal gas compressed adiabaticallyexperiences:

A . a decrease in temperature and an increase ininternal energy.

B . a decrease in temperature and a decrease ininternal energy.

C . an increase in temperature and an increase ininternal energy.

D . no change in either the temperature orinternal energy.

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

A . increases.B . decreases.C . remains the same.D . cannot be determined.

3 A 1.0 kg object moving east with a velocity of10 m/s has a head-on collision with a 0.5 kgobject which is at rest. Neglecting friction, whatis the momentum of the system after collision?

A . 15 kg m/s; eastB . 15 kg m/s; westC . 10 kg m/s; eastD . 10 kg m/s; west

4 . An object is moving counterclockwise in a circleat constant speed. Which of the followingdiagrams correctly indicates both its velocityvector and its acceleration vector?

v

a

A.

var

B.

v

aC.

v

a

D.at

5 . An ambulance originally at rest has its sirengoing. The ambulance starts accelerating awayfrom a boy standing nearby. He will hear:

A . a decrease in the pitch of the siren.B . no change in the pitch of the siren.C . an increase in the pitch of the siren.D . an increase, and then a decrease in the pitch

of the siren.

6 . The diagram shows energy levels E1, E2, E3, andE4 in an atom. If the transition from energy E3to E2 gives rise to a photon of radiation of aparticular wavelength, which of the followingtransitions could give rise to a photon ofradiation of longer wavelength?

Energy (eV)E 4E 3

E 2

E 1

A . E4 to E1B . E3 to E1C . E2 to ElD . E4 to E3

7 . The activity of a radioactive source falls to one-sixteenth of its original value in 32 minutes.What is the half-life, in minutes, of this decayprocess?

A . 2 minB . 4 minC . 8 minD . 16 min

8 . A parallel plate air capacitor is connected across abattery of constant emf. If the separation of theplates is decreased, which of the followingincreases?

I. Charge on the plates of the capacitorII. Potential difference between the plates

of the capacitorIII. Capacitance of the capacitor

A . III onlyB . I and II onlyC . I and III onlyD . II and III only

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

A.

B.

C.

D.

f f'

O

f f'

O

f

f'O

f f'

O

II

I

I

1 0 . When the 2/3 resistor is replaced by an 8/3 resistor, the voltage across the resistor changesfrom:

9 V

2/3

1/3

A . 4 V to 6 V.B . 4 V to 8 V.C . 6 V to 4 V.D . 6 V to 8 V.



KAPLAN 5

1 1 . From the data shown below, estimate the half-life of element X.

10 x 105

7.5 x 105

5 x 105

2.5 x 105

0 1.5 3.0 4.5 6.0

Element X

Num

ber o

f Und

ecay

ed N

ucle

ii

Years elapsed

A . 1.5 yearsB . 3.0 yearsC . 4.5 yearsD . 6.0 years

1 2 . A planet has a diameter one half of the Earthsdiameter and a mass that is one half of theEarths mass. What would be the acceleration dueto gravity on the planet?

A . 2.45 m/s2B . 4.90 m/s2C . 9.80 m/s2D . 19.6 m/s2

1 3 . In the diagram below a stream of electrons leavesthe electron gun at G and strikes the fluorescentscreen at P. When the current is switched on atS, it flows through the wire coils in ananticlockwise direction as seen by the observer atO; the observer sees the spot of light at P:

G P

S

O

A . move downwards.B . move to the left.C . move to the right.D . remain still.

1 4 . Object A of mass M is released from height Hwhile object B of mass 0.5 M is released fromheight 2H. What is the ratio of the velocity ofobject A to the velocity of object B immediatelybefore they hit the ground? (Note: Assume thatair resistance is negligible.)

A . 1:1B . 1: 2 C . 1:2D . 1:4

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

AB

A . decreased by a factor of 4.B . increased by a factor of 2.C . increased by a factor of 4.D . increased by a factor of 8.

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

P

VR

S

TQ

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.

1 7 . A fluid flows through a pipe such that thevelocity at point A is 1/4 that at point B. Whatis the ratio of the radius of the pipe at point A tothat at point B?

A . 1:4B . 2:1C . 4:1D . 1:2

1 8 . A cube of wood whose sides are each 10 cmweighs 16 N in air. When half submerged in anunknown liquid it weighs only 10 N. What is thedensity of the liquid? (Note: Assume g = 10m/s2.)A . 600 kg/m3B . 800 kg/m3C . 1,000 kg/m3D . 1,200 kg/m3

1 9 . Two fixed charges of +Q and +3Q repel eachother with a force, F. If an additional charge of2Q is now added to each of the two charges,what is the new force between them?

A . ZeroB . F/6 attractionC . F/3 repulsionD . F/3 attraction

2 0 . If the binding energy of a nucleus is 186.2 MeV,what is its mass defect? (Note: 1 amu yields 931MeV of energy, 1 amu is equivalent to 1.66 1027 kg.)A . 3.3 1028 kgB . 6.6 1028 kgC . 8.3 1027 kgD . 3.1 1025 kg



KAPLAN 7

2 1 . A block starts from rest and slides down alongone quadrant of a frictionless circular track whoseradius is one meter. Which of the following istrue?

1 m

A . The speed of the block at the bottom will bethe 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.

2 2 . Which one of the following is a scalar quantity?

A . VelocityB . ForceC . ImpulseD . Kinetic energy

2 3 . A fireboat is fighting a fire to its right but alsohas its hoses going full blast into the water to itsleft. Which law of physics best explains thereason for this?

A . Newtons law of gravityB . Newtons second lawC . Newtons third hewD . Conservation of energy

2 4 . In a frictionless horizontal mass spring system,as the mass moves from the point of maximumexpansion through the equilibrium position tothe point of maximum compression, thepotential energy of the system:

A . decreases continually .B . increases then decreases.C . decreases then increases.D . remains the same.

2 5 . Which of the following diagrams accuratelydepicts the electric field lines between 2 chargesof equal magnitude but opposite sign?

A. C.

B. D.

MCAT

2 6 . Four positively charged spheres (+1 C each) andfour negatively charged spheres (1 C each) aredistributed at equally spaced intervals around acircular hoop whose radius is 1 m as shownbelow. What is the electrostatic potential at thecenter of the hoop? (Note: k is the electrostaticconstant.)

+

+

+

+

A . 0 VB . 4k VC . 8k VD . 16k V

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

I. TorqueII. Work

III. Energy

A . I onlyB . I and II onlyC . II and III onlyD . I, II, and III

2 8 . In an attempt to restore the natural rhythm of aheart attack victim, electric shock plates aresometimes used. The shock plates are run off a10 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 . 0.1 secB . 0.3 secC . 1.0 secD . 3.0 sec

2 9 . A bat locates obstacles in its flight path byemitting sound in the same direction as it travels.When the sound encounters an obstacle, it isreflected back to the bat. Assuming that the batsvelocity and the frequency of the emitted soundare constant, and assuming the obstacle isstationary, which of the following describes thefrequency observed by the bat?

A . It is greater than the frequency emitted by thebat.

B . It is less than the frequency emitted by thebat.

C . It is the same as the frequency emitted by thebat.

D . Its character cannot be determined.

3 0 . An object is moved from 2f towards f, where f isthe focal length of this converging lens. Theimage formed on the other side of the lens willmove from:

2 f f f 2 f

A . f to 2f.B . 2f to f/2.C . 2f to f.D . 2f to infinity.

END OF TEST


KAPLAN 9

THE ANSWER KEY IS ON THE NEXT PAGE

MCAT

10

ANSWER KEY:1. C 11. B 21. A2. A 12. D 22. D3. C 13. D 23. C4. A 14. B 24. C5. A 15. B 25. A

6. D 16. A 26. A7. C 17. B 27. D8. C 18. D 28. C9. B 19. D 29. A

10. C 20. A 30. D


KAPLAN 11

EXPLANATIONS

1 . CThis 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 thatQ 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 thereforenegative. Another way we could have reached this conclusion from the formula W = PV for an ideal gas. Since inthis 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 cantherefore be eliminated. Now all that is left is to decide whether an increase in internal energy would increase ordecrease the temperature. Since temperature increases with the average kinetic energy of the gas molecules, as theinternal energy increases, so would the temperature.

2 . AIn 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) toredevelop 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 ishowever not an isolated system (energy is flowing out of it into the cold rod) and therefore it doesnt violate what wehave just said. The increase in entropy of the cold hot upon warming is larger in magnitude than the decrease inentropy of the hot rod as it cools, thus leading to a net increase in entropy.

3 . CThe 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 themomentum immediately before a collision equals the momentum immediately after the collision. Before thecollision, the 1.0-kg object has a momentum of 1.0 kg 10 m/s = 10 kgm/s. This is the total initial momentumsince the other object is at rest (v = 0, hence p = mv = 0). This initial momentum of the system points in the samedirection as the velocity of the moving object, i.e. towards the east. This must then also be the momentum of thesystem after collision.

4 . AThe 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 beeliminated. For uniform circular motion, there is no tangential component to the acceleration. The acceleration istherefore entirely radial and is called the centripetal acceleration, with magnitude equal to v2/r. It points towards thecenter of the circle.

5 . AThis 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 + vs)

where f is the observed frequency, f is the actual frequency (from the point of view of the source), v the velocity ofsound in the medium, and vs the velocity of the source. We do not need to remember this equation specificallythough 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 lowerfrequency or pitch. The faster the source is moving away, the lower the perceived pitch would be. Here theambulance is accelerating from the boy; its speed is therefore increasing. The pitch would then be decreasing.

6 . D

MCAT

12

We are asked for a transition that would give rise to a photon of a longer wavelength. When a particle (forexample an electron) relaxes from a higher energy state to a lower energy state, the excess energy is released in theform of a photon. The energy of a photon is directly proportional to its frequency via E = hf, but is inverselyproportional to the wavelength: E = hc/. We are therefore looking for the transition that corresponds to a smallerenergy difference between the initial and final states than between E3 and E2. The energy difference between levels iseasily read off from the figure as the vertical distance between the energy levels involved. Out of all the choices, onlychoice D, a transition from level 4 to level 3, represents a transition that releases less energy than the transition fromlevel 3 to level 2.

7 . CWhen 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 passedthere is one sixteenth of the sample left. This means that in 32 minutes 4 half-lives have passed. This corresponds to8 minutes per half-life.

8 . CThe capacitance of a parallel-plate capacitor is approximately given by:

C = 0Ad

where is the dielectric constant of the medium between the plates, 0 is the permittivity of free space, A is the areaof overlap between the two plates, and d is the separation of the two plates. From this equation we can see that if theseparation of the plate decreases, the capacitance will increase.

A more general relationship involving capacitance, which applies to capacitors that are not necessarily ofthe parallel-plate variety, is C = Q/V, where Q is the charge stored on one of the pair of conductors making up thecapacitor, and V the voltage across the two. We are told in the question stem that the capacitor is connected across abattery of constant emf. Applying Kirchhoffs law, this means that the voltage across the plates will have toconstant as wellin fact, its magnitude will be the same as the emf of the battery. Since C increases while V has toremain constant, Q, the charge, will have to increase as well.

9 . BThis 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 thelens, erect and enlarged. The converging lens is then a magnifying lens. If you didn't know this, you could haveselected some sensible numbers and plugged them into the equation 1f =

1i +

1o . If we assume that the focal

length is 3 centimeters and the object distance is 2 centimeters, we can calculate an approximate value for the imagedistance. Rearranging the equation and putting in the numbers, we get:

1i =

13

12 =

16

The image distance is then 6 centimeters. Since the value is negative, the image is virtual. For a lens thismeans that the image is located on the same side of the lens as the object. There is only one answer choice wherethis is the case, and that is answer choice B.

1 0 . DIn both cases the two resistors are in series. Case 1 has a 2/3- resistor and a 1/3- resistor in series,

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

1 1 . B


KAPLAN 13

This question requires you to interpret a graph. On the x-axis you are given the time in years, and on the y-axis 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 105nuclei. When one half-life elapses, one-half of the sample decays, and that implies that there is one-half of thesample remaining. One-half of 10 x 105 is 5 x 105. At what time do we have 5 x 105 nuclei remaining? From thegraph, 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 5x 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 thegraph, that occurs in 6 years. So another 3 years have passed. Therefore, the half-life is 3 years, answer choice B.

1 2 . DAny 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 m2 are the two masses, and R isthe distance between their centers. Let's look at the case when an object on the Earth's surface is attracted to theEarth. 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 alsoknow that this gravitational force on m2 is equal to m2g, where g is the acceleration due to gravity at the Earth'ssurface.If we equate these two expressions for the force of attraction on m2 and cancel m2 which appears in both, wesee that the acceleration due to gravity g is equal to Gm1/R2. Moving to another planet, the analysis would be thesame but m1 and R would be different for the new planet. In this problem the new m1 and R are one-half of the Earthvalues. So the mass of the new planet equals m1/2, and the radius of the new planet equals R/2. This gives us thatthe new acceleration a is

a = G m1/2(R/2)2 = G

m1/2R2/4 = 2G

m1

R2

Now Gm1/R2 is the acceleration due to gravity on Earth so substituting in gives us that a = 2g. Thegravitational acceleration due to gravity on the Earth is 9.8 m/s2 which would make the acceleration on the otherplanet two times that or 19.6 m/s2.

1 3 . DThis 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 electronbeam, causing the beam to be deflected. In order to find the direction of the magnetic force, we first need to know thedirection of the magnetic field. To find the direction of the field, we put the thumb of our right hand in the directionof the current. Our remaining fingers curl around the wire until we get to the point in question. The direction of thefield 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 imagineyourself at O looking at the tube, the current travels from your right to your left at the top of the tube. Therefore asseen from O, the thumb of your right hand should point towards the left across the top of the tube. Now curl yourfingers around. Inside the tube, where the electron beam is, your fingers go along the dashed line from G to P. Interms of the electron beam, your fingers and therefore the magnetic field are going in the same direction as theelectron beam.

Now whenever the magnetic field is in the same direction or in the opposite direction to the motion of thecharged particle, the magnetic force on the particle is zero. This can be seen from the equation F=qvBsin. If themagnetic field is in the same, or opposite direction, then is either 0 or 180. Therefore the sine of is 0, so themagnetic force is also zero. When no magnetic force is present, there is no deflection of the beam.

1 4 . BWe 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 = vo2 + 2as where v is thefinal 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 thatthey had no initial velocity, so vo = 0. This simplifies our equation to v2 = 2gs, or v = (2gs)1/2. Notice that this

MCAT

14

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 = 2gh . For object B, vb = 2g2h = 4gh . Now taking the ratio we find that

va

vb =

2gh4gh = 1/2 = 1/ 2

1 5 . BThe 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 Ahas a diameter of 2 meters, and that the diameter of piston B is 4 meters. But notice that the areas of the pistons donot change. The information that the question stem gives you about the diameters of the two pistons is not needed toanswer the problem. The point is that if the areas remain constant, the force on piston A is directly proportional tothe force on piston B, so if the force on A is doubled, then the force on B must also double.

1 6 . AAn 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 ofthe work done by the gas is expressed as PV. P is the pressure that it expands against and V is the increase involume. Now since under adiabatic conditions no heat flows into the system, the energy needed to do this work ofexpansion must come from the internal energy of the gas itself. So the internal energy of the gas after adiabaticexpansion will have to be less than the internal energy before the expansion. From points Q to R the volume of thegas has expanded and work has been done by the gas so that the internal energy of the gas must be greater at point Qthan it is at R.

1 7 . BTo answer this question we must use the continuity equation, which states that the product of the cross-

sectional area and the velocity is a constant. This means that we equate the product of the cross-sectional area and thevelocity 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 = vB/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 atpoint A as being rA2, and the cross sectional-area at point B as being rB2. Substituting into the continuityequation, we get that rA2vA equals rB2vA. Canceling out the 's and the vA's and rearranging the terms, we are leftwith rA2/rB2 = 4. Taking the square root of this we get that rA/rB = 2. Therefore, the ratio of the radius at point A tothat at point B is 2 to 1, which is answer choice B.

1 8 . DTo 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 itdisplaces. In other words, the buoyant force, Fb, equals the weight of the displaced fluid, which is the mass of thedisplaced fluid times g, the acceleration due to gravity. But we're looking for the density of the fluid, so we need todefine the buoyant force in terms of density. The equation for density is = m/V, where is the density, m is themass, and V is the volume. We can rewrite this equation to get an equation for m, and we get that m = V. So nowwe have the buoyant force Fb = Vg. Solving for the density of the liquid, we find 1 = Fb/V1g, where the subscript1 denotes liquid.

The problem tells you to assume that g equals 10 m/s2. Now we have to calculate the buoyant force and thevolume of liquid displaced. The buoyant force is the difference between an object's weight in air and its weight in thefluid. We're told that the cube weighs 16 newtons in air and only 10 newtons in the liquid. So, the buoyant force isequal to the difference, or 16 10 = 6 N. We're told that the block is only half submerged, so, the volume of liquiddisplaced V1 equals half of the volume of the cube. Since the volume of any cube is its side cubed and each side is 10cm or 0.1 m, we find that the volume of the block is 103 m3. And so the volume of the water displaced is one-halfthe volume of the block, or 103/2, or 5 x 104 m3.


KAPLAN 15

In sum, we found the buoyant force, Fb = 6 newtons, the volume of the liquid displaced, V1 = 5 x 105 m3,and the acceleration due to gravity, g = 10 m/s2. Now we just have to substitute into the equation, 1 = Fb/V1g.Doing the math we find that the density of the liquid equals 1,200 kg/m3, which is answer choice D.

1 9 . DThis 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 thetwo 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 thecharges are fixed, we can assume that r has not changed. Let's plug these values into Coulomb's law. The forceequals kQ(Q)/r2 which equals kQ2/r2. Remember that the original force F was equal to 3kQ2/r2. So this new forceis 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 particlesattract, 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 otherbecomes Q. Since they have opposite signs, the force is attractive. Therefore, answer choice D, F/3 attraction iscorrect.

2 0 . AThe 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 docalculations, you are more likely to make mistakes. We are given the fact in the question stem that 1 amu yields 931MeV of energy, and so we have:

E = mc2931 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, butall we are after is some proportionality constant between E and m and as long as we keep our units consistent weare fine.

In the problem we are given the binding energy as 186.2 MeV. hence, the mass defect, expressed in amu, is

m = Ec2

= 186.2 MeV x 1 amu931 MeV = 0.2 amu

But we want the answer =in kilograms. We are also told that in 1 amu there are 1.66 x 1027 kilograms.Multiplying 0.2 amu by 1.66 x 10-27 kg/amu, we find the mass defect is 0/33 x 1027 or 3.3 x 1028 kilograms.

2 1 . AHere 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 isthe 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 trackis 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 bemg. 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 theheight h, is v2 = vo2 + 2ah, where a is the acceleration, which in this case is g, the acceleration due to gravity. Nowwe know that the initial velocity is zero since it starts from rest, and we also know that the height is 1 meter since itstarts from the same height as the block that slides down the track. So our equation becomes v2 = 2g. This is exactlythe same as the result that we obtained from the block sliding down the track, which tells us that the final velocityin both cases is equal. Therefore answer choice A is correct.

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Let's go through the other answer choices and see why they are incorrect. Answer choice B states thatmechanical energy will not be conserved. This statement is false since we have already said that because the track isfrictionless there are no non-conservative forces acting on the block, and so the sum of the potential and kineticenergy 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 havethought that this answer was correct since the acceleration due to gravity remains constant. However, you mustremember that the block is not in free fall but is following a circular path, and so the direction of the accelerationvector is continually changing. In order for a vector to be constant, both the magnitude and the direction must remainconstant, 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 non-conservative forces acting on the block, so no extra work has to be done by gravity. Thus answer choice D isincorrect, and once again the correct answer is choice A.

2 2 . DA 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 is3 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 tothe left with a force of 5 newtons. Now, the magnitude of the force is 5 newtons and the direction is to the left. Soforce 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 avector you get a vector, so impulse is a a vector. Impulse is also equal to the change in momentum and changes inmomentum 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 thespeed, which is the magnitude of the velocity vector, is a scalar. Kinetic energy is a scalar, and answer choice D iscorrect.

2 3 . CHere 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 fireboatis fighting the fire, it has its hoses going to the right, and it is exerting a force to the right on that water. Now, byNewton's third law, the water provides a reactive force that is equal in magnitude but opposite in direction towardsthe left, and this causes the fireboat to be pushed towards the left. In order to prevent the boat from moving towardsthe left, the fireboat must also have its hoses pushing water to the left, creating a second reactive force on it which isto the right.

The other answer choices don't imply that the boat would have to blast water in the opposite direction tostay in place. Newton's law of gravitation, choice A, states that there exists a force of mutual attraction between allmatter which is proportiomal to the product of the masses and inversely proportiomal to the square of the distanceseparating 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 finallyconservation 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.

2 4 . CThis 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 inany system, including a mass-spring system, is defined as mv2/2. Potential energy does not have such a universalformula that applies for every case; the potential energy function for gravity is different from the potential energy forthe 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 maximumcompression . Let's consider the energy at maximum extension. The words maximum expansion should tell you that


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the spring will not extend any more. At this point the speed should be zero. If it were not, then the spring wouldcontinue to expand. If the speed is zero at this point, the kinetic energy must also be zero. Now we know fromconservation of energy that the total mechanical energy is a constant. So when the kinetic energy equals, zero, thepotential energy must be a maximum. If potential energy is a maximum at this point, it cannot increase. Thiseliminates answer choice B.

Now we are told that the spring moves from this position through the equilibrium point. At theequilibrium point there is no expansion or compression. Here, the speed of the mass is a maximum, and thereforethe kinetic energy is a maximum. Now from conservation of energy if kinetic energy is a maximum, potentialenergy 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 asat the point of maximum expansion. At the point of maximum compression the speed is zero, otherwise the springwould continue to compress. If the speed is zero, the kinetic energy must be also. By conservation of energy whenthe kinetic energy is a minimum, potential energy must be a maximum. So the potential energy has gone from itsmaximum value at maximum extension, to its minimum value at the equilibrium point, to its maximum value atmaximum compression. In other words, the potential energy has decreased and then increased. Therefore, answerchoice 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 sameconclusion. At maximum compression or expansion, x is a maximum, and therefore the potential energy is amaximum. At the equilibrium point, x equals zero so the potential energy is zero. Therefore, the potential energygoes from maximum, to zero to maximum. This means that it decreases and then increases.

2 5 . AAn 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 positiveparticle. We know that because opposite charges attract and like charges repel in electrostatics. So we are looking foran 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 negativecharge. Answer choice B shows the opposite situation. So it is wrong. Answer choices C and D are the kind of fieldpatterns you would expect from two similarly charged particles. They repel each other. Therefore, the lines don'tmeet and, in fact, they diverge from each other. If you didn't recognize the pattern, you could have eliminated answerchoice C because it depicts field lines going away from, not toward, a negative particle. Similar reasoning could beused for answer choice D -- for a positive charge the field lines point away from, not toward, the charge.

2 6 . AWe 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 arrangedin a circle of radius 1 meter.

What is the electrostatic potential? The electrostatic potential is defined as the work per unit charge requiredto move a test charge from infinity to a particular point. Mathematically, we can say V = W/q, where V is thepotential, 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. Pluggingthat 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 thedistance 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 eachpositively charged sphere, we use q = +1 coulomb, and r = 1 meter. We find that each positively charged sphereproduces 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.

2 7 . DIn 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 sameunits. 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

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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 theformula ML2/T2. Distance has units of length, L. What about force? From Newton's second law, force is masswhich is M and acceleration, which has units of meters over seconds-squared or L/T2. So force is ML/T2, and forcetimes 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 orenergy, it has the same units of ML2/T2. Therefore all three Roman numeral statements are correct, which meansthat the correct answer is choice D.

2 8 . CIn 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 30amps, 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 thework or energy, and t is the time. We don't directly have a value for power, but we can figure it out. In any electricalsystem the power is the product of the current, i, and the voltage, V. In other words, P = iV. We have values forboth of those quantities, so we can equate our two expressions for power, and find that W/t = iV. We want to solvethis 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 thework or energy drawn is 300 joules, the current is 30 amps, and the voltage is 10 volts, so the time that the shocklasts is 300/(10 x 30), or 300/300 -- and that's just 1 second, answer choice C.2 9 . A

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

3 0 . DTo 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 theobject is at f are all different. So we can just find that distance; that will be enough to locate the correct answer. Sowhen 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. 1/0is 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 of2f. Our equation becomes 1/i = 1/f 1/(2f), which equals 1/(2f). And taking the reciprocal we find that i = 2f. So, asan object is moved from 2f to f, the image formed will move from 2f to infinity, answer choice D.

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Kaplan MCAT Physics - Discretes Test