I85PREP
⌘ K
Bremen Abitur Physik study materials
Bremen Abitur Physik study resources

Bremen Abitur Physik practice questions

One question at a time. Build your Bremen Abitur Physik understanding.

First 50 of 190 questions. Each one lists the choices, the correct answer, and an explanation. The checker at the top of this page saves progress in this browser when it is available.

Continue in the practice tool above to load more questions, 50 at a time.

Question 1Electricity Magnetism

Two point charges +Q and +Q are fixed a distance r apart in vacuum. The electrostatic force between them has magnitude F. If both charges are doubled and the separation is halved, the new force magnitude is

  1. F
  2. 4F
  3. 16F Correct answer.
  4. 32F

Explanation

Coulomb’s law: F ∝ q₁q₂/r². Doubling both charges multiplies the product by 4; halving r multiplies by 4 via 1/r² → overall factor 16.

Question 2Electricity Magnetism

An electron (charge −e) moves with velocity v perpendicular to a uniform magnetic field B. The magnetic force magnitude is

  1. zero because electrons have negative charge
  2. e v B Correct answer.
  3. e v / B
  4. e B / v

Explanation

F = |q| v B sinθ with θ = 90°, so F = e v B. The negative sign affects direction (opposite a positive charge), not the magnitude formula.

Question 3Electricity Magnetism

If the plate separation of an isolated parallel-plate capacitor (constant Q) is increased, what happens to the potential difference U and the stored energy W?

  1. U decreases and W decreases
  2. U increases and W increases Correct answer.
  3. U increases and W decreases
  4. U and W both stay constant

Explanation

For fixed Q, C decreases as d increases (C∝1/d), so U = Q/C increases. Stored energy W = Q²/(2C) also increases as C falls.

Question 4Electricity Magnetism

An electron is accelerated from rest through U = 100 V. Its kinetic energy is

  1. 100 J
  2. 1.6 × 10⁻¹⁷ eV
  3. 0 J because mass is tiny
  4. 100 eV Correct answer.

Explanation

A charge e falling through potential difference U gains kinetic energy eU, which is 100 eV by definition of the electronvolt.

Question 5Electricity Magnetism

The self-induced emf in a coil is ε = −L ΔI/Δt. For L = 0.20 H and a current rising uniformly from 0 to 5.0 A in 10 ms, |ε| is

  1. 100 V
  2. 1.0 V
  3. 10 V Correct answer.
  4. 0.010 V

Explanation

|ε| = L ΔI/Δt = 0.20 × (5.0)/(0.010) = 100 × 0.20 = 10 V.

Question 6Electricity Magnetism

Faraday’s law in integrated form for N loops is ε = −N dΦ/dt. If the flux through one loop increases by 0.040 Wb in 0.020 s and N = 50, the average |ε| is

  1. 40 V
  2. 100 V Correct answer.
  3. 0.080 V
  4. 1.6 V

Explanation

|ε| = N |ΔΦ|/Δt = 50 × 0.040 / 0.020 = 50 × 2.0 = 100 V.

Question 7Electricity Magnetism

Lenz’s law states that an induced current

  1. always flows clockwise in the laboratory frame
  2. opposes the change in magnetic flux that produces it Correct answer.
  3. maximizes the change in magnetic flux
  4. exists only in open circuits

Explanation

Lenz’s law: the induced current’s magnetic effect opposes the flux change that causes it-energy conservation / direction rule for induction.

Question 8Electricity Magnetism

In a region with crossed uniform fields E and B (E ⊥ B), a charged particle with v = E/B perpendicular to both feels zero net Lorentz force. This is the principle of a

  1. cyclotron resonance heater
  2. pure electrostatic accelerator only
  3. transformer core
  4. velocity filter (Wien filter) Correct answer.

Explanation

For F_E + F_B = 0 when v ⊥ B and appropriate directions, qE = q v B ⇒ v = E/B: the Wien / velocity filter selects that speed.

Question 9Electricity Magnetism

The capacitance of a parallel-plate capacitor in vacuum is C = ε₀ A/d. If A is doubled and d is doubled, C

  1. doubles
  2. halves
  3. stays the same Correct answer.
  4. quadruples

Explanation

C ∝ A/d; doubling A and d leaves the ratio unchanged, so C is unchanged.

Question 10Electricity Magnetism

The electric potential difference between two points is 12 V. Moving a +3.0 µC charge quasistatically from the lower to the higher potential requires work of magnitude

  1. 4.0 µJ
  2. 36 J
  3. 0 because electrostatic force is conservative
  4. 36 µJ Correct answer.

Explanation

W_ext = q ΔV = (3.0×10⁻⁶ C)(12 V) = 3.6×10⁻⁵ J = 36 µJ when moving to higher potential against the field for a positive charge.

Question 11Electricity Magnetism

A circular loop of radius r = 0.10 m lies in a plane perpendicular to a uniform B = 0.20 T. The magnetic flux through the loop is

  1. 2.0 × 10⁻² Wb approximately without π
  2. 0 because the field is uniform
  3. 6.3 × 10⁻³ Wb Correct answer.
  4. 0.020 Wb

Explanation

Φ = B A = B π r² = 0.20 × π × 0.010 ≈ 0.20 × 0.0314 = 6.3×10⁻³ Wb.

Question 12Electricity Magnetism

Two long parallel wires 4.0 cm apart carry equal currents of 5.0 A in the same direction. The force per unit length is attractive with magnitude (use μ₀/(2π) ≈ 2.0 × 10⁻⁷ T·m/A)

  1. 1.0 × 10⁻⁴ N/m
  2. 1.25 × 10⁻⁴ N/m Correct answer.
  3. 2.5 × 10⁻⁵ N/m
  4. 5.0 × 10⁻⁵ N/m

Explanation

F/ℓ = (μ₀/(2π)) I₁ I₂ / d ≈ (2.0×10⁻⁷)(25)/(0.040) = (5.0×10⁻⁶)/0.040 = 1.25×10⁻⁴ N/m. Same-direction currents attract.

Question 13Electricity Magnetism

The electric field lines of a positive point charge

  1. form closed loops around the charge
  2. begin on negative charges only and end on the positive charge
  3. are always parallel and equally spaced everywhere
  4. begin on the charge and radiate outward Correct answer.

Explanation

By convention, field lines emanate from positive charges and terminate on negative charges; for an isolated +q they radiate outward.

Question 14Electricity Magnetism

A series RC circuit is driven by a DC source after a long time (steady state). The current is

  1. maximum and limited only by R
  2. zero because the capacitor acts as an open circuit Correct answer.
  3. oscillating at the source frequency
  4. limited only by C with R shorted

Explanation

In DC steady state the capacitor is fully charged and i = 0; it behaves like an open circuit.

Question 15Waves Optics

Two coherent sources produce constructive interference when the path difference is

  1. (m + 1/2) λ only for all m
  2. always zero regardless of λ
  3. equal to the amplitude
  4. m λ (integer number of wavelengths) Correct answer.

Explanation

Constructive interference: Δs = m λ (m = 0, ±1, ±2, …) for sources in phase.

Question 16Waves Optics

Light of wavelength 600 nm illuminates a double slit with d = 0.20 mm. Screen distance L = 2.0 m. The fringe spacing is

  1. 1.2 mm
  2. 6.0 mm Correct answer.
  3. 0.60 mm
  4. 6.0 cm

Explanation

Δy = λ L / d = (600×10⁻⁹)(2.0)/(0.20×10⁻³) = (1.2×10⁻⁶)/(2.0×10⁻⁴) = 6.0×10⁻³ m = 6.0 mm.

Question 17Waves Optics

Standing waves form from

  1. interference of two waves of equal frequency traveling in opposite directions Correct answer.
  2. a single traveling wave in free space with no reflection
  3. incoherent noise only
  4. electrostatic induction only

Explanation

Standing waves arise when counter-propagating waves of the same f interfere (e.g., incident + reflected on a string).

Question 18Waves Optics

Polarisation is a property of

  1. transverse waves (e.g., EM waves) Correct answer.
  2. longitudinal sound waves in air
  3. DC currents in resistors only
  4. scalar pressure fields without direction

Explanation

Polarisation describes the orientation of the transverse oscillation; longitudinal waves in fluids lack a transverse polarisation degree of freedom.

Question 19Waves Optics

An electromagnetic wave in vacuum has E and B fields that are

  1. perpendicular to each other and to the propagation direction Correct answer.
  2. parallel to each other and to the propagation direction
  3. electrostatic only with B = 0
  4. longitudinal oscillations of E along the ray

Explanation

Plane EM waves: E ⊥ B ⊥ k (propagation), with |E|/|B| = c in vacuum.

Question 20Waves Optics

Electromagnetic waves carry energy; the Poynting vector magnitude in vacuum for a plane wave scales as

  1. E B / μ₀ (order of |S|) Correct answer.
  2. E / B only
  3. B² / ε₀ only without E
  4. zero for all EM waves

Explanation

|S| = |E × B|/μ₀ = E B / μ₀ for perpendicular E and B in vacuum (instantaneous).

Question 21Oscillations

In undamped SHM, the total energy is constant and equals

  1. (1/2) k x² at every instant with x the displacement
  2. m g h only
  3. zero at maximum displacement
  4. (1/2) k A² where A is the amplitude Correct answer.

Explanation

E = (1/2) k A² = (1/2) m ω² A²; at maximum displacement kinetic energy is zero but potential (elastic) is maximum and equals total E.

Question 22Oscillations

Resonance in a driven damped oscillator occurs when the driving frequency is near the system’s natural frequency. Then the amplitude is

  1. minimized
  2. maximized (for velocity resonance / typical displacement near ω₀) Correct answer.
  3. exactly zero
  4. independent of damping

Explanation

Near resonance the steady-state amplitude is largest (with the precise peak slightly shifted by damping for displacement).

Question 23Oscillations

A pendulum of length L = 1.0 m approximates a simple pendulum on Earth (g ≈ 9.8 m/s²). Its small-angle period is about

  1. 0.50 s
  2. 6.3 s
  3. 2.0 s Correct answer.
  4. 1.0 s

Explanation

T = 2π √(L/g) ≈ 6.28 × √(1/9.8) ≈ 6.28 × 0.32 ≈ 2.0 s.

Question 24Oscillations

In an ideal LC oscillation, when the capacitor charge is maximum,

  1. current is maximum and energy is all magnetic
  2. current is zero and energy is all electric in the capacitor Correct answer.
  3. both current and charge are zero
  4. inductor energy equals capacitor energy at that instant always only if R > 0

Explanation

At maximum |Q|, i = dQ/dt = 0 and energy is stored as (1/2)Q²/C in the capacitor.

Question 25Oscillations

A mass m = 0.50 kg on a spring with k = 200 N/m oscillates with amplitude 4.0 cm. The total energy is

  1. 0.080 J
  2. 4.0 J
  3. 0.0040 J
  4. 0.16 J Correct answer.

Explanation

E = (1/2) k A² = 0.5 × 200 × (0.040)² = 100 × 0.0016 = 0.16 J.

Question 26Oscillations

Critical damping means the oscillator

  1. oscillates forever without energy loss
  2. returns to equilibrium as quickly as possible without oscillating Correct answer.
  3. oscillates with amplitude growing in time
  4. has zero spring constant

Explanation

Critical damping is the boundary between underdamped oscillation and overdamped return; it returns fastest without crossing into oscillation.

Question 27Oscillations

For SHM, maximum acceleration has magnitude

  1. A / ω²
  2. g only
  3. ω² A Correct answer.
  4. ω A

Explanation

Differentiating x = A cos(ωt + φ) twice shows that the maximum acceleration magnitude is a_max = ω² A.

Question 28Oscillations

Quality factor Q of a resonance is higher when

  1. the oscillator cannot store energy
  2. damping is weaker (lighter damping) Correct answer.
  3. resistance is larger in a series RLC at fixed L,C
  4. energy is dissipated faster per cycle

Explanation

High Q means sharp resonance and light damping-energy decays slowly relative to the oscillation period.

Question 29Oscillations

Hooke’s law restoring force F = −k x implies SHM because

  1. force is constant
  2. force is proportional to velocity
  3. force is proportional to x²
  4. acceleration is proportional to −x (constant mass) Correct answer.

Explanation

F = m a = −k x ⇒ a = −(k/m) x, the defining SHM differential relation.

Question 30Modern Physics

In the photoelectric effect, increasing the intensity of monochromatic light above threshold mainly increases

  1. the number of photoelectrons per second (photocurrent), not the maximum KE per electron Correct answer.
  2. the maximum kinetic energy per photoelectron without changing rate
  3. the work function of the metal
  4. the threshold frequency of the metal

Explanation

More photons per second ⇒ higher electron emission rate; K_max = hf − W depends on frequency, not intensity (Einstein model).

Question 31Modern Physics

A hydrogen atom transitions from n = 3 to n = 2. The photon energy is

  1. 1.89 eV Correct answer.
  2. 13.6 eV
  3. 3.40 eV
  4. 10.2 eV

Explanation

E₃ = −13.6/9 ≈ −1.51 eV; E₂ = −13.6/4 = −3.40 eV; ΔE ≈ 1.89 eV.

Question 32Modern Physics

A photon’s momentum has magnitude

  1. h / λ Correct answer.
  2. h λ
  3. m c with m the electron mass
  4. zero because E = p c implies p = 0

Explanation

For photons E = p c and E = h c / λ ⇒ p = h/λ.

Question 33Modern Physics

The ground-state energy of positronium (electron–positron atom) is −6.8 eV in a simple Bohr-like model (reduced mass μ = m_e/2). Compared with hydrogen’s −13.6 eV, this illustrates that level energies scale with

  1. reduced mass Correct answer.
  2. only nuclear charge ignoring mass
  3. photon spin only
  4. sound speed in air

Explanation

Bohr energies ∝ μ; half the reduced mass halves the binding scale to −6.8 eV.

Question 34Modern Physics

A free electron has kinetic energy 4.0 eV. Its de Broglie wavelength compared with a 1.0 eV electron (same mass, non-relativistic) is

  1. half as large Correct answer.
  2. twice as large
  3. one-fourth as large
  4. four times as large

Explanation

K = p²/(2m) ⇒ p ∝ √K ⇒ λ ∝ 1/√K. For 4× kinetic energy, λ is halved.

Question 35Experimental Reasoning

In a GCE A-Level physics practical, a student records the diameter of a wire as 0.52 mm using a micrometer screw gauge with a zero error of +0.03 mm. What is the corrected diameter?

  1. 0.52 mm
  2. 0.58 mm
  3. 0.49 mm Correct answer.
  4. 0.55 mm

Explanation

A positive zero error means the instrument reads a value when the jaws are closed, so the error must be subtracted from every reading. Here 0.52 − 0.03 = 0.49 mm. Adding the error would double-count the fault rather than remove it.

Question 36Experimental Reasoning

A digital stopwatch displays elapsed time in increments of 0.01 s, for example 12.33 s,12.34 s and 12.35 s. What is the resolution of this displayed time?

  1. 12.34 s
  2. 0 s
  3. 0.005 s
  4. 0.01 s Correct answer.

Explanation

The smallest stated difference between adjacent displayed times is 0.01 s, so the display resolution is 0.01 s. This does not establish the uncertainty of a manually timed event: reaction time, instrument behavior and the measurement method can contribute additional uncertainty.

Question 37Experimental Reasoning

A metre rule is used to measure the length of a pendulum. Which practice best reduces the uncertainty in this length measurement?

  1. Align the rule with the measured length, view the relevant graduation with the line of sight perpendicular to the scale, and use a clear reference mark Correct answer.
  2. Measure the length only once to avoid confusion
  3. Hold the rule at an angle to the string for a firmer grip
  4. Round the reading to the nearest centimetre

Explanation

Aligning the rule with the measured length avoids an oblique length comparison. Viewing the graduation perpendicularly reduces parallax caused by separation between the indicator and scale. Use clear reference marks and measure the pendulum length from its pivot to the bob center. Resolution and repeatability still limit the result.

Question 38Experimental Reasoning

In an experiment to verify Ohm's law, a student plots a graph of current I against potential difference V. What does the gradient of this graph represent?

  1. The reciprocal of the resistance Correct answer.
  2. The resistance of the resistor
  3. The power dissipated
  4. The charge passed per second

Explanation

Ohm's law gives I = V/R, so plotting I on the vertical axis against V gives a straight line through the origin with gradient 1/R. The resistance itself would be the gradient only if V were plotted against I.

Question 39Experimental Reasoning

A student takes six readings of the time for 20 oscillations of a simple pendulum. Which quantity is the most appropriate to report as the period?

  1. The mean of the six times divided by 20 Correct answer.
  2. The mean of the six times multiplied by 20
  3. The difference between the largest and smallest times
  4. The largest of the six times divided by 20

Explanation

The period is the time for one oscillation, so the mean time for 20 oscillations is divided by 20. Averaging repeated timings reduces the effect of random reaction-time errors.

Question 40Experimental Reasoning

Which instrument is most suitable for measuring the internal diameter of a narrow glass tube?

  1. Metre rule
  2. Vernier calipers Correct answer.
  3. Micrometer screw gauge
  4. Travelling microscope

Explanation

Vernier calipers have internal jaws that can be inserted into a bore to measure internal diameter. A micrometer screw gauge is designed for external dimensions of thin objects and has no internal jaws.

Question 41Experimental Reasoning

A student measures the mass of a block five times and obtains values that agree closely with one another, but the accepted value is 20% higher. Which description fits these results?

  1. Accurate but not precise
  2. Neither accurate nor precise
  3. Both accurate and precise
  4. Precise but not accurate Correct answer.

Explanation

Precision describes how closely repeated readings agree with each other, while accuracy describes closeness to the true value. Consistent readings that all differ from the accepted value by the same amount indicate a systematic error, so the results are precise but inaccurate.

Question 42Experimental Reasoning

In a practical on the refraction of light through a glass block, which set of apparatus is essential for measuring the angles of incidence and refraction?

  1. Ammeter, voltmeter and rheostat
  2. Ray box, protractor and plain paper Correct answer.
  3. Stopwatch, metre rule and retort stand
  4. Thermometer, beaker and balance

Explanation

Angles are found by tracing the incident and refracted rays on paper and measuring them with a protractor, with a ray box providing a narrow beam. The other sets belong to timing, thermal and electrical experiments.

Question 43Experimental Reasoning

A spring's loading graph deviates from its initial straight line. Which additional observation most directly indicates permanent deformation? Assume the temperature is unchanged, the load is fully removed, the spring has settled, and the length measurements are checked.

  1. The loading graph alone is nonlinear, with no observation after unloading
  2. The unloaded spring remains longer than its original unloaded length Correct answer.
  3. The load axis is relabelled from newtons to kilonewtons
  4. The extension is reported in metres instead of centimetres

Explanation

A residual increase in the checked unloaded length under the stated conditions indicates permanent deformation. A departure from a straight loading graph establishes a failure of the initial linear relation, but does not by itself prove irreversible change. The limit of proportionality and the limit of elastic recovery are different concepts.

Question 44Experimental Reasoning

In an experiment to determine the specific heat capacity of a metal block, which improvement most directly reduces the largest systematic error?

  1. Insulating the block to reduce heat loss to the surroundings Correct answer.
  2. Increasing the mass of the block
  3. Using a thermometer with a finer scale
  4. Repeating the heating and averaging

Explanation

The main systematic error is energy lost from the block to the air, which makes the calculated specific heat capacity too high. Insulation reduces this loss; a finer thermometer or repeats address random reading errors, which are usually smaller.

Question 45Experimental Reasoning

A student measures the resistance of a wire using an ammeter and voltmeter and obtains R = 4.0 Ω. If the voltmeter reading is 2.0 V with an uncertainty of ±0.1 V and the ammeter reading is 0.50 A with an uncertainty of ±0.02 A, what is the percentage uncertainty in R?

  1. 9% Correct answer.
  2. 5%
  3. 2%
  4. 4%

Explanation

For a quotient, percentage uncertainties add: 0.1/2.0 gives 5% and 0.02/0.50 gives 4%, so the total is 9%. The absolute uncertainty in R is then 0.09 × 4.0 = 0.36 Ω.

Question 46Experimental Reasoning

In a Young's double-slit experiment, a student measures the fringe separation by counting across many fringes rather than measuring one fringe. Why is this done?

  1. It increases the wavelength of the light used
  2. It eliminates the effect of the screen distance
  3. It removes the need to measure the slit separation
  4. It reduces the percentage uncertainty in the fringe separation Correct answer.

Explanation

Measuring across, say, ten fringes and dividing gives a larger length, so the fixed reading uncertainty becomes a smaller fraction of the measured value. It does not change the physics of the interference pattern.

Question 47Experimental Reasoning

A student plots a graph of the square of the period T² against the length l of a simple pendulum. What does the gradient of this graph equal?

  1. g/4π²
  2. 2π/g
  3. 4π²/g Correct answer.
  4. g/2π

Explanation

From T = 2π√(l/g), squaring gives T² = (4π²/g)l, so a plot of T² against l is a straight line through the origin with gradient 4π²/g. Rearranging this gradient gives g.

Question 48Experimental Reasoning

When reading a burette or a thermometer, a student consistently views the meniscus from above. What type of error does this introduce and how is it corrected?

  1. A calibration error, corrected by using a different instrument
  2. A random error, corrected by taking more readings
  3. A zero error, corrected by subtracting a constant
  4. A systematic parallax error, corrected by aligning the eye with the meniscus Correct answer.

Explanation

Viewing from a fixed wrong angle shifts every reading in the same direction, so the error is systematic. The remedy is to place the eye level with the meniscus so the line of sight is perpendicular to the scale.

Question 49Experimental Reasoning

A student investigating the I–V characteristic of a filament lamp finds that the graph curves so that the gradient decreases as V increases. What conclusion follows?

  1. The lamp obeys Ohm's law at all voltages
  2. The filament has a constant resistance
  3. The resistance of the filament increases as its temperature rises Correct answer.
  4. The current is independent of the potential difference

Explanation

A decreasing gradient on an I–V graph means the ratio V/I rises, so the resistance increases. This happens because the filament heats up and its lattice ions scatter electrons more strongly.

Question 50Experimental Reasoning

In a moment experiment, a metre rule is balanced on a pivot and a known mass is hung at a measured distance. Which precaution most improves the reliability of the result?

  1. Finding the balance point of the rule before adding the masses Correct answer.
  2. Using the heaviest available mass
  3. Reading the distances to the nearest 10 cm
  4. Balancing the rule on a knife edge held by hand

Explanation

The rule's own weight acts at its centre of gravity, so its balance point must be located and allowed for before moments are taken. Holding the pivot by hand allows movement and adds random error.