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Post by Professor Dan Fullerton on March 27, 2015

Correct link:

1998 AP Practice Exam: Multiple Choice Questions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • 1998 AP Practice Exam Link 0:11
  • Multiple Choice 36 0:36
  • Multiple Choice 37 2:07
  • Multiple Choice 38 2:53
  • Multiple Choice 39 3:32
  • Multiple Choice 40 4:37
  • Multiple Choice 41 4:43
  • Multiple Choice 42 5:22
  • Multiple Choice 43 6:00
  • Multiple Choice 44 8:09
  • Multiple Choice 45 8:27
  • Multiple Choice 46 9:03
  • Multiple Choice 47 9:30
  • Multiple Choice 48 10:19
  • Multiple Choice 49 10:47
  • Multiple Choice 50 12:25
  • Multiple Choice 51 13:10
  • Multiple Choice 52 15:06
  • Multiple Choice 53 16:01
  • Multiple Choice 54 16:44
  • Multiple Choice 55 17:10
  • Multiple Choice 56 19:08
  • Multiple Choice 57 20:39
  • Multiple Choice 58 22:24
  • Multiple Choice 59 22:52
  • Multiple Choice 60 23:34
  • Multiple Choice 61 24:09
  • Multiple Choice 62 24:40
  • Multiple Choice 63 25:06
  • Multiple Choice 64 26:07
  • Multiple Choice 65 27:26
  • Multiple Choice 66 28:32
  • Multiple Choice 67 29:14
  • Multiple Choice 68 29:41
  • Multiple Choice 69 31:23
  • Multiple Choice 70 31:49

Transcription: 1998 AP Practice Exam: Multiple Choice Questions

Hello, everyone, and welcome back to

In this lesson, we are going to take an AP Practice Exam and go through the multiple choice portion.0003

The link of that is available right here.0011

I highly recommend that you take a minute, go through, and print it out, and try and take the test on your own first.0013

Then come back and we will go through the walk through this together.0020

As we take a look here, let us start on the E and M portion of this practice test from 1998.0025

In question 36 is where we are going to begin.0031

As we look at number 36, let us see.0036

We a resistor and capacitor acted in series to a battery.0040

I always like to draw these out.0044

There is V0 ±, we have got our resistor R, we got our capacitor C, and there is our circuit.0048

I will define the direction for positive current flow there I.0059

Which in the following questions relating the current and the circuit and the charge describes the circuit?0065

If we use Kirchhoff’s voltage law and we go around, let us start from here and go around clockwise.0071

I could have write my KVL that we have -V0 + IR + the voltage across our capacitor VC = 0.0076

But since capacitance is charge / voltage, voltage is charge/ capacitance so we could write this as - V0 + IR + Q/ C = 0.0091

Or V0 - IR -Q/ C = 0 and that will match the form of answer B.0108

37, which of those following combinations of 4 ohm resistors would dissipate 24 W connected to a 12V battery?0126

First, it would be nice know what that resistance is.0135

If power is V² / R, then resistance is V²/ our power which is going to be 12 V²/ 24W.0139

144/ 24 will be 6 ohms.0149

We need a combination of resistors that will give us 6 ohms.0152

If those are all 4 ohms resistors, it looks to me like E gives us a 4 ohm resistor in series with 2/ 4.0156

The 2/ 4 in parallel gives us 2 ohms.0164

4 + 2 is 6, correct answer must be E.0166

38, we have 2 uncharged conductors 1 and 2 on insulating stands, they are in contacts.0175

Let us draw those and we are going to bring a negatively charged rod near them, there it is.0181

When we do that, the electrons are closest to the rod and they are screaming away.0187

They are repelled so they are all going to hang out over here as much as possible, leaving this one slightly positively charged.0192

We separate those spheres, what is now true of conductor number 2, this one over here.0199

It is negatively charged so the answer must be C.0206

Taking a look at 39, we have a couple of charges in a test charge + Q, 0212

what is the direction of force on the test charge due to the other two charges?0223

We are going to get a force from the top one that is down the force F.0227

We are going to get a force to the left F from the one on the right because they repel.0232

The net is going to be down into the left or E.0237

40, relates to the same question, if F is the magnitude of the force on the test charge due to 1, what is the net force acting on it?0244

We can figure this out with the Pythagorean Theorem.0253

That is going to be in that direction, the combination of those two.0257

The total is going to be √ F² + F² which is √ 2F² or √ 2 × F.0260

The answer is D. 0274

Taking a look at 41, Gauss’s law provides a convenient way to calculate the electric field outside and each of the following accept what?0278

A large plate, we did that.0298

A sphere, we did that.0301

A cubed, it is going to be difficult with Gauss’s law because that is symmetry piece.0303

The long solid and a long hollow cylinder, our lectures on Gauss’s law we did A, B, D, and E.0307

The only one that is going to be difficult to use Gauss’s law for that purpose is going to be C.0314

Let us take a look, moving on here to 42, have a wire resistance R dissipating some power when the current passes through it.0321

The wire is replaced by another wire with resistance 3R, what is the power dissipated with the same current?0332

Power is I² R so our final power, once we triple our resistance is going to be I² × 3R, which is just going to be 3 I² R or 3 × our initial power.0338

The answer must be D, 3P.0354

Moving on 43, we got a narrow beam of protons producing that current of 1.6 ma.0361

They are 10⁹ proton in each meter, which is the best estimate of the average speed of the protons in the beam?0369

I like it, a little more challenging.0375

Current is 1.6 × 10⁻³ amps.0378

We have got 10⁹ protons / m and we are trying to find the average speed.0383

The way I might go about this is, I’m first going to recognize the current as charge per unit time.0394

We need to also somehow get velocity in here so I'm going to look at velocity being distance /time.0400

Therefore, time is distance / velocity and I'm going to plug that in for T here to get that my current is going to be Q/ D × V.0406

Let us also look and see if we can figure out the charge / m, since we got a Q/ D factor here.0421

We have 10⁹ elementary charges, the charge on each proton / m.0427

Let us convert that into Coulomb’s, that is 1.6 × 10 ⁻19 C/ elementary charge, which gives us 1.6 × 10 ⁻10 C / m.0433

Putting all of this together now, we said current is Q/ D × V.0447

Therefore, V must be current × D/ Q which is going to be our current was 1.6 × 10⁻³ amps Q/ D, that is what we have here C/ m.0454

Q/ D in the denominator would be 1.6 × 10 ⁻10 C/ m, which is going to leave us with 10⁷ m/ s.0469

Is it one of our answer? It is, answer D.0482

Very good, let us take a look at 44.0487

Which of the following describes the line of magnetic field in the vicinity of the beam due to the beam's current?0493

Either way, concentric circles around the beam.0499

Absolutely, that is got to be at 44 A.0502

45, we have got a couple of charges on X axis and we want to know where the electric field strength could be 0.0508

That is got to be between the two so that they cancel out.0518

And where is it going to be between the two?0525

That has got to be a little bit closer to that + 2Q charge so that the distance factor 0526

to make up for the fact that the charge on the right has less charge, so less force.0532

I would say it has to be between 2 and 3, answer C.0536

And 46, relating to the same problem.0544

The electric potential is negative at some points on the line in which of the following ranges?0546

That does not make sense, with only positive charges we are not going to have negative electric potentials.0551

I'm going to go with E on that one.0557

There are no places where you are going to have a negative potential at least in that current charge configuration.0560

Let us take a look here at number 47, we have got a graph showing the electric potential and region of space as a function of position.0570

Where would the charged particle experience the force of greatest magnitude?0580

You are going to have the greatest force where you have the greatest electric field strength.0584

Remember that the electric field is - DV DR so if that relates to the slope, 0592

we are going to have the strongest field where we have the strongest, the greatest slope on our voltage position graph.0598

That to me looks like it is going to be position D, the greatest slope is going to have the greatest electric field.0604

Therefore, the greatest force on a charged particle.0612

48, work that must be done by an external agent to move a point charge of 2 µc from the origin 2, 3 m away as 5 J.0619

Find the potential difference.0628

Potential difference is work per unit charge, that is going to be 5 J / 2 mc or 0.002 C.0631

It is just 2500 V, the answer is C.0639

49, we have got a wire loop, we have got a wire, and we are looking for net force on the loop.0648

We have done similar things here in our lectures.0655

Here is our current I2, let us draw our loop up here.0658

Where this is I1 going that direction and the first thing I'm going to do if I want to know the net force on the loop is I'm going to figure out the direction of the magnetic field due to I2.0667

From the right hand rule, it looks like that is going to be into the plane of the page 0677

so I'm just going to draw that with some axis up here to help me remember that.0680

They should be uniform but I have poor drawing skills here.0687

The direction of the force, by the right hand rule, if we have positive current, 0695

point the fingers of my right hand in the direction current is flowing, bend in the direction of the magnetic field.0700

You are going to see a force that is up from that section of wire.0705

Let me use red here.0708

Up from that section of wire, over here it is going to be to the right.0710

Here it is going to be down and here it is going to be the left.0715

To where it is going to be that these two are going to cancel, these are not however 0719

because the closest wire is going to be in the stronger magnetic field because it is closer to the source.0723

The smaller distance from the source of that magnetic fields.0728

This magnetic field is stronger here than here.0731

We will have a greater force down so the answer must be A, toward the wire.0733

Number 50, one of the trickier problems on the test.0746

A uniform magnetic field B is parallel to the XY plane as shown and we have got a proton initially moving with velocity V at that angle θ.0750

How will that follow that kind of path?0758

In here you have really got to think and visualize in 3 dimensions.0760

How the right hand rule is going to work?0764

Point your fingers in the direction it is moving, bend in the direction of the magnetic field, 0766

and you are going to see an evolving, twisting path as you get a force but keep bringing it around in the helical path.0772

That is going to be D, A helical path with its axis parallel to the Y axis in the direction of the magnetic field.0780

51, taking a look here, we have a parallel plate capacitor.0791

I like drawing parallel plate capacitors, let us do that.0798

Parallel plate capacitor + Q on one plate -Q on the other and that is separated by some distance D.0805

Good so far.0816

Let us see, if each plate has an area A.0818

A single proton of charge + E is released from rest of the surface of the positively charged plate.0822

There is our proton charge + E. 0829

What kinetic energy be or what will it be proportional to when it hits the other plate?0831

Let us go back to my definition of potential is work / charge, which implies that the work done which is QV.0841

When we get down here that is going to be the kinetic energy, 100% efficient in our problem.0851

But we also know that if C = Q/ V, then V = Q/ C.0857

We can write this as our charge of our proton × a charge on one of our plates ÷ the capacitance is equal to QV.0862

Since capacitance is proportional to area/ the separation of the plates,0876

we can say that the change in kinetic energy is going to be, not equal to, proportional to EQ/ C, which is A/ D.0882

DQ D/ A, do we have anything that looks like that, the answer is A.0896

Moving on to 52, in which of the following case does there exist a non 0 magnetic field that can be conveniently determined using Ampere’s law.0906

Outside the point charge if it is at rest.0919

It cannot be A, that is not going to work.0921

You do not get a magnetic field from a charge at rest.0923

Inside a stationary cylinder carrying current, no it is not moving again.0926

No charge, no magnetic field.0929

How about C, inside a very long current carrying solenoid?0931

We actually did that in the Ampere’s law video.0935

We know the answer is C, but let us take a look at the next 2 just for kicks.0938

At the center of the current carrying loop of wire, we did that but not with Ampere’s law.0942

That was pretty good, be pretty tough to integrate.0947

We did that with the Biot-Savart law and outside the square current carrying loop of wire.0950

That would be pretty tough to integrate too.0954

C is going to be the nice simple easy one.0956

53, positive beam of protons moves parallel to the X axis in the positive X direction as shown.0961

The magnetic field is pointed in the positive Y direction, in what direction must the electric field be pointed?0970

But let us se, the magnetic field is pointing in the positive Y and it is going to go with 0 deflection.0976

Then the magnetic force has to exactly balance the electrical force.0983

By the right hand rule then, that means that the direction of the electric field must be in the -Z direction 0986

in order to balance the force from the magnetic field.0994

That is got a be E, negative Z direction.0998

Taking a look at 54, a vertical length of copper wire moves to the right with a steady velocity V 1005

in the direction of constant horizontal magnetic field B.1012

To the following describes the induce charges on the ends of the wire.1016

54, V cross B is going to be 0, they are in the same direction.1020

I'm going to go with E on that one.1025

55, suppose an electron charge -E orbit a proton in a circular orbit of constant radius R.1031

If we assume the proton stationary, we are only worried about the electrostatic forces.1043

Find the kinetic energy of the two particle system which is going to be the kinetic energy of the electron.1048

The force on the electron, the electrical force is going to be the centripetal force, 1056

which implies then our electrical force by Coulomb’s law is 1/ 4 π ε₀, the charge on each of these proton and electron.1062

The magnitude is E so that is going to be E² ÷ R and that has to equal our centripetal force MV²/ R.1070

Q1 Q2/ R, get rid of square there.1087

We got to solve then for MV² is going to be equal to, we multiply that by R.1093

We are going to get a E²/ 4 π ε₀ R.1101

And MV² is awfully close to the kinetic energy.1111

If we want the kinetic energy, that is kinetic energy so I just divide that by 2.1114

Our kinetic energy then is going to be E²/ 8 π ε₀ R.1120

Because that is one of our choices, 1/ 8 π ε₀ R.1133

Choice B gives us the correct answer.1139

56, we have got a square loop of wire with side L and resistance R.1148

It is held at rest in a uniform field but the field decreases with time according to that formula.1154

Find the induced current in the loop.1161

Faraday’s law, E = –D φ BDT which is - the derivative of BA because our area is not changing.1164

Which is going to be - the derivative with respect to time of, our magnetic field strength B is given by the formula A – BT.1175

And our area is just L², it is a square.1188

L² can come out, it is going to be L² × the derivative of A – BT, which implies that induced EMF then is just going to be BL².1191

The current then is E/ R just going to be BL² / R.1207

And how about direction, it looks like with time but that magnetic field is getting weaker.1218

Dense law as we wanted to oppose that change that will give us a counterclockwise current by the right hand rule.1224

BL²/ R counterclockwise, looks like that is going to be choice E.1230

57, a negatively charged particle in a uniform magnetic field moves in a circular path as shown,1240

which of the following graphs to picks have the frequency of revolution F depends on the radius R.1249

Let us start, we know that the velocity is distance /time or in this case 2 π R ÷ T, the time for once around is the period.1258

We can also look at this from a centripetal force perspective.1268

The centripetal force is MV² / R which has to be equal to whatever is causing that force.1272

What is causing the centripetal force is the magnetic force here, which is going to be Q VB.1281

Therefore, we could solve for V and say that that is going to be Q BR/ M.1287

Putting these together, Q BR/ M is our velocity has to equal 2 π R / T.1297

But by the way, 1/ the period is the frequency so I'm going to write that as 2 π R F.1309

And just solve for frequency is going to be Q BR/ M × 2 π R or Q B/ 2 π M.1315

Notice here that we have no dependence on the radius.1331

The correct answer must be A, does not depend on radius at all.1334

Let us take a look at 58, the only force acting on an electron is due to a uniform electric field.1343

The electron moves at constant.1353

It has got to be a constant acceleration because it is a constant force.1356

The direction is opposite that of the field because the field points in the direction of the force a positive charge would feel.1363

58 right away, A.1369

59, in a region of space we got a spherical symmetric electric potential given by the function V of R = KR², where K is a constant.1375

What is the magnitude of the electric field when your distance R from the origin?1387

The electric field from potential is - DV DR which is just going to be -2 KR.1392

Therefore, the magnitude of the electric field is just going to be 2K and we are looking at R₀ instead of R.1400

2 K R0 is going to be our answer C.1408

In the following question 60, what is the direction of the electric field at RO and the direction of the force on electron.1414

Notice we had negative here, so that means that the electric field is going to be toward the origin and 1422

because it is a force on the electron, it is going to the opposite direction of the field.1428

The answer there B, it is toward the origin with the force, the electric field toward the origin of the force in the opposite direction.1433

The answer is B because of this -2 KR that we just found.1440

61, 2 charged particles each with a charge of +Q are located along the X axis at X = 2 and X =4.1450

Which of the following shows the graph of the magnitude of the electric field along the X axis?1459

As I look at 61, it is pretty obvious right in between that has to be 0 so that gets rid of choice C and choice B.1464

D and E do not make any sense, it is going to be A.1475

62, positive electric charge moved at constant speed between two locations in an electric field1481

with no work done by or against the field, how can that occur?1490

Right away, I’m thinking that the only way that happens is if you are on equal potential.1493

D, the charge is moved along the equal potential line.1498

There it is, nice and simple.1501

63, we got a non conducting hollow sphere of radius R carrying a large charge + Q.1507

A small charge + Q is located at point P and what must be the work done in moving the charge +Q 1513

from P through the hole to the center of the sphere?1521

We are only going to have to do work to get to the surface and that amount of work is going to be charge × the potential difference, 1525

which is going to be charge × the final potential - the initial potential, which is going to be our charge × 1534

our final K Q/ R the radius - K Q/ r, we can do some factoring Kq Q 1 / R -1 / r.1543

It looks like it is going to be the choice E.1561

64, we have got a couple capacitors altering µf connected to the circuit with a 12 V battery.1568

Find the equivalent capacitance between X and Z.1577

The first thing I'm going to do is see that those 2, 3 µf capacitor is in parallel.1580

Those add up so that is going to be 6 µf.1585

Then we have got to go through A 3 µf capacitor.1590

Those two in series, if we found that equivalent should tell us our answer.1596

We can do that a bunch of ways but right away we know that that is going to be less than the smallest so less than 3.1601

As I do that, 1/ C equivalent is 1/6 µf + 1/3 µf is going to be, 1/ C equivalent is going to being 1/6 µf + 2/6 micro F is 3/6 µf.1615

Therefore, C equivalent is 2 µf.1634

Our choice there for 64 must be B.1640

And 65, the follow on question.1646

Potential difference between Y and Z is going to be what?1650

We are looking for the potential difference if we can put a vault meter right there between Y and Z.1654

If C = Q/ V then potential = Q/ C, we know our charge is going to be VC, is going to be 12V × 2 µf or 24 µc.1662

The charge on each plate is 24 µc.1685

Once we know that, the potential between Y and Z is just going to be Q / C or we will have 24 µc on the plate ÷ that capacitance 3 µf is 8V.1689

65 should be D. 1705

66, in the figure contains two identical light bulbs in series with a battery.1713

At first they are at equal brightness when switch S is closed, which of the following occurs to the bulbs?1720

Once switch S is closed, right away bulb 2 is not going to have any electricity going through.1727

All the electricity is going to go right through S, it has shorted out.1734

Our choice is either B or E.1737

If we short out bulb 2, all of that potential is now going to go through bulb 1 so it is going to expand more power, 1740

the answer has to be B, bulb 1 gets brighter and bulb 2 goes out.1748

67, the bar magnet and a wire loop current carrying I are arranged in which direction is the force on the current loop due to the magnet?1755

As I look at that one, let us say that that current loop creates a south pole1765

but nears the magnet due to the right hand rule so that is going to attract it.1770

Your force then has to be A, toward the magnet.1774

68, a wire loop of area A is placed in a time varying that is spatially uniform magnetic field perpendicular to the plane of the loop.1782

The induced EMF was given by that function.1791

The time varying magnetic field could be given by, EMF is – D φ B DT is going to be B AT ^½.1794

Or which implies that E = - the derivative of the integral of B ⋅ DA which is - D/ DT of BA or - A DB DT.1807

Which implies then, we have – B T ^½ must equal DB DT, which implies that the integral of -BT ^½ DT must equal the integral of DB.1828

Or integral of DB is just B integral of - BT ^½ is going to be – B T³/2 / 3/2 + some constant, 1837

implies then that B is going to be equal to -2/3 B T³/2, which is we are looking for that and not worry about direction.1859

2/3 BT³/2, the answer choice E.1875

That is more involved multiple choice in there. 1880

69, talking about a capacitor with two identical conducting plates parallel to each other separated by a distance D,1884

the stand is connected if you ignore effects what is the electric field between the plates?1892

Electric field between the plates are parallel capacitor is constant as long as you are not near the edges.1897

Just B/ D, the answer is B.1904

70, an insulating plastic material is inserted between the plates without otherwise disturbing the system.1909

What does that do to the capacitance?1916

Capacitance is ε A/ D.1919

If you increase ε by putting in an insulator, your capacitance has to go up because A and D are changing.1923

Capacitances must increase, the answer has to be A.1930

Alright that is the end of the multiple choice section of this practice test.1937

Hopefully that gives you a good feel for where you are strong and some opportunities for some more work.1940

Thank you so much for watching

Come on back in the next lesson and we will do the free response portion of this test.1948

Make it a great day everyone.1952