Experimental Solutions. Language: English

IPhO 2022

E1: Planet - SOLUTION

Plotting

A.1: The free-fall acceleration g can be found by dropping the ball from low heights such that the air frictionand effects from the curvature of the planet are minimized. We also choose the radius and density of the ball
to be as big as possible to minimize the effect of air friction, i.e. setting r = 50 cm, ρ = 10 g/cm3 . The drop heightis then given by h = gt2 /2, and so we can find g from theslope of t2 vs h. From the graph, we measure the slope
2/g = 0.127 s2 /m and its error ∆(2/g) = 0.004 s2 /m and so
g = 15.7 m/s2 with an error of ∆g = 0.5 m/s2 .
r = 50 cm, ρ = 10 g/cm3
h(m) s(m) t(s) t2 (s2 )0
0.0
0.0 0.020
0.0
1.7 2.940
0.0
2.2 4.860
0.0
2.8 7.880
0.1
3.2 10.2
100
0.1
3.5 12.2120
0.2
3.9 15.2
140
0.0
4.1 16.8
160
0.1
4.6 21.2
180
0.1
4.8 23.0200
0.1
5.1 26.0

Fit line

Valuesoverall plot
points don’t cover 60% of thearea
missing axis labels
missing axis units
one plotting mistake
two or more plotting mistakes
drawn on graphline passes through origin
slope computed with units
uncertainty of slope computed
15.0 m/s2 ≤ g ≤ 16.4 m/s2
14.3 m/s2 ≤ g ≤ 17.1 m/s2
units for value∆g ≤ 0.7 m/s2
∆g ≤ 1.4 m/s2
units for error
sum

0.30 pts-0.10 pts
-0.05 pts
-0.05 pts
-0.05/-0.10-0.10/-0.10
0.05 pts
0.05 pts
0.10 pts
0.10 pts
0.20/0.20
0.10/0.20
0.05 pts
0.20/0.20
0.10/0.20
0.05 pts
2.0 pts

Points are added additively (including negative points),
except for blocks of grey background, where the option with maximal points should be chosen (in absolutevalue)
A.2: How far one can see from on top of the tower canbe related to the radius of the planet via the right triangle shown in the figure below. Applying the Pythagoras
theorem on the triangle, one gets (R + H)2 = L2 + R2 andsoL2 − H 2
R== 13 200 km.
2H

25

t2 , s 2

20
15
10
5
0
0

50

100
h, m

150

200

Marking scheme:
TheoryData2h = gt /2
varying only hmaximising r
maximising ρtable has units
h distributed roughly uniformly
hmax < 300 m
hmax − hmin > 100 mcorrect calculations of derived quantities
7 or more measurements
6 measurements
5 measurements
4 or fewer measurements

Marking scheme:
0.20 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.30/0.30
0.25/0.30
0.20/0.30
0.10/0.30

TheoryValuescorrect geometry (either a figure or implicitly assumed)
correct formula
correct value
sum

0.20 pts
0.20 pts
0.10 pts
0.5 pts

A.3: From Newton’s law of gravity, g = GM /R2 . Hence,M=

gR2= 4.2 × 1025 kg.
G

By adding the errors in quadrature, we find the error∆M =∆g
M = 0.2 × 1025 kg.
g

Our estimation of free-fall acceleration has a contribution from the centrifugal force caused by the rotation

Experimental Solutions. Language: English

IPhO 2022of the planet. This serves to reduce the acceleration on
the surface and hence decrease our estimation of theplanet’s mass.

15

s, m

Marking scheme:

TheoryValuescorrect formula
correct phenomena
3.9×1025 kg ≤ M ≤ 4.5×1025 kg
3.6×1025 kg ≤ M ≤ 4.8×1025 kg
missing units for value∆M ≤ 0.3 × 1025 kg∆M ≤ 0.6 × 1025 kg
missing units for error
sum

0.10 pts
0.20 pts
0.10/0.10
0.05/0.10-0.05 pts
0.10/0.10
0.05/0.10-0.05 pts
0.5 pts

The student can’t get overall negative points for value
nor error (for example when the value is completely outof range and the units are wrong).

10

5

0
0

r = 5 cm, ρ = 0.1 g/cm3
h(m) s(m) t(s)
0
0.0
0.0
20
0.3
1.7
40
0.8
2.6
60
1.5
3.4
80
2.4
4.2
100
3.2
5.0
120
4.1
5.7
140
5.0
6.5
160
5.8
7.3
180
6.9
7.9
200
7.4
8.5
240
9.4
10.0
280
11.7
11.6
320
13.4
12.9
360
15.4
14.4
400
16.4
15.4

10

15

t, s
Marking scheme:
TheoryData

B.1: In general, if the variations in gravitational acceleration are small (as is the case here as H ≪ R), as a
response to air drag, objects tend to terminal velocity
where they experience no net acceleration. In the reference frame of air, this corresponds to the object fallingstraight down with some terminal speed vt . In the labframe, the object then has horizontal and vertical speedsof u and vt respectively.
In order to find u, we can choose to drop an object thatreaches terminal velocity as fast as possible and then observe how the displacement s relates to the fall time t.
When terminal velocity is reached, we expect s = s0 + ut,
where s0 captures the displacement related to reachingterminal velocity. To maximize the effects of air drag, we
minimize radius and density, i.e. setting ρ = 0.1 g/cm3 ,and r = 5 cm. Plotting s vs t, we measure the slope to beu = 1.31 m/s with an error of ∆u = 0.04 m/s.

5

Plotting

Fit line

Valuesidea of reaching terminal velocity as fast as possibles = s0 + ut
varying only hminimising r
minimising ρtable has units
h distributed roughly uniformly
hmax ≥ 300 m
hmax − hmin ≥ 300 m
7 or more measurements
6 measurements
5 measurements
4 or fewer measurements
overall plot
points don’t cover 60% of thearea
missing axis labels
missing axis units
one plotting mistake
two or more plotting mistakes
drawn on graphslope computed with units
uncertainty of slope computed
1.25 m/s ≤ u ≤ 1.37 m/s
1.19 m/s ≤ u ≤ 1.43 m/sunits for value∆u ≤ 0.06 m/s∆u ≤ 0.12 m/sunits for error
sum

0.15 pts
0.10 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.30/0.30
0.25/0.30
0.20/0.30
0.10/0.30
0.30 pts-0.10 pts
-0.05 pts
-0.05 pts
-0.05/-0.10-0.10/-0.10
0.10 pts
0.10 pts
0.10 pts
0.20/0.20
0.10/0.20
0.05 pts
0.20/0.20
0.10/0.20
0.05 pts
2.0 pts

B.2: By keeping the measurements close to the surface,
we can assume to a good approximation uniform air
density. Then, using similar reasoning as before, we expect h = h0 + vt0 t, where h0 captures the part of reachingterminal velocity.
At terminal velocity, the drag force balances out gravitational acceleration:
mg = 0.24Aρa vt2 .

Experimental Solutions. Language: English

IPhO 2022
Using m = 4πρr3 /3 and A = πr2 , we get√
4ρrg
vt (ρa ) =.
3 · 0.24ρasum

On the surface, vt0 = vt (ρa = ρa0 ). Using the measurements from the last subtask, we can plot t vs h and measure the slope to be 1/vt0 = 0.037 s/m with an error of∆(1/vt0 ) = 0.002 s/m. Hence, vt0 = 27.0 m/s, ∆vt0 =2∆(1/vt0 )/vt0= 2 m/s. Now,
ρa0 =

4ρrg
3
2 = 0.60 kg/m .
3 · 0.24vt0

and the error is∆ρa0 =

2∆vt0
ρa0 = 0.07 kg/m3 .
vt0

8

t, s

6
4
2
0
0

50

100
h, m

150

200

Marking scheme:
TheoryData

Plotting

Fit line

Valuesh = h0 + vt0 t
formula for terminal velocity
final expression for ρa0
reusing the data from the last
subparthmax ≤ 200 m
6 or more measurements
overall plot
points don’t cover 60% of thearea
missing axis labels
missing axis units
one plotting mistake
two or more plotting mistakes
drawn on graphslope computed with units
uncertainty of slope computed
0.52 kg/m3 ≤ ρa0 ≤ 0.68 kg/m3
0.44 kg/m3 ≤ ρa0 ≤ 0.76 kg/m3∆ρa0 ≤ 0.08 kg/m3∆ρa0 ≤ 0.16 kg/m3
no units for both value and error

0.05 pts
0.10 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.25 pts-0.05 pts
-0.05 pts
-0.05 pts
-0.05/-0.10-0.10/-0.10
0.05 pts
0.05 pts
0.10 pts
0.10/0.10
0.05/0.10
0.10/0.10
0.05/0.10-0.05 pts

1.0 pts

B.3: Due to the adiabatic profile of the atmosphere, thefurther up you go, the more the temperature and air density decreases, but the terminal velocity increases. We
can estimate the terminal velocity of the ball at different
heights by comparing the dropping time of a ball withthe smallest possible terminal velocity (so minimal density and radius). This hence gives a direct probe for the
air density and thus the height of the atmosphere.
If the ball reaches terminal velocity instantly, then the
difference in falling time between dropping the ball atheights h1 and h2 > h1 comes simply from h1 < h < h2 .
This is because in both cases the ball falls for the sameamount of time at h < h1 (because the terminal velocity
only depends on height). Then, if h2 − h1 ≪ h1 , we canestimate)(h2 − h1
h1 + h2
≈
.
(1)
vt
2
t(h2 ) − t(h1 )In reality, the ball doesn’t reach the terminal velocity
instantaneously. However, it turns out we can, to a good
approximation, neglect this effect. As a rough order ofmagnitude estimation, on the ground level, the ball experiences a time difference of vt0 /(2g) = 0.8 s comparedto the instantaneous case. This difference will increaseas the ball is dropped from further up, but as long as theatmosphere isn’t too much sparser in the upper parts ofthe tower (we can verify this later), the difference will
be insignificant compared to the total falling time of theball. Hence, we approximate the terminal velocity via
equation (1).
Because the calculated velocities are very sensitive on
the measured quantities, we do repeated measurements
throughout the whole height of the tower.

h(m)
200
400
600
800
1000
1200
1400
1600
1800
2000

r = 5 cm, ρ = 0.1 g/cm3
s1 (m) t1 (s) s2 (m) t2 (s) s3 (m)
7.6
8.4
7.8
8.6
7.8
17.0
15.7 16.9
15.6 17.3
26.1
22.6 25.4
22.2 26.2
33.6
28.5 34.6
29.2 34.3
41.1
34.3 43.0
35.7 43.3
51.1
41.9 50.2
41.2 50.0
57.9
47.2 58.8
47.8 58.7
65.5
53.0 65.1
52.8 65.3
70.9
57.1 72.2
58.2 71.4
78.5
62.9 79.6
63.8 79.5

t3 (s)
8.6
15.7
22.7
29.1
35.8
41.1
47.8
52.9
57.5
63.7

Using equation (1) we make a separate table withvelocities, while also adding the ground level velocity
found in one of the earlier part (we set it at h = 100 m because that was the centre of the range of measurements).
We find air density using
ρa =

4ρrg
.
3 · 0.24vt2

From the density profile of an adiabatic atmosphere,
(
)
h
0.4
ρaγ−1 = ρ0.4
=
ρ
1
−
.
a
a0
H0
Hence, we find H0 by plotting ρ0.4
a0 against h and fitting astraight line.

Experimental Solutions. Language: English

IPhO 2022
r = 5 cm, ρ = 0.1 g/cm3
3 0.4)v(m/s) ρa (kg/m3 ) ρ0.4
a ((kg/m )
27.0
0.599
0.814
28.0
0.556
0.791
29.3
0.510
0.764
31.1
0.452
0.728
31.6
0.438
0.719
32.6
0.411
0.701
32.3
0.420
0.707
37.7
0.307
0.624
42.6
0.241
0.566
34.1
0.376
0.676

h(m)
100
300
500
700
900
1100
1300
1500
1700
1900

From the plot, we measure the slope a = −ρ0.4
a0 /H0 =−4
3 0.4−1.1 × 10 (kg/m ) /m and the intercept b = ρ2.5
a0 =0.4
0.82 (kg/m3 ) so H0 = −b/a = 7500 m. We calculate theerror from two reasonably chosen lines that correspondto maximal and minimal estimates for H0(0.40.80 (kg/m3 )
1
∆H0 ≈
−
0.4
2
−8.4 × 10−5 (kg/m3 ) /m)
0.4
0.83 (kg/m3 )≈ 2000 m.
+
0.4−1.4 × 10−4 (kg/m3 ) /m

)
1
h (
1−
h
v t0
4H0 (γ − 1)
t
1
1
≈
−
h.
h
v t0
4vt0 H0 (γ − 1)t≈Plotting t/h vs h and calculating H0 similarly to before(by calculating the intercept and the slope), we get H0 ≈6300 m, which falls within the error range. However,
because of the approximations, this approach will beawarded a maximum of 2.0 out of 3.0 points (the following grading scheme still applies, but is capped out at 2.0).
Marking scheme:
Theoryapproximating vt0 via finite
difference
reasoning why the ballreaches terminal velocity
effectively instantaneously
linearising vt0 vs h
expressing H0 in terms of theslope/intercept
varying only hminimising r
minimising ρtable has units
h distributed roughly uniformly
hmax − hmin ≥ 1800 m
calculating derived quantities
15 or more measurements(can be repeat)
10 - 14 measurements
1 - 9 measurements
overall plot
points don’t cover 60% of thearea
missing axis labels
missing axis units
one plotting mistake
two or more plotting mistakes
drawn on graphslope computed with units
uncertainty of slope computed
5500 m ≤ H0 ≤ 9500 m
3500 m ≤ H0 ≤ 11 500 m
units for value∆H0 ≤ 2000 m/s∆H0 ≤ 4000 m/s
units for error
sum

Data

We can also confirm that our assumption about thedensity of the atmosphere not dropping significantly inthe upper parts of the tower holds true.

3ρ0.4
a , (kg/m )

0.4

0.80

Plotting

0.75
0.70
0.65
0.60

Fit line

0.55
0

500

1000
h, m

1500

2000
ValuesAlternative, less accurate solution
In this approach, it’s assumed that when the air drag ismaximised, the ball falls at the terminal velocity vt0 for
the whole duration of the fall. This gives1(dhh )− 2(γ−1)= vt (h) = vt0 · 1 −.
dt
H0Rearranging and integrating,∫1(1h ) 2(γ−1)
t≈dh 1 −.
v t0
H0
So far this is exact and differs from the exact solution
by the “speeding up” term which is a constant and has
a smaller relative contribution the higher up one goes.
In order to approximate this integral, we can do a firstorder binomial expansion to get

0.30 pts
0.15 pts
0.25 pts
0.10 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.10 pts
0.20 pts
0.45/0.45
0.30/0.45
0.15/0.45
0.30 pts-0.10 pts
-0.05 pts
-0.05 pts
-0.05/-0.10-0.10/-0.10
0.10 pts
0.15 pts
0.15 pts
0.20/0.20
0.10/0.20
0.05 pts
0.20/0.20
0.10/0.20
0.05 pts
3.0 pts

B.4: From the expression for adiabatic atmosphere we
have
RT0 γ
H0 =
µg γ − 1
so
µ=

RT0 γ
−1
−1
= 72 g mol ≈ 70 g mol
H0 g γ − 1

and

√
∆µ =

∆g 2
∆H02
−1
+
µ = 20 g mol .
H02
g2

Experimental Solutions. Language: English

IPhO 2022

r = 50 cm, ρ = 10 g/cm3
h(m) s(m) h1.5 (m1.5 )
0
0.0
0
200
0.2
2800
400
0.5
8000
600
0.8
14700
800
1.2
22600
1000 1.8
31600
1200 2.341600
1400 2.852400
1600 3.4
64000
1800 4.076400
2000 4.889400

From ideal gas law,
p0 =
and

√
∆p0 =

ρa0 RT0= 20 000 Pa
µ

∆µ2
∆ρ2a0
+
p0 = 6000 Pa.
µ2
ρ2a0

Marking scheme:

Valuescorrect expression for µ
correct expression for p0−1−1
45 g mol ≤ µ ≤ 95 g mol−1∆µ ≤ 25 g mol
12 000 Pa ≤ p0 ≤ 28 000 Pa∆p ≤ 8000 Pa
sum

0.15 pts
0.15 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.5 pts

C.1: Our goal is to find the rotation speed Ω of the planet.
The rotation of the planet affects the ball’s trajectory
via centrifugal and Coriolis force. The centrifugal force,
however, due to H ≪ R is impossible to disentangle fromgravitational acceleration. Coriolis force affects the ball⃗ × ⃗v . This is perpendicular
via acceleration ⃗acor = −2Ω
to both the velocity of the ball and rotation axis of theplanet. Hence, it’s directed along the equator, and increases linearly with the falling speed. Thus, the horizontal acceleration is given by ax = 2Ωvy + adrag .

5
4
s, m

Theory

3
2
1
0
0

20000

40000
60000
1.5
1.5
h ,m

80000

The procedure is then to minimize the effect of air drag(maximal radius and density) and hope that the CorioAlternative solution.
lis effect contributes enough to the horizontal displacement.∫ If we neglect air drag, ∫then ax = 2Ωvy = 2Ωgt so
An alternative approach is to consider the system in
vx = ax dt = Ωgt2 and x = vx dt = Ωgt3 /3. The final the non-rotating frame (where we don’t have to deal
displacement will then be s = gΩt3f /3, where the falling with fictitious forces). In there, the ball starts off withspeed v0 = Ω(R + H). Due to the conservation of angulartime satisfies H = gt2f /2. Putting them together, we get
momentum, as the ball drops towards the ground, the√ball’s angular speed will start increasing and the ground
3
2Ω 2H
s=
.
will start lagging behind (the ground rotates with Ω). At
3gheight h, when the ball moves with angular speed ω, theconservation of angular momentum reads ω(R + h)2 =By varying the radius/density, we do indeed confirm thatΩ(R + H)2 and so the angular lag between the ball andthe effect of Coriolis force is significant, on the order ofthe ground iscouple of meters. By doing a suitable number of mea(())2
surements in the range 0 to 2000 m and plotting s vs h1.5 ,R+H
H −hwe measure the slope∆ω = ω − Ω = Ω− 1 ≈ 2Ω
.
R+h
R
√
2Ω 2
a=
= 5.3 × 10−5 m−1/2
The positional velocity shift along the ground is then3gvx = ∆ωR = 2Ω(H − h) = Ωgt2 . We recover the same
expression as for Coriolis force, and from there we proand the error
ceed the same way as before.
−6−1/2∆a = 1.1 × 10 m
Marking scheme:
Theorysuch that
T =
and

4π
2π
=
Ω
3a
√(

∆T =

0.5

√

∆g
g

2
= 28 000 s ≈ 8h
g

)2
+

∆a2
T = 0.2 h.
a2

Data

Deriving s(h)linearising s vs hvarying only hminimising r and ρ
table has units
h distributed roughly uniformly

0.80 pts
0.10 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts

Experimental Solutions. Language: English

IPhO 2022

Plotting

Fit line

Valueshmax − hmin ≥ 1800 m
calculating derived quantities
7 or more measurements
6 measurements
5 measurements
4 or fewer measurements
overall plot
points don’t cover 60% of thearea
missing axis labels
missing axis units
one plotting mistake
two or more plotting mistakes
drawn on graphslope computed with units
uncertainty of slope computed
27 000 s ≤ T ≤ 29 000 s
26 000 s ≤ T ≤ 30 000 s
missing units for value∆T ≤ 1000 s∆T ≤ 2000 s
missing units for error
sum

0.05 pts
0.05 pts
0.30/0.30
0.25/0.30
0.20/0.30
0.10/0.30
0.30 pts-0.10 pts
-0.05 pts
-0.05 pts
-0.05/-0.10-0.10/-0.10
0.10 pts
0.10 pts
0.10 pts
0.20/0.20
0.10/0.20-0.05 pts
0.20/0.20
0.10/0.20-0.05 pts
2.5 pts

Experimental Solutions. Language: English

IPhO 2022

E2: Cylindrical Diode - SOLUTION
Original solutions distributed on day of exam; this isshown in black.
Corrections and significant changes are shown in red
below; as of July 15, 12:30 PM China time.
Pay particular attention to the bounds on G (C)in B.1, and the bounds on B in C.3, as these corrections were made fairly late in the grading cycle, andmight not have been caught by all markers!
Take the logarithm of Equation 1,
log I∞ = log C + α log Rc + β log Le + γ log V
A.1: Collect data by varying V . To minimize error, select maximum values for all fixed variables, this means
Le = 99 cm, Rc = 10 cm, and Re = 1.0 cm. Distribute thevoltages logarithmically between 10 and 2000
V (V)
10
20
50
100
200
500
1000
2000

I (mA)
5
13
52
147
415
1620
4630
12900

log V
1.0
1.3
1.7
2.0
2.3
2.7
3.0
3.3

log I
0.70
1.11
1.72
2.17
2.62
3.21
3.67
4.11

Plot this on a graph; the best fit line islog I = 1.490 log V − 0.8095

4

log I

3

2

1
1

1.5

2
2.5
log V

3

3.5

so γ = 1.49.
A statistical analysis of the uncertainty in the slope
yields γ = 1.490 ± 0.005.
Assessing the slope by visually fitting lines through theerror bars on the points requires considering that error
bars on a log axis are given by
(
)
ln y
1 δy
δ(log y) = δ=ln 10
ln 10 ySince the largest relative error is in the smallest valued
quantity, the focus is on δV /V for V = 10V and δI/I for

I = 5mA. The error bars associated with the log-log plotat that point are then
(1 ± 0.02, 0.70 ± 0.04)
The other error bars are smaller; focusing on that point
alone we can fit two extreme lines and get
γ = 1.485 ± 0.025
Either approach is acceptable.
Marking scheme:Data

vary only V
Re ≥ 1cm
Rc ≥ 5Re
Le ≥ 90cm
table has units
V distributed as log
Vmax ≥ 1000V
Vmin ≥ 10V
Vmin ≤ 50V
Correct calculations
7 or more points
6 points
5 points
4 or fewer points
Plotting covers > 50% of area
Axis labels
Axis units correct
one plotting mistake
two or more plotting mistakes
Fitline drawn on graphslope correctly computedwith units
1.45 < γ < 1.55
uncertainty of slope computed
δγ ≤ 0.03
sum

0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.30/0.30
0.25/0.30
0.20/0.30
0.10/0.30
0.10 pts
0.05 pts
0.05 pts-0.05/-0.10-0.10/-0.10
0.10 pts
0.10 pts
0.10 pts
0.10 pts
0.10 pts
1.5 pts

Measured data should be entered into spreadsheetthat will calculate results; if deviation is too large, data
point should not count.
Evidence of reverse engineering should result inzero points for the entire section
A.2: Collect data by varying Le . To minimize error, selectmaximum values for all fixed variables, this means V =2000 V, Rc = 10 cm, and Re = 1 cm.
Le (cm)
99
90
80
70
60
55
50

I (mA)
13000
11800
10700
9170
7960
7310
6700

log Le
1.996
1.954
1.903
1.845
1.778
1.740
1.699

log I
4.144
4.072
4.029
3.962
3.901
3.864
3.826

Plot this on a graph; the best fit line islog I = 0.9767 log Le + 2.1649

Experimental Solutions. Language: English

IPhO 2022

Rc (cm)
20
19
18
16
14
12
10

4.1

log I

4

I (mA)
6640
6970
7380
8240
9390
11000
13100

log Rc
1.301
1.279
1.255
1.204
1.146
1.079
1.000

log I
3.822
3.843
3.868
3.916
3.973
4.041
4.117

Plot this on a graph; the best fit line is

3.9

log I = −0.9816 log Rc + 5.1000
3.8

1.7

1.75

1.8

1.85 1.9
log Le

1.95

2
4.1

A statistical analysis of the uncertainty in the slope
yields β = 0.98 ± 0.02.
Graphical fitting of the steepest and shallowest lines
yields β = 0.97 ± 0.02.
Marking scheme:

log I

so β = 0.9767.
4

3.9

3.8
1

Data

vary only Le
Re ≥ 1cm
Rc ≥ 5ReV ≥ 100V
table has units
Le distributed evenly
Le,max ≥ 90cm
Le,min ≥ 3Rc
Le,min ≤ 50cm
Correct calculations of derived quantities
7 or more points
6 points
5 points
4 or fewer points
Plotting covers > 50% of area
Axis labels
Axis units correct
one plotting mistake
two or more plotting mistakes
Fitline drawn on graphslope correctly computedwith units
0.97 < β < 1.03
uncertainty of slope computedδβ ≤ 0.03
sum

0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.30/0.30
0.25/0.30
0.20/0.30
0.10/0.30
0.10 pts
0.05 pts
0.05 pts-0.05/-0.10-0.10/-0.10
0.10 pts
0.10 pts
0.10 pts
0.10 pts
0.10 pts
1.5 pts

A.3: Collect data by varying Rc . To minimize error, selectmaximum values for all fixed variables, this means V =2000 V, Le = 99 cm, and Re = Rc /10 cm.

1.05

1.1

1.15 1.2
log Rc1.25

1.3

so α = −0.9824.
A statistical analysis of the uncertainty in the slope
yields β = −0.98 ± 0.01.
Graphical fitting of the steepest and shallowest lines
yields β = 0.97 ± 0.02.
Marking scheme:

Experimental Solutions. Language: English

IPhO 2022
Data

vary only Rc , Re , keeping ratio constant.
Re ≥ 1cm
Le ≥ 50cm
V ≥ 100V
table has units
Rc distributed evenly
Rc,max ≥ 15cm
Rc ≥ 5Re
Rc,min ≤ 10cm
Correct calculations of derived quantities
7 or more points
6 points
5 points
4 or fewer points
Plotting covers > 50% of area
Axis labels
Axis units correct
one plotting mistake
two or more plotting mistakes
Fitline drawn on graphslope correctly computedwith units−1.03 < α < −0.97
uncertainty of slope computedδα ≤ 0.03
sum

0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.30/0.30
0.25/0.30
0.20/0.30
0.10/0.30
0.10 pts
0.05 pts
0.05 pts-0.05/-0.10-0.10/-0.10
0.10 pts
0.10 pts
0.10 pts
0.10 pts
0.10 pts
1.5 pts

B.1: Use all three sets of data, and the exponents fromall three, and then average the results

TheoryFit

clear statement
Used Rc = 10Re
C computed
More than 9 data points
8 or 9 data points
7 or 8 data points
5 or 6 data points
C has correct units
14.3 ≤ C ≤ 14.9
14.0 ≤ C ≤ 15.2
uncertainty computed
0.1 < δC ≤ 0.3
0 < δC ≤ 0.5
sum

Clear statement of theory means that somewhere
there is a justification for the data they are collecting andusing. This can be in the form of the log formula; words
are not necessary. Reusing data is okay.
The dimensions of C (G in the question paper) should
be sensible; writing irrational exponents on various
units is inconsistent with the physics.
The values for C (G in the question paper) have been
updated, they show as red.
C.1: Start by assuming that Le matters, and look at values near Rc . Repeat for other variables. Remember that
C depends on the ratio between Rc /Re , so change thesetogether!
Using nearest half integers, we have for the first equation
Le
I∞ = C V 3/2
Rc
so thatlog C = log I − 1.495 log V − 0.9854 log Le + 0.9781 log Rc
which gives
C = (0.0146 ± 0.0003)mA/V3/2
Using exponents of 1.5, 1, and -1 are acceptable.
The theoretical value is approximately:
8πϵ0
9

√

2e
≈ 1.47 × 10−5 A/V3/2 .
m

Note that there is a nasty correction (the texts usuallycall it β, which is not the same as our exponent), that we
use in the code, but aren’t expecting students to find, because of this correction, we don’t expect the theoreticalvalue to hold. Students who try to solve the theoreticalproblem will be vexed by this.
For space reasons, we write numerical C below without explicit units, but using the units of µA/V3/2 , that is
C = 14.6µA/V3/2
Students must have clear units!
Marking scheme:

0.20 pts
0.10 pts
0.10 pts
0.20/0.20 pts
0.15/0.20 pts
0.10/0.20 pts
0.05/0.20 pts
0.10 pts
0.10/0.10 pts
0.05/0.10 pts
0.10 pts
0.10 pts
0.05/0.10 pts
1.0 pts

Rc
cm
10
12
8
10
10
12
8
10
10

Re
cm
1
1.2
0.8
1
1
1.2
0.8
1
1

Le
cm
10
10
10
12
8
12
8
10
10

F =

Imeasured
Le 3/2
V
CR
c

V
V
1000
1000
1000
1000
1000
1000
1000
1100
900

I
mA
535
470
647
630
451
537
537
617
457

I∞
mA
500
416
624
599
400
500
500
576
426

F
1.071
1.129
1.036
1.051
1.129
1.075
1.075
1.071
1.072

From this we conclude that if Rc ↑, F ↑; if Le ↑, F ↓; if
V ↑, F doesn’t change.
Also, we notice that the ratio Rc /Le seems to be the important quantity.
Marking scheme:
Data
Data

clearly collected
Rc ↑ =⇒ F ↑Le ↑ =⇒ F ↓Vc ↑: F no significant change
Re ↑: F ↑ or doesn’t change
sum

0.10 pts
0.10 pts
0.10 pts
0.10 pts
0.10 pts
0.5 pts

For Re increasing, either F increasing or F no significant change is acceptable.

Experimental Solutions. Language: English

IPhO 2022

If a student keeps Rc /Re ratio constant to do this, or ifthey keep Re constant while varying Rc , they will haveno penalty. Either approach is fine; we didn’t ask them
to work out the behavior of the function G(Rc /Re ) except
at the ratio of 10.
C.2: We propose
Rc
F =A+BLe
with x = Rc /Le .
There were changes in the problem statement, so
Marking scheme:
Checked x = Rc /Le
sum

0.50 pts
0.5 pts

If a student does not check anything, and in their working space claims that x = Le /Rc , then they would get+0.25 pts.
C.3: It is important to collect data that varies Rc and Le
independently, so as to not bias our hypothesis. We will
also keep the ratio with Rc /Re = 10, in order to avoid
other effects with the constant in part B. The potential
was kept at a constant 2000 V in all of the measurements
below.
What follows is a corrected data table that fixes theG error from part B; it provides different limits on theacceptable values for the slope!
Rc (cm)
20
20
20
20
20
6
8
10
15
20

Le (cm)
10
15
20
30
40
10
10
10
10
10

I (mA)
898
1210
1520
2160
2810
2420
1840
1520
1100
902

I∞
654
981
1308
1962
2616
2180
1635
1308
872
654

x
2.000
1.333
1.000
0.667
0.500
0.600
0.800
1.000
1.500
2.000

F
1.380
1.237
1.172
1.111
1.081
1.099
1.135
1.172
1.272
1.380

We plot the results below; blue are the values of fixed
Rc while green are the values of fixed Le .
1.4

F

1.3

1.2

1.1
0.5

1

1.5
x

The result is about
F (x) = 0.974 + 0.202x
Marking scheme:

2

Data

vary Levary Rc
Re ≥ 0.5cm
Rc ≥ 5ReV ≥ 500V
table has units
Le ≥ 10cm
Le,max ≤ 40cm
Le well distributed
Rc well distributed
Correct calculations of derived quantities
10 or more points
9 points
8 points
6 or 7 points
5 or fewer points
Plotting covers > 50% of area
Axis labels
Axis units correct
one plotting mistake
two or more plotting mistakes
Fitline drawn on graphslope correctly computedwith units
0.19 < B < 0.21
sum

0.10 pts
0.10 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.05 pts
0.10 pts
0.30/0.30
0.25/0.30
0.20/0.30
0.10/0.30
0.05/0.30
0.10 pts
0.05 pts
0.05 pts-0.05/-0.10-0.10/-0.10
0.10 pts
0.10 pts
0.10 pts
1.5 pts

A student can keep Re fixed, or the ratio Rc /Re fixed
when doing this.