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  3. Paper 1A: multiple choice

Paper 1A: multiple choice

How Paper 1A multiple-choice questions are built, the strategies that answer them quickly and reliably, the traps hidden in the wrong options, and ten worked questions.

  • IB Physics
  • Exam skills
  • SL and HL

17 min readLast reviewed 14 Oct 2026

Paper 1A is the multiple-choice part of Paper 1. Every question has four options, labelled A to D, and exactly one of them is correct. Each question earns one mark, and there is no penalty for a wrong answer, so no question should ever be left blank. A calculator and a clean copy of the data booklet are available, but the numbers are chosen so that the arithmetic takes seconds: most questions test whether you can choose the right physics, not whether you can push buttons.

The skill being tested is different from that of a written paper. In Paper 2 a sound method earns marks even when the final answer slips; in Paper 1A only the letter counts. That makes two habits essential: a fast, reliable route to the answer, and a check that the answer is not one of the carefully designed wrong options.

Key points

  • Every wrong option (a distractor) is the answer a student gets after one specific, common slip.

  • Working out your own answer before reading the options protects you from being led by them.

  • Proportional reasoning, units, estimation and limiting cases often answer a question faster than a full calculation.

  • About half of the questions need a calculation or a ratio; the rest test concepts, graphs and diagrams.

  • Questions at SL test SL content only; HL papers add questions on the additional higher level content.

How the questions are built

A typical stem is three or four lines long: a short physical situation, sometimes a diagram, and one question. Most stems end with a direct question ("What is the tension in the cable?"). Some end mid-sentence, and the options complete it ("The internal energy of an ideal gas is"); those options start with a lower-case word and end with a full stop.

Format

What it looks like

How to approach it

Numbers with units

Four values, almost always in ascending order

Calculate or estimate first; the order lets you spot factor-of-2 and power-of-ten neighbours

Expressions or ratios

W22W​, 2FMRMR2F​, 22​, "What is Ek at YEk at XEk​ at XEk​ at Y​?"

Work with symbols and ratios; check units of each expression

Words or phrases

Four short statements or sentence endings

Decide what you would write before reading them

Two-quantity table

Four rows, each giving a pair of values ("What are the radius and the period?")

Decide each column separately, then find the row that matches both

Graph or diagram options

Four sketches: "Which graph shows the variation of … with …?"

Check intercept, gradient, curvature and asymptotes one at a time

Statement combinations

Statements I, II, III, then A "I and II only", B "I and III only", C "II and III only", D "I, II and III"

Judge each statement true or false independently

The options are not random. Each distractor is built by applying one plausible mistake to the correct method: forgetting the ½ in kinetic energy, using a diameter as a radius, leaving a temperature in degrees Celsius, not squaring a ratio. A student who makes that slip finds their answer waiting among the options, which is why "my answer is there, so it must be right" is not a safe check.

Definition

A distractor is a wrong option in a multiple-choice question, designed so that a specific mistake in reasoning or calculation leads to it.

Stems also contain cue words that carry physics. "Negligible internal resistance" means the terminal pd equals the emf. "Released from rest" sets u=0u=0. "Frictionless" or "in a vacuum" removes energy losses. "Ideal gas" allows pV=nRTpV=nRT and means there are no intermolecular forces, so the internal energy is all kinetic. A question that seems to lack information often has the missing piece in one of these phrases.

A routine for every question

  1. Read the whole stem, including the last line. Mark the quantity asked for and any cue words. Many marks are lost by answering a slightly different question, such as giving the distance when the displacement was asked for.

  2. Cover the options and predict. Decide the physics and, for a calculation, get a value or at least a rough size. For a conceptual question, say in a few words what the answer should be.

  3. Match and check. Find your prediction among the options. Then ask which slip each neighbouring option represents. If you can explain every distractor, you can be confident in the key.

  4. If stuck, eliminate. Remove options with the wrong unit, the wrong direction, the wrong order of magnitude or an impossible limiting case. Guessing between two options is far better than guessing between four.

  5. Mark and move on. If a question resists you for more than a couple of minutes, choose the best remaining option, mark the question, and return to it at the end.

Tip

Write on the question paper. A quick sketch of the forces, a ratio written as fractions, or a crossed-out option takes seconds and prevents the most common errors, which are slips of attention rather than gaps in knowledge.

Strategies that save time

Proportional reasoning

Many questions ask how a quantity changes when others are scaled: "The distance is doubled and the current halved. What is the new force?" There is no need to find any actual value. Write the relationship as a proportionality, then replace each variable with its scale factor.

For example, the power dissipated in a resistor connected to a fixed pd is P=V2RP=RV2​, and R=ρLAR=AρL​. A wire of the same material with twice the length and twice the diameter has L→2LL→2L and A→4AA→4A (area goes as diameter squared), so R→24R=R2R→42​R=2R​ and P→2PP→2P.

PnewPold=RoldRnew=Anew/LnewAold/Lold=42=2Pold​Pnew​​=Rnew​Rold​​=Aold​/Lold​Anew​/Lnew​​=24​=2

Common mistake

Scaling a diameter as if it were an area. When a diameter or radius doubles, a cross-sectional area becomes four times as large and a volume eight times as large. Distractors almost always include the answer that treats area as proportional to the diameter.

Units as a check and an eliminator

Every correct physics equation is homogeneous: both sides have the same units. When the options are algebraic expressions, working out the units of each one often leaves only one candidate. When the options are numbers, the unit of the quantity asked for confirms that the right equation has been used: an answer in kg m s−1kg m s−1 is a momentum, not an energy.

The SI base units used for this are kg, m, s, A, K and mol. A newton is kg m s−2kg m s−2, a joule is kg m2 s−2kg m2 s−2 and a pascal is kg m−1 s−2kg m−1 s−2.

Estimation and orders of magnitude

When the options differ by factors of ten or more, a rough calculation with one-significant-figure numbers is enough. Estimation questions ("What is the order of magnitude of …?") expect you to know rough values of everyday quantities: a person has a mass of about 70 kg, a storey of a building is about 3 m high, the density of air is about 1 kg m⁻³ and of water 1000 kg m⁻³, a car on a motorway travels at about 30 m s⁻¹.

Tip

Round every value to one significant figure and work in powers of ten: 3.2×104×6.8×10−33.2×104×6.8×10−3 becomes 3×7×101≈2003×7×101≈200. The exact answer is 218, which is enough to choose between options that differ by a factor of ten.

Limiting cases

An expression that is correct must behave sensibly at the extremes. If one option for the acceleration of a block on a frictionless slope at angle θθ is gcos⁡θgcosθ, setting θ=0θ=0 gives an acceleration of gg on a horizontal surface, which is impossible: that option is wrong. The correct gsin⁡θgsinθ gives 0 on the flat and gg for a vertical drop. Testing an option at θ=0θ=0, at 90∘90∘, at zero mass or at infinite resistance takes seconds.

Reading graphs

Graph questions test a small set of skills. For any graph, identify what the gradient represents and what the area under the line represents, and read the values from the axes with their units and prefixes.

Graph

Gradient

Area under the line

Displacement–time

Velocity

(no useful meaning)

Velocity–time

Acceleration

Displacement

Acceleration–time

(rate of change of acceleration)

Change in velocity

Force–time

(rate of change of force)

Impulse (change in momentum)

Force–extension

Spring constant kk

Elastic potential energy stored (work done)

Power–time

(rate of change of power)

Energy transferred

Potential difference–current

Resistance only for a straight line through the origin

(no useful meaning)

Common mistake

Taking the resistance of a component from the gradient of a VV–II graph that curves. Resistance is defined as R=VIR=IV​ at a point, so read one pair of values and divide; the gradient of the tangent is a different quantity.

For "Which graph shows …?" questions with four sketches, use the features in turn. Does the line pass through the origin? Is it straight or curved, and which way does it curve? Does it approach an asymptote? Is there a maximum, and where? Usually two of these features eliminate three options.

Statement questions

In a question with statements I, II and III, judge each statement on its own and write T or F beside it. Only then look at the options. Because the options are always the same four combinations, deciding that one statement is false removes two or three options at once. If statement II is false, only B ("I and III only") can be correct, as long as I and III are true.

Traps hidden in the distractors

The same slips appear in every theme. Learning to recognise them makes most distractors easy to reject.

Trap

Example of the distractor

A missing factor of 2

Forgetting the ½ in Ek=12mv2Ek​=21​mv2 or in s=v22as=2av2​; using the final speed where the average speed is needed

A wrong power

Not squaring in F∝1r2F∝r21​, P=I2RP=I2R or Ek∝v2Ek​∝v2; using a square root where a cube root is needed

An inverted ratio

Giving oldnewnewold​ when newoldoldnew​ was asked

Celsius for kelvin

Doubling 27 °C to 54 °C instead of 300 K to 600 K in the gas laws or in L=σAT4L=σAT4

Radius from the wrong point

Using the height above a planet’s surface instead of the distance from its centre

Diameter for radius

Using the full diameter of a wire, planet or circular path

One component instead of the combination

Giving the current in one branch, or one resistor’s value, instead of the total

Linear thinking for exponential decay

Assuming two half-lives remove the whole sample

Direction errors

A force on the wrong body in a Newton’s third law pair; a field drawn the wrong way

Unit and prefix slips

mm taken as m, MeV taken as J, μCμC taken as mC

A missed detail in the stem

"From rest", "the other end is closed", "a neutron is absorbed", "the background count is 20 counts per minute"

Exam tip

When your answer matches an option, look at the options either side of it. If one of them is exactly half, double, a square or a power of ten away from yours, check the step that would produce it. This takes ten seconds and catches most careless errors.

Worked multiple-choice questions

Each question is followed by its answer, the reasoning, and the slip behind each distractor. Try each one before opening the solution.

Question 1: reading a velocity–time graph

Velocity–time graph of a car. The velocity rises in a straight line from 0 at t = 0 to 12 m s⁻¹ at t = 4 s, then stays constant at 12 m s⁻¹ until t = 10 s.Velocity–time graph of a car. The velocity rises in a straight line from 0 at t = 0 to 12 m s⁻¹ at t = 4 s, then stays constant at 12 m s⁻¹ until t = 10 s.
The car accelerates uniformly for 4.0 s and then travels at a constant 12 m s⁻¹. The distance travelled is the area under the line.

Worked example

The velocity–time graph shows the motion of a car for 10 s. What is the distance travelled by the car in the 10 s?

  • A. 60 m

  • B. 72 m

  • C. 96 m

  • D. 120 m

Show the solution

Answer C. The distance is the area under the velocity–time graph: a triangle for the first 4.0 s and a rectangle for the next 6.0 s.

s=12×4.0×12+6.0×12=24+72=96 ms​=21​×4.0×12+6.0×12=24+72=96 m​

Distractors. A treats the whole graph as one triangle (12×10×1221​×10×12). B is only the constant-velocity part. D multiplies the top speed by the whole time, as if the car had moved at 12 m s⁻¹ throughout.

Question 2: a collision

Worked example

A trolley of mass mm moving with speed vv collides with a stationary trolley of mass 2m2m. The trolleys stick together. What is the speed of the trolleys immediately after the collision?

  • A. v33v​

  • B. v22v​

  • C. 2v332v​

  • D. vv

Show the solution

Answer A. Momentum is conserved in the collision. The total momentum before is mvmv, and afterwards the combined mass 3m3m moves with speed v′v′:

mv=3m v′⇒v′=v3mv=3mv′⇒v′=3v​

Distractors. B assumes equal masses. C swaps the masses, as if the 2m2m trolley were the moving one. D ignores the second trolley. A quick check: the combined speed must be less than vv and should be small, because the moving trolley is the lighter one.

Question 3: an ideal gas and temperature

Worked example

A fixed mass of an ideal gas is held at constant volume at a temperature of 27 °C. The gas is heated until its pressure doubles. What is the new temperature of the gas?

  • A. 54 °C

  • B. 327 °C

  • C. 600 °C

  • D. 873 °C

Show the solution

Answer B. At constant volume p∝Tp∝T, where TT is the absolute temperature. Convert first: T=27+273=300 KT=27+273=300 K. Doubling the pressure doubles the absolute temperature to 600 K, which is 600−273=327 ∘C600−273=327 ∘C.

Distractors. A doubles the Celsius value. C is the correct kelvin value written with the wrong unit. D adds 273 to 600 K instead of subtracting it.

Question 4: a series and parallel circuit

Circuit: a 12 V cell in series with a 4.0 Ω resistor and a parallel pair of resistors, 6.0 Ω and 3.0 Ω.Circuit: a 12 V cell in series with a 4.0 Ω resistor and a parallel pair of resistors, 6.0 Ω and 3.0 Ω.
The 6.0 Ω and 3.0 Ω resistors share the same potential difference; together they are in series with the 4.0 Ω resistor.

Worked example

The circuit contains a cell of emf 12 V and negligible internal resistance, connected to three resistors. What is the current in the 3.0 Ω resistor?

  • A. 0.67 A

  • B. 0.92 A

  • C. 1.3 A

  • D. 2.0 A

Show the solution

Answer C. The parallel pair has a combined resistance found from 1R=16.0+13.0R1​=6.01​+3.01​, so R=2.0 ΩR=2.0 Ω. The total resistance is 4.0+2.0=6.0 Ω4.0+2.0=6.0 Ω and the cell current is 126.0=2.0 A6.012​=2.0 A.

The pd across the parallel pair is 2.0×2.0=4.0 V2.0×2.0=4.0 V, so the current in the 3.0 Ω resistor is 4.03.0=1.3 A3.04.0​=1.3 A.

Distractors. A is the current in the 6.0 Ω branch. B adds all three resistances in series (12131312​). D is the total current from the cell. A check: the two branch currents, 0.67 A and 1.3 A, add up to the 2.0 A total.

Question 5: choosing an expression by its units

Worked example

The speed vv of a transverse wave on a stretched string depends on the tension FF in the string and on its mass per unit length μμ. Which expression for vv could be correct?

  • A. FμFμ​

  • B. FμμF​

  • C. μFFμ​

  • D. FμμF​​

Show the solution

Answer D. In SI base units, FF is kg m s−2kg m s−2 and μμ is kg m−1kg m−1. Then

Fμ→kg m s−2kg m−1=m2 s−2,Fμ→m s−1μF​→kg m−1kg m s−2​=m2 s−2,μF​​→m s−1

Only D has the units of a speed. B has units of speed squared, C is the reciprocal of that, and A has units of kg s−1kg s−1. The units cannot fix a dimensionless number in front, but here no option has one, so units alone decide.

Question 6: field strength on another planet

Worked example

A planet has the same mean density as the Earth and twice the Earth's radius. The gravitational field strength at the Earth's surface is gg. What is the gravitational field strength at the surface of the planet?

  • A. g44g​

  • B. g22g​

  • C. gg

  • D. 2g2g

Show the solution

Answer D. The field strength at the surface is g=GMR2g=R2GM​. With the same density, the mass scales with volume, so M∝R3M∝R3. Then g∝R3R2=Rg∝R2R3​=R, and doubling the radius doubles gg.

Distractors. A keeps the mass the same and applies only the inverse square. B inverts the correct ratio. C assumes that the same density means the same surface field.

Question 7: half-life

Worked example

The activity of a radioactive sample falls from 800 Bq to 100 Bq in 36 hours. What is the half-life of the sample?

  • A. 4.5 h

  • B. 9.0 h

  • C. 12 h

  • D. 18 h

Show the solution

Answer C. Halve the activity repeatedly: 800 → 400 → 200 → 100 Bq. That is three halvings, so three half-lives take 36 hours and one half-life is 12 hours.

Distractors. A divides 36 by the ratio 8. B counts the four values in the chain instead of the three steps between them. D divides the total time by 2, as if the activity had halved only once.

Question 8: a standing wave on a string

Worked example

A string of length 0.90 m is fixed at both ends. It vibrates in its third harmonic at a frequency of 240 Hz. What is the speed of waves on the string?

  • A. 144 m s⁻¹

  • B. 216 m s⁻¹

  • C. 324 m s⁻¹

  • D. 432 m s⁻¹

Show the solution

Answer A. The third harmonic on a string fixed at both ends has three loops, each half a wavelength long, so L=3λ2L=23λ​ and λ=2×0.903=0.60 mλ=32×0.90​=0.60 m. Then v=fλ=240×0.60=144 m s−1v=fλ=240×0.60=144 m s−1.

Distractors. B takes the wavelength equal to the string length. C inverts the fraction (λ=3L2λ=23L​). D uses the first-harmonic wavelength 2L2L with the third-harmonic frequency.

Question 9: an order-of-magnitude estimate

Worked example

A classroom measures 8 m by 6 m by 3 m. What is the order of magnitude of the mass of the air in the room?

  • A. 100100 kg

  • B. 102102 kg

  • C. 105105 kg

  • D. 108108 kg

Show the solution

Answer B. The volume is 8×6×3≈150 m38×6×3≈150 m3. The density of air is about 1.2 kg m−31.2 kg m−3, so the mass is about 180 kg, of order 102102 kg.

Distractors. C uses the density of water (1000 kg m−31000 kg m−3). A would need a density of about 0.01 kg m−30.01 kg m−3, a hundred times too small, and comes from treating air as almost massless. D combines a volume in litres (1 m3=10001 m3=1000 L) with the density of water. The real answer surprises many students: the air in a classroom has a mass greater than two adults.

Question 10: statements about a pendulum

Worked example

Three statements about a simple pendulum oscillating with small amplitude are:

  • I. The period does not depend on the mass of the bob.

  • II. The period doubles when the length is doubled.

  • III. The period would be greater on the Moon than on the Earth.

Which statements are correct?

  • A. I and II only

  • B. I and III only

  • C. II and III only

  • D. I, II and III

Show the solution

Answer B. The period is T=2πlgT=2πgl​​. Statement I is true: mass does not appear. Statement II is false: doubling ll multiplies TT by 22​, not 2. Statement III is true: gg on the Moon is smaller, so TT is larger.

Once II is known to be false, options A, C and D are all eliminated, so only I and III need checking to confirm B.

Using the whole paper well

Paper 1A rewards steady accuracy across all the questions more than brilliance on a few. The questions follow the order of the course, so the difficulty does not rise steadily: a hard question may be followed by an easy one. A question that takes far longer than the rest is still worth only one mark.

  • Answer every question on the first pass, even with a provisional guess, so nothing is left blank if time runs short.

  • Mark questions you are unsure of and return to them at the end with fresh eyes.

  • Change an answer only when you find a definite error in your reasoning; a vague doubt is not a reason.

  • Keep the data booklet open at the right page for each theme; checking a symbol or a constant is quicker than recalling it.

  • For negative stems (questions with NOT in capitals), circle the word before answering.

Exam tip

When an option reads "cannot be determined", check carefully whether the stem really gives enough data. It is occasionally the key, when a quantity such as a mass or a distance is genuinely missing, but more often a cue word in the stem supplies what seemed to be absent.

Common mistake

Spending the time saved on easy questions re-reading them instead of working on the hard ones. Once a question has a confident answer with every distractor explained, leave it.

Self-check

  • To do:

    I can recognise the six formats of multiple-choice question and approach each one.

  • To do:

    I can predict an answer before reading the options and explain the slip behind each distractor.

  • To do:

    I can use proportional reasoning to find how a quantity changes when others are scaled.

  • To do:

    I can use units to check an expression and eliminate options.

  • To do:

    I can make a quick order-of-magnitude estimate using everyday values.

  • To do:

    I can test an expression with a limiting case.

  • To do:

    I can state what the gradient and the area represent for the common physics graphs.

  • To do:

    I can judge I/II/III statements independently and use one false statement to eliminate options.

  • To do:

    I can name the common distractor traps: factors of 2, wrong powers, inverted ratios, Celsius for kelvin, radius errors and missed details.

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