IB Physics Paper 1B · HL data-based questions practice papers
HL · 20 marks · 2 questions on experimental data. Each generated paper follows the SL & HL Paper 1B · HL data-based questions format, with the same number and style of questions as the real exam, and comes with an IB-style mark scheme.
120 × 100 = 5.00% ✓ compared with 5371 × 100 = 1.35% for p, so V is more significant ✓
2
(e)
gradient falls
n decreases at high pressure (small V, large 1/V) ✓ so points at high 1/V fall below the line; the graph curves downwards ✓
2
Question 2Structured · 10 marksA* challenge
A steel ball is released from an electromagnet. An electronic timer starts when the current is switched off and stops when the ball passes through a light gate a height h below. The results are shown.
h / m ± 0.002
0.20
0.40
0.60
0.80
1.00
t / s ± 0.001
0.213
0.297
0.358
0.413
0.462
Table 2.1
(a)
Show that h = 12gt2 if the ball starts from rest and air resistance is negligible.
[1]
(b)
Calculate the percentage uncertainty in t2 for h = 0.20 m.
[2]
Fig. 2.1
(c)
Use the graph of h against t2 to determine g.
[3]
(d)
The magnet releases the ball a short time after the current is switched off. Explain the effect of this on the graph and on the value of g.
[3]
(e)
Suggest one other source of systematic error.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
s = ut + ½at², u = 0
s = ut + ½at2 with u = 0 and a = g ✓
1
(b)
0.941%
%Δt = 0.470% ✓ %Δ(t2) = 0.941% ✓
2
(c)
9.53 m s−2
gradient = g2✓ gradient = 4.77 m s−2✓ g = 9.53 m s−2 (accept ±5%) ✓
3
(d)
intercept, g unaffected
every time is too large by the same delay ✓ the graph is no longer a straight line through the origin (it has an intercept) ✓ the gradient, and so g, is only approximately unaffected; plotting √h against t removes the problem ✓
3
(e)
h measured to wrong point
h measured to the bottom/top of the ball rather than consistently (or the light gate beam is not horizontal) ✓
1
Question 3Structured · 10 marks
A student investigates how the resistance R of a wire varies with its length L. The diameter of the wire is measured with a micrometer as 0.28 ± 0.01 mm. R is measured with a digital ohmmeter.
L / m ± 0.001
0.20
0.40
0.60
0.80
1.00
R / Ω ± 0.1
1.7
3.3
4.8
6.4
8.0
Table 3.1
(a)
Calculate the percentage uncertainty in the cross-sectional area of the wire.
[2]
Fig. 3.1
(b)
Use the graph to determine the resistivity of the metal.
[3]
(c)
The line has a positive intercept on the R-axis. Suggest a reason for this.
[2]
(d)
The uncertainty in the gradient is estimated to be 4%. Determine the absolute uncertainty in the resistivity.
[2]
(e)
Suggest one way the student could reduce the percentage uncertainty in the cross-sectional area.
gradient = 7.82 Ω m−1✓ A = π(d2)2 = 6.16 × 10−8 m2✓ ρ = gradient × A = 4.8 × 10−7 Ω m (accept ±5%) ✓
3
(c)
contact resistance
a resistance that is present even when L = 0 ✓ e.g. resistance of the leads/contacts or a zero error of the ohmmeter (systematic error) ✓
2
(d)
± 5 × 10−8 Ω m
%Δρ = 4% + 7.14% = 11.1% ✓ Δρ = 5 × 10−8 Ω m ✓
2
(e)
repeat diameter readings
measure the diameter at several points and orientations and average, or use a thicker wire ✓
1
Question 4Structured · 11 marks
A student heats a 0.5 kg block of iron with an electrical heater. The potential difference across the heater is 12.0 ± 0.1 V and the current is 4.2 ± 0.1 A. A thermometer in the block is read every 60 s.
t / s
60
120
180
240
300
360
θ / °C ± 0.5
27.3
39.6
52.0
64.2
76.9
88.9
Table 4.1
(a)
Calculate the power of the heater and its percentage uncertainty.
[3]
(b)
Use the data to determine the specific heat capacity of the block.
[3]
(c)
The accepted value for iron is 450 J kg−1 K−1. Explain why the experimental value is larger.
[2]
(d)
Suggest why the temperature does not rise during the first part of the heating.
[1]
(e)
Suggest two improvements to the experiment.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
50.4 W ± 3.21%
P = VI = 50.4 W ✓ percentage uncertainties add ✓ 0.112.0 + 0.14.2 → 3.21% ✓
3
(b)
490 J kg−1 K−1
rate of temperature rise = 0.206 K s−1 from the linear region ✓ c = Pm × (Δθ/Δt)✓ c ≈ 490 J kg−1 K−1 (accept ±8%) ✓
3
(c)
thermal energy lost
some energy is transferred to the surroundings / not all goes into the block ✓ so the temperature rise is smaller than expected, giving a larger calculated c ✓
2
(d)
heater warms up first
the heater and thermometer take time to warm up / thermal energy takes time to conduct to the thermometer ✓
1
(e)
insulate; oil in holes
insulate the block (lagging) ✓ put oil/thermal paste in the holes for good thermal contact (or start below room temperature and finish the same amount above) ✓
2
Question 5Structured · 10 marks
A student measures the terminal potential difference V across a cell for different currents I, by changing a variable resistor in series with the cell and an ammeter.
I / A ± 0.01
0.08
0.17
0.25
0.33
0.42
V / V ± 0.02
1.38
1.29
1.20
1.08
1.00
Table 5.1
Fig. 5.1
(a)
Write down the equation relating V, I, the emf ε and the internal resistance r.
[1]
(b)
Use the graph to determine ε and r.
[4]
(c)
Calculate the maximum power that the cell can deliver to an external resistor.
[2]
(d)
The student suggests that a larger range of currents would improve the experiment. Discuss one advantage and one disadvantage.
[2]
(e)
Explain why the switch should be opened between readings.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
V = ε − Ir
V = ε − Ir ✓
1
(b)
ε ≈ 1.48 V, r ≈ 1.17 Ω
line of best fit drawn ✓ ε = y-intercept = 1.48 V ✓ r = −gradient ✓ r = 1.17 Ω (accept ±10%) ✓
4
(c)
0.470 W
maximum when R = r, so P = ε24r✓ 0.470 W ✓
2
(d)
range vs heating
a wider range spreads the points and reduces the uncertainty in the gradient ✓ larger currents heat the cell, changing r, or drain it so ε falls ✓
2
(e)
avoid heating / discharge
so the cell does not heat up or run down, keeping ε and r constant ✓
1
Question 6Structured · 9 marks
A student shines a ray of light into a semicircular glass block and measures the angle of incidence i in air and the angle of refraction r in the glass with a protractor (± 1°).
i / °
20
30
40
50
60
70
r / °
13
19
25
31
35
39
Table 6.1
(a)
Calculate the percentage uncertainty in r for i = 20°, and explain why small angles give less reliable results.
[2]
Fig. 6.1
(b)
The graph shows sin i against sin r. Use it to determine the refractive index of the glass.
[2]
(c)
Calculate the critical angle for the glass.
[2]
(d)
Calculate the speed of light in the glass.
[2]
(e)
Explain why a semicircular block is used, with the ray aimed at the centre of its flat face.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
7.69%
113 × 100 = 7.69% ✓ the same ±1° is a larger fraction of a small angle ✓
2
(b)
1.47
refractive index = gradient ✓ n = 1.47 (accept 1.41 to 1.53) ✓
2
(c)
42.8°
sin c = 1n✓ c = 42.8° ✓
2
(d)
2.04 × 108 m s−1
v = cn✓ 2.04 × 108 m s−1✓
2
(e)
no refraction at curved face
the ray leaves along a radius, so it is not refracted at the curved surface ✓
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Original IB-style practice questions written for mentorriyaz.in. These are not IB past paper questions, and this site is not affiliated with or endorsed by the International Baccalaureate Organization. Updated 2026-11-04. Paper generator · mentorriyaz.in