12+11+05+Kogumik+FINAL

Proceedings of the 43rd International Physics Olympiad15th – 24th of July 2012 • Tallinn and Tartu, Estonia

PhysIcsIs LOvE

4

Estonian Ministry of Education and Research

The Estonian Information Technology Foundation

Designed by loremipsum.ee

Edited by OÜ Komadisain

Published in 2012

www.ipho2012.ee

IPhO illustrations by Toom Tragel

Cover photo by Henry Teigar

Proceedings of the 43rd International Physics Olympiad is licensed under a Creative

Commons Attribution 3.0 Estonia License.

5

ContentssPEEchEs 13Opening Ceremony 13

Closing Ceremony 19

PEOPLE 25Participants 25

Organizers 54

PrOgrams 63Students 63

Leaders and Observers 67

PrObLEms and sOLuTIOns 73Theoretical Competition 73

Experimental Competition 107

rEsuLTs 131Gold Medalists 131

Silver Medalists 133

Bronze Medalists 135

Honorable Mentions 137

Special Prizes 139

Statistics of the competition results 140

6

InTErnaTIOnaL bOard 151Minutes 151

Statutes 157

Syllabus 168

aPPEndIcEs 175Tartu – The World Capital of Physics 175

Circulars 183

Newsletter 197

7

ForewordEstonia had the honour of hosting the 43rd International Physics Olympiad (IPhO),

which took place from the 15th to the 24th of July 2012. On this occasion the dele-

gations of 80 countries came together, making it one of the largest international

events ever hosted by Estonia. The success of the venue became possible owing to

the long-term efforts of a large number of people – the devoted members of the

Steering Committee, Organizing Committee, Markers, Academic Committee, and

the numerous volunteers – the Guides, Media Team, etc. Of equal importance was

the comprehensive support of the Estonian Ministry of Education and Research,

as well as the aid of the University of Tartu, Tallinn University of Technology,

and the main sponsors. Last but not least, the success of the Olympiad stems

from the support and cooperation of all the participants – contestants, leaders,

observers, and visitors! Organizing such a huge event was a challenging but highly

rewarding task!

The International Physics Olympiad has a long history, and many of its tradi-

tions originate from an era which was very different from the current one; perhaps

the most important difference lies in the communication and information tech-

nologies. A dozen years ago at an IPhO, a group of team leaders had a discussion

about the prospects of the IPhO in the internet era. Will it remain, or will it be

superceded by online competitions? During the subsequent years, the IPhO has not

only stayed but also grown. The fun of face-to-face discussions with new friends

and time spent together can never be substituted by meetings in virtual space.

However, the advent of new technologies requires and also makes possible certain

changes in the organization of the IPhO. The organizers of the 43rd IPhO ventured

to introduce several innovations, most of which received praising feedback from

the participants. In particular, it was the first Olympiad where (a) the students

and team leaders stayed in different cities; (b) the team leaders were asked to

provide only digital copies of the problem translations (thus bypassing the hard

copies); (c) there was a fully automated scanning of the students’ solutions, which

made use of barcodes; (d) the distribution of the graders’ marks and submitting of

the leaders’ marks was made digitally online; (e) the competition city became the

8

World Capital of Physics; (d) there was a “career day” during which the students

had an opportunity to visit the information booths of some of the leading uni-

versities in the world; (f) the Olympiad was preceded by a 10-month-long online

competition “Physics Cup – IPhO2012”.

Another innovation of this Olympiad – perhaps not as prominent but by no

means less important than the ones listed above – lies in the style of the problems.

To be more specific, it was not so much an innovation, but rather reverting back to

the style of the problems used 20 or more years ago. During the last two decades,

the problem texts had become longer and longer, with numerous sub-questions,

often with multiple nested structures. The reasoning behind such a trend was

simple: the number of contestants (and the number of languages in which the

solutions were written) became bigger and bigger, making the grading process

increasingly difficult. A larger number of smaller tasks make it easier to achieve a

fair grading. However, there were also serious implications: the students needed

to waste a lot of time on reading the texts; there was less room for creativity – most

often, the approach to the solution was already written into the problem text.

It has been argued that the style of the IPhO problems from the last decade

was closer to the real tasks of the science of physics. This claim, however, is not

entirely correct. Indeed, a good physical research involves several stages: (a) find-

ing a challenging and important topic; (b) making a solvable model – neglecting

marginal effects and keeping only the qualitatively important components (d)

solving the problem which was formulated at the previous stage; (e) analyzing

the implications of the solution. So, it is true that an immediate problem solving

makes up only a small part of a physical study. Nevertheless, a good insight into

physical phenomena, which is developed by solving creative physical problems,

is also very important during stages (a) and (b). Furthermore, in order to advance

with really important and innovative topics where a new physics is to be devel-

oped, creativity is unavoidable. While the technical skills can be easily developed

later, typically during university studies, creativity needs to be developed from

the very beginning of physics studies. Creative problems also make physics fun,

which is what attracts talented young people. This was the reasoning behind the

decision of the Academic Committee of the 43rd IPhO to take the risk of making

shorter and more creative problems, the solutions of which were more difficult to

grade. Now, while looking back, one can say that the risk paid off: the feedback

was nothing but positive. Let us cite here a contestant from China, Hengyun Zhou:

9

“I liked this year’s IPhO problems very much. Having

gone through most of the past papers, I think that this

year’s problems are the best to date. First, they consisted

of difficult enough problems, and left most of the think-

ing process for the students so that we had to use all our

knowledge and skills to figure out the correct approach

to the problems. Many of the problems in the past paved

the entire way for the students, so all students had to

do was follow the instructions, but this year we had to

come up with a method of our own. Additionally, this

year’s problems emphasised the physics rather than

mathematic skills. The most difficult part in the prob-

lems was building an appropriate model, and that part

really intrigued me, although I failed to build a correct

model in many problems in the end.”

Jaak kIkas and Jaan kaLdaAcademic Committee of the 43rd IPhO

10

11

Sir Harold Kroto1996 Nobel Prize winner,

the honorary guest of

the IPhO2012

“The job of the scien-

tists is to check, not

to believe everything.

The freedom to doubt

is a privilege. If you

know you are unsure,

you have a chance to

change the situation.”

From the academic lecture by Sir Harold Kroto July 20th Tartu – the World Capital of Physics

12

13

SpeechesOpening CeremonyEnE ErgmaPresident of the Parliament of Estonia

Dear participants of the International Physics Olympiad!

Dear delegation leaders, guests, ladies and gentlemen!

It is a great pleasure for me to greet you in Tallinn, the

capital of the Republic of Estonia.

Welcome!

Already in 1996, physicist and Member of the

Academy of Sciences Jaak Aaviksoo, who was then the

Minister of Education of the Republic of Estonia, submit-

ted the application to hold the IPhO in Estonia in 2012.

This took place only five years after Estonia had restored

its independence, but already then we were convinced

that the motor of progress in the 21st century would be an

economy that is based on science and high technologies

and needs specialists with a strong science education.

Since then, young people of Estonia have success-

fully participated in several Olympiads, and have not

returned home from them empty-handed. The fact that

the International Physics Olympiad is held in Estonia

became the reason why it was decided to declare this aca-

demic year the Year of Science in Estonia, and why Tartu

becomes the World Capital of Physics on the 20th of July.

But why is the year 2012 so special for physicists?

I believe there is no sense in asking this question in

this hall, because everybody knows the answer – Higgs

Photo by Henry Teigar

14

boson does exist! The almost half-a-century-long saga of experimental discovery

of this particle shows how complicated it is to get nature to reveal its secrets. But

physicists are purposeful people and they are not deterred by difficulties.

Dear future young colleagues!

I would like to tell you that an education in physics is the best thing in the

world. Having studied physics at Moscow University and worked for many years

as an astrophysicist both in science institutions and as a university lecturer, I can

tell you that dealing with physics and science is really fun. And if the laboratories

on Earth become too narrow for you, turn your glance towards the Universe, where

the great Creator, Nature, has built the most perfect physics laboratory, the con-

ditions of which scientists will never be able to reproduce on Earth.

I wish you all a successful Olympiad, a nice stay in Estonia and a successful

future career in science.

And many thanks to the instructors of the young people and to the organizers

of the Olympiad!

15

hans JOrdEnsThe President of the International Physics Olympiad

Your Excellency,

Distinguished guests,

Dear participants,

For all those who love physics, participation in the

International Physics Olympiad is an exciting event that

cannot be overestimated for its importance. Large-scale

events like this one will have a profound impact on the

lives of all participants whatever they will do later on.

The IPhO is like a watershed: there is one life before the

Olympiad and another one afterwards. And the two are

very different.

For that reason, we are grateful to Estonia as the

country that hosts the Olympiad today. I had the priv-

ilege to visit Estonia one month ago to confer with the

organizers about the progress of the organization and

I was really impressed by the work they had done. You

must be aware that Estonia is a young and rather small

country with an even smaller population. To organize a

large-scale event like the Physics Olympiad is something

that is not done overnight. On top of this relatively dif-

ficult job Estonia wanted to have the competition take

place in two cities almost 200 km apart. The solution to

that was the introduction of modern techniques, much

more than we have ever used before in the IPhO. You

should not be surprised that Skype, being developed in

Estonia, will be one of them.

I hope you will enjoy your stay in Estonia. Estonia is

Photo by Siim Pille

16

a lovely country with very interesting cities, with ancient histories the traces of

which are still very visible. But most of the country is covered by a pastoral and in

some places also rough and wild nature. You will have the opportunity to enjoy

all that during the excursions.

The students stay in the city of Tartu which is situated more to the south-

east. You will like this small town for her nice old centre. But you will most of

all remember Tartu as the city where you took the theoretical and experimental

tests, which, I can assure you, pose interesting physics problems that demand all

of your skills to solve them.

It is a privilege to take part in the Physics Olympiad. It is our aim to have as

many countries as possible participating in the competition. I, therefore, regret

that from the 104 invited countries 24 could not send a team for financial, organ-

izational or other reasons. In the world today we should be able to provide free

access for someone who wants to participate in this event, no matter where he or

she lives. As long as I am president of the International Physics Olympiad I take

it as my duty to promote that.

But still, 80 teams are present. You find yourself amongst some 400 others who

have the same fascination for physics as you have. That already by itself is a great

experience. I hope you take the opportunity to make friends. In this generation

more than ever before it is easy to stay in touch. Science and especially physics is

an international activity. And due to that physicists are able to tackle and solve

problems like the proof of the existence of the Higgs boson as they just did in

CERN. But while some questions are solved, new ones pop up. We are truly living

in exciting times.

I, therefore, wish you all the best in this competition and I hope that it may

be your first step towards a career in Physics.

17

Jaak kIkasChairman of the Academic Committee

Honorable President of the Estonian Parliament,

Honorable President of the International Physics

Olympiad,

Dear participants of the 43rd International Physics

Olympiad,

Honorable guests, Ladies and gentlemen,

On behalf of the Academic Committee of the 43rd

International Physics Olympiad it is my pleasure to

welcome you to Estonia. The IPhO is a big event for

Estonia – never before have we hosted an international

gathering with so many nations participating, not to

mention the number of brilliant young minds – the

future of world physics.

Estonia has been participating in the IPhOs since

1992, and since then we have enjoyed the hospitality of

21 countries. We very much hope that we can, in turn,

provide a bit of the warmth that we ourselves have

enjoyed all over the world. The Academic Committee

has been working hard to make the coming Olympiad

a memorable event for you. First of all, we have tried

to prepare a balanced set of problems, including both

simple and truly challenging questions, rich in physical

content and relatively simple mathematically. We also

hope that beside the examinations the Olympiad is a

chance for you to find new friends, have a good time,

and learn a bit about our country and the people living

here. Well, yesterday you already learned something

Photo by Siim Pille

18

about the Estonian weather. To be honest, it can be better. Sometimes J

And don’t worry that after they have discovered the Higgs there’s nothing left

for you to discover. Because if you raise your eyes to the skies, as academician

Ergma recommended, what do you see there? Almost nothing! Because about

95% of what is up there is hidden from our sight – I mean the dark matter and

dark energy. What these really are – it may well be your chance to find out. And

on your road to big discoveries, participation in the IPhO is a pretty big step.

On behalf of the Academic Committee I wish you every success in the forth-

coming competition.

Enjoy the Olympiad!

19

Jaak aavIksOOMinister of Education and Research

Dear friends of physics, I would like to start with thank-

ing you, all of you. Firstly because of your love towards

physics. As a physicist I know what it means. It is a

personal pleasure, it is a contribution to technological

development, and I hope very much it is also a contri-

bution to a better world.

I would also like to thank you for making Estonia,

Tallinn and Tartu the Capital of Physics for at least 7 days.

It is your work, it is your love towards physics that has

made it possible. Thank you for coming! Thank you for

your efforts, solving complicated theoretical problems,

being skilled in experiments. And last but not least,

thank you for joining the physics family here in Estonia.

Of course I would like to congratulate all of you who

have made their way to the prizes, be it silver, bronze or

gold. I would also like to thank all of your teachers. All of

those people who have helped you to achieve what you have

achieved so far. I would like to thank your families. And I

would like to wish you, every one of you, success in your

future endeavours, in physics and in your personal lives.

And last but not least, try to make this world a bet-

ter place through your love towards physics, through

your love towards friends, different people, developing

ways and means to contribute to social progress and a

peaceful world.

Thank you for coming and congratulations on your

achievements!

Photo by Karl Veskus

Closing Ceremony

20

hans JOrdEnsThe President of the International Physics Olympiad

Your Excellency,

Distinguished guests,

Dear participants.

First of all, I want to thank our hosts from Estonia

who made this Olympiad a great success. Much work is

needed to have an Olympiad run well, especially when

you take into account the relatively large burden on the

shoulders of a small country like Estonia. Hundreds of

people, staff, guides and volunteers, were involved in

the organization to make the competition work for some

400 participants.

So please join me in a big round of applause for the

organizers and all those who made this Olympiad possible!

The Physics Olympiad is, first and foremost, a com-

petition and I’d, therefore, like to say a few words,

especially to those who participated: the competitors.

I really hope you enjoyed your stay here in Estonia. I

hope you are satisfied with your results in the competi-

tion. I hope you enjoyed the company of your peers from

all over the world and I hope you could make friends

with some of them. All this is important. But I also hope

that you feel privileged that you could take part in the

International Physics Olympiad. Taking into account

the millions of young people of your age, you now belong

to a very select group. The French have a nice phrase for

that: “Noblesse oblige”, meaning “Nobility obliges”, or

in other words, privilege entails responsibility.

Photo by Karl Veskus

21

There is an aspect of being excellent of which you must be aware. From now on

you will be regarded as a role model, and that gives you a certain responsibilities. Not

only your peers but most of all those who are younger than you will look up to you.

You have the possibility to inspire youngsters much more than anyone else, simply

because you are young yourself and because you have already achieved so much.

So don’t spend all your time just studying science. Participate in outreach

activities as well and use the abilities you have obtained from being here and the

position of the role model you have become. I would say: show the world that

science is fascinating and exciting. That it can be understood and that it should

be understood in order to make proper decisions. There is a significant amount of

scientific illiteracy amongst people who rule the world. Science not only teaches

you about the laws of nature but it stimulates a critical attitude towards what

you observe. It teaches you to distinguish between facts and fiction. It is that

critical attitude that keeps people’s eyes open in the quest for truth. And you are

the ambassadors to advocate that.

I’d like to conclude by wishing you all the best in your future lives and let us

hear from you, if not as a Nobel Prize Laureate then at least as someone who made

a difference. Try to become a happy person and let physics help you to achieve that.

22

nIELs chrIsTIan harTLIngPresident of Organising Committee IPhO 2013

Distinguished guests, ladies and gentlemen,

Denmark has participated in the International

Physics Olympiad since Oslo 1996. A few years later,

in 1999, it was decided that Denmark should host the

International Physics Olympiad in 2013.

The year 2013 was mainly chosen in memory of the

100th anniversary of Niels Bohr’s theory of the hydro-

gen atom. One may say that this theory marks the very

beginning of quantum mechanics. Therefore, it seemed

natural to celebrate this year with young physics stu-

dents from all over the world. And to be quite honest:

We also hoped that the anniversary would help us

gather the necessary funds.

Back in 1999, we had a feeling that the year 2013 was

a distant future. But as time flies, we now realize with

some surprise that there is only one year left.

Estonia and Denmark have a lot in common. We are

both small nations, more or less the same size, with

a coastline on the Baltic Sea, and we have the same

weather. What you may not know is that, according to

legend, Denmark got its flag in Estonia. During a bat-

tle against Estonia on the 15th of June 1219, the Danish

army was about to lose. Then suddenly a flag fell from

the sky, and a voice said, “Under this flag you will win”!

The Danish Army did win, and we got our red and white

flag, which is the oldest in the world. So we may say

that we got our national flag from Estonia! [And today,

Photo by Merily Salura

23

once more, we received a flag in Estonia.]

This year we’ve had a wonderful time here in Estonia, and finally I want to

thank our hosts for these amazing days, which we will never forget. You have

made a fantastic arrangement. Now Estonia hands over the baton to Denmark,

but it will not be easy to match this marvelous organization.

We look very much forward to welcoming you in Copenhagen, Denmark, to

the 44th International Physics Olympiad in 2013.

24

25

PeopleParticipantsInTErnaTIOnaL PhysIcs OLymPIad cOmmITTEEPresident Hans Jordens

Executive Secretary Ming Juey Lin

aLbanIaStudent Geri Emiri

Arled Papa

Leader Antoneta Deda

armEnIaStudent Vardan Avetisyan

Aram Mkrtchyan

Virab Gevorgyan

Aleksandr Petrosyan

Razmik Hovhannisyan

Leader Gagik Grigoryan

Bilor Kurghinyan

azErbaIJanStudent Haji Piriyev

Valeh Farzaliyev

Ramazan Ramazanov

Farid Mammadov

Ahmad Mehribanli

Leader Mirzali Murguzov

Rana Mammadova

Observer Rashadat Gadmaliyev

26

ausTraLIaStudent Eric Huang

Jonathan Lay

Nicholas Salmon

Siobhan Tobin

Christopher Whittle

Leader Matthew Verdon

Bonnie Zhang

Observer Alix Verdon

ausTrIaStudent Oliver Edtmair

Christoph Weis

Tobias Karg

Maximilian Ruep

Martin Stadler

Leader Helmuth Mayr

Engelbert Stuetz

bangLadEshStudent Kinjol Barua

Wasif Ahmed

Shinjini Saha

Shovon Biswas

Ahmed Maksud

Leader Fayez Ahmed Jahangir Masud

M.Arshad Momen

27

bELarusStudent Ihar Lobach

Albert Samoilenka

Aliaksey Khatskevich

Aliaksandr Yankouski

Vadzim Reut

Leader Anatoli Slabadzianiuk

Anton Mishchuk

Observer Leonid Markovich

bELgIumStudent Romain Falla

Leandro Salemi

Basile Rosen

Basile Vermassen

Mathias Stichelbaut

Leader Bernadette Hendrickx

Philippe Leonard

Observer Sophie Houard

Matthieu Dontaine

bOLIvIaStudent Cesar Tapia

Leader Veronica Subieta

bOsnIa and hErzEgOvInaStudent Selver Pepic

Amer Ajanovic

Nudzeim Selimovic

Sladjan Veselinovic

Stefan Gvozdenovic

Leader Rajfa Musemic

Rodoljub Bavrlic

28

brazILStudent Luis Gustavo Lapinha Dalla Stella

Guilherme Renato Martins Unzer

Lara Timbo Araujo

Ivan Tadeu Ferreira Antunes Filho

Jose Luciano De Morais Neto

Leader Euclydes Marega Junior

Fernando Wellysson de Alencar Sobreira

Observer Leonardo Bruno Pedroza Pontes Lima

Ronaldo Fogo

Antonio Giacomo Pedrine

buLgarIaStudent Katerina Naydenova

Yordan Yordanov

Veselin Karadzhov

Kaloyan Darmonev

Konstantin Gundev

Leader Victor Ivanov

Miroslav Abrashev

canadaStudent Sepehr Ebadi

Yun Jia (Melody) Guan

Tristan Downing

Henry Honglei Wu

Simon Blouin

Leader Andrzej Kotlicki

Jean-François Caron

Visitor Chantal Haussmann

29

cOLOmbIaStudent Daniel Eduardo Fajardo Fajardo

Andres Rios Tascon

Andres Zorrilla Vaca

Leader Fernando Vega Salamanca

Eduardo Zalamea Godoy

crOaTIaStudent Samuel Bosch

Bruno Buljan

Luka Skoric

Karlo Sepetanc

Grgur Simunic

Leader Nikolina Novosel

Ticijana Ban

czEch rEPubLIcStudent Ondřej Bartoš

Jakub Vošmera

Stanislav Fořt

Martin Raszyk

Lubomír Grund

Leader Jan Kříž

Bohumil Vybíral

cyPrusStudent Nicolas Shiaelis

Anastasios Stylianou

Marios Ioannou

Marios Maimaris

Fidias Ieridis

Leader Demetrios Philippou

Emmanouil Lioudakis

30

dEnmarkStudent Molte Emil Strange Andersen

Jakob Lass

Christian Aamand Witting

Kasper Tolborg

Nikolaj Theodor Thams

Leader Jens Ulrik Lefmann

Christian Thune Jacobsen

Observer Niels Christian Hartling

Marianne Hartling

Niels Østergård

Maja Lehmann Jacobsen

Henrik Bruus

EL saLvadOrStudent Bryan Alexander Escalante Castro

Valerie Argentina Dominguez Rivera

Julio Carlos Chorro Huezo

Leader Jose Roberto Dimas Valle

Raúl Alvarenga

EsTOnIaStudent Jaan Toots

Tanel Kiis

Kaur Aare Saar

Kristjan Kongas

Andres Erbsen

Leader Mihkel Kree

Taavi Pungas

31

FInLandStudent Iiro Lehto

Matias Mannerkoski

Jyri Maanpää

Arttu Yli-Sorvari

Tapio Hautamäki

Leader Heikki Mäntysaari

Lasse Franti

FrancEStudent Jonathan Dong

Jean Douçot

Paul Kirchner

Theodor Misiakiewicz

Simon Pirmet

Leader Christian Brunel

Nicolas Billy

Observer Solene Chevalier-Thery

gEOrgIaStudent Jemal Shengelia

Giorgi Kobakhidze

Sergi Chalauri

Saba Kharabadze

Sandro Maludze

Leader Vladimir Paverman

Kakhaber Tavzarashvili

Observer Mariami Rusishvili

32

gErmanyStudent Qiao Gu

Sebastian Linß

Vu Phan Thanh

Lorenz Eberhardt

Georg Krause

Leader Stefan Petersen

Gunnar Friege

Observer Jochen Kröger

grEEcEStudent Stavros Efthymiou

Fotios Vogias

Emmanouil Sakaridis

Emmanouil Vourliotis

Michalis Halkiopoulos

Leader George Kalkanis

Panagiotis Tsakonas

hOng kOngStudent Lam Ho Tat

Lai Kwun Hang

Chan Cheuk Lun

Fung Tsz Chai

Lo Hei Chun

Leader Dik Wai Yin

Wong Kwok Yee

Observer Ng Siu Cho

Sun Ke

33

hungaryStudent Péter Juhász

Áron Dániel Kovács

Zoltán Laczkó

Roland Papp

Attila Szabó

Leader Péter Vankó

Máté Vigh

Observer Ferenc Sarlós

IcELandStudent Hólmfríður Hannesdóttir

Atli Thor Sveinbjarnarson

Freyr Sverrisson

Pétur Rafn Bryde

Stefan Alexis Sigurðsson

Leader Ingibjörg Haraldsdóttir

Martin Swift

IndIaStudent Bijoy Singh Kochar

Jeevana Priya Inala

Kunal Singhal

Pulkit Tandon

Rahul Trivedi

Leader Patrick Dasgupta

Raghavendra Maigur Krishna

Observer Shirish Pathare

34

IndOnEsIaStudent I Made G.N. Kumara

Luqman Fathurrohim

Ramadhiansyah Ramadhiansyah

Werdi Wedana Gunawan

Adrian Nugraha Utama

Leader Mohammad S. Rosid

Kamsul Abraha

Observer Bobby Eka Gunara

Visitor Bambang Hartono

IrELandStudent John Cristopher Horatio Mulholland

Dale Alexander Hughes

Liam Tomas Mulcahy

Thomas Sherlock Wyse Jackson

Leader Eamonn Cunningham

David Rea

IsLamIc rEPubLIc OF IranStudent Mohamad Ansarifard

Mehrdad Malak Mohammadi

Amir Yousefi

Ramtin Yazdanian

Sajad Khodadadian

Leader Mehdi Saadat

Ayoub Esmailpour

Observer Seyyed Nader Rasuli

35

IsraELStudent Itay Knaan Harpaz

Chen Solomon

Yigal Zegelman

Ittai Rubinstein

Eden Segal

Leader Eli Raz Somech

Igor Lisenker

Observer Yoav Merhav

ITaLyStudent Federica Maria Surace

Roberto Albesiano

Michele Fava

Martin Vlashi

Federico Re

Leader Dennis Luigi Censi

Paolo Violino

Observer Giorgio Busoni

Francesco Minosso

JaPanStudent Yuichi Enoki

Tasuku Omori

Kazumi Kasaura

Kohei Kawabata

Hiromasa Nakatsuka

Leader Fumiko Okiharu

Kazuo Kitahara

Observer Tadao Sugiyama

Masashi Mukaida

Yuto Murashita

Katsuhiko Shinkaji

Masao Ninomiya

36

kazakhsTanStudent Ivan Senyushkin

Ilya Vilkoviskiy

Kaisarbek Omirzakhov

Nurzhas Aidynov

Mussa Rajamov

Leader Askar Davletov

Guliya Nurbakova

Observer Yernur Rysmagambetov

kuwaITStudent Nour Alsajari

Rawan Alsenin

Suad Alasfoor

Maryam Ramadan

Zainab Busalehhah

Leader Tareq Abdullah

Eman Hamad

Observer Anoud Alkandari

Visitor Yousef Alsenin

Muhammad Alsajari

Huda Esmail

Khaled Malak

Nozhah Almatrood

kyrgyzsTanStudent Fedor Ignatov

Salizhan Kylychbekov

Zakirbek Mamatayir uulu

Ermek Belekov

Meder Kutbidin uulu

Leader Raia Sultanalieva

Abdymanap Tashmamatov

Observer Ainagyl Osmonalieva

37

LaTvIaStudent Luka Ivanovskis

Georgijs Trenins

Kristaps Znotins

Andris Gerasimovics

Maris Serzans

Leader Vyacheslavs Kashcheyevs

Andris Muznieks

LIEchTEnsTEInStudent Benedikt Kratochwil

Lukas Lang

David Hälg

Leader Fritz Epple

Daniel Oehry

LIThuanIaStudent Mantas Abazorius

Tomas Čerškus

Daumantas Kavolis

Marius Kerys

Žygimantas Stražnickas

Leader Pavelas Bogdanovičius

Edmundas Kuokštis

Observer Indrė Grigaitytė

macaO, chInaStudent Chan Lon Wu

Wai Pan Si

Ka Fai Chan

Wai Hong Lei

Wai Hei Hoi

Leader Iat Neng Chan

38

macEdOnIaStudent Vesna Bacheva

Filip Simeski

Biljana Mitreska

Ljupcho Petrov

Leader Stanisha Veljkovikj

maLaysIaStudent Lee Yuan Zhe

Yeoh Chin Vern

Ooi Chun Yeang

Koay Hui Wen

Imran Ariffin

Leader Wan Mohd Aimran Wan Mohd Kamil

Chin Mai Ying

Observer Rosman Md Ajis

mExIcOStudent Eduardo Acosta-Reynoso

Javier Mendez-Ovalle

Kevin Bustillos-Barrera

Alberto Trejo-Avila

Jorge Torres-Ramos

Leader Victor Romero-Rochin

Raul Espejel-Morales

mOLdOvaStudent Cristian Zanoci

Ion Toloaca

Nicoleta Colibaba

Dinis Cheian

Ilie Popanu

Leader Victor Paginu

Igor Evtodiev

39

mOngOLIaStudent Tsogt Baigalmaa

Battushig Myanganbayar

Munkhtsetseg Battulga

Battsooj Bayarsaikhan

Bilguun Batjargal

Leader Batsukh Garmaa

Baatarchuluun Tsermaa

Observer Sharavsuren Byamba

Soyolmaa Dorjyanjmaa

mOnTEnEgrOStudent Petar Tadic

Marko Petric

Nikola Potpara

Vladimir Pejovic

Janko Radulovic

Leader Jovan Mirkovic

Nevenka Antovic

Observer Tatijana Carapic

nEThErLandsStudent Koen Dwarshuis

Troy Figiel

Ruben Doornenbal

Thijs van der Gugten

Martijn van Kuppeveld

Leader Ad Mooldijk

Enno van der Laan

40

nIgErIaStudent Ayomide Andrae Bamidele

Musa Muhammed Damina

John Nwankwo Chijioke

Anthony Okonkwo

Michael Tari Charles Okhide

Leader Ayhan Yaman

Lewis Obagboye

Observer Alaba Aminat Agbaje

Okey Junior Chikezie

nOrwayStudent Tiantian Zhang

Håkon Tásken

Oda Lauten

Anders Strømberg

Marius Leiros

Leader Torbjørn Mehl

Joakim Bergli

PakIsTanStudent Muhammad Suhaib Qasim

Usman Ayyaz

Usman Ali Javid

Muhammad Taimoor Iftikhar

Leader Shahid Qamar

Muhammad Aftab Rafiq

41

PEOPLE’s rEPubLIc OF chInaStudent Wenzhuo Huang

Yijun Jiang

Hengyun Zhou

Siyuan Wei

Chi Shu

Leader Xiaolin Chen

Kun Xun

Observer Liangzhu Mu

Chunling Zhang

Feng Song

POLandStudent Bartlomiej Zawalski

Michal Pacholski

Kacper Oreszczuk

Filip Ficek

Jan Rydzewski

Leader Jacek Jasiak

Jan Mostowski

POrTugaLStudent Francisco Machado

Pedro Paredes

Manuel Cabral

Matheus Marreiros

Simão João

Leader Fernando Nogueira

Rui Travasso

PuErTO rIcOStudent Logan Abel

Leader Hector Jimenez

42

rEPubLIc OF kOrEaStudent Woojin Kweon

Sooshin Kim

Wonseok Lee

Jaemo Lim

Suyeon Choi

Leader Sung-Won Kim

Ki Wan Jang

Observer Chan Ju Kim

Hyun Joo Lee

Yuri Kang

Weon Kyun Mok

Kug-Hyung Lee

rEPubLIc OF sIngaPOrEStudent Ding Yue

Huan Yan Qi

Kuan Jun Jie, Joseph

Soo Wah Ming, Wayne

Ang Yu Jian

Leader Rawat Rajdeep Singh

Chung Keng Yeow

Observer Chng Chia Yi

Berthold-Georg Englert

Visitor Aleksandra Englert

rOmanIaStudent Tudor Giurgică-Tiron

Dan - Cristian Andronic

Sebastian Florin Dumitru

Tudor Ciobanu

Roberta Răileanu

Leader Delia-Constanţa Davidescu

Adrian Dafinei

Observer Victor Păunescu

43

russIaStudent Alexandra Vasilyeva

Nikita Sopenko

Ivan Ivashkovskiy

Lev Ginzburg

David Frenklakh

Leader Valery Slobodyanin

Dmitry Aleksandrov

Observer Mikhail Osin

saudI arabIaStudent Sulaiman Almatroudi

Abdullah Alsalloum

Ali Alhulaymi

Mohammad Alhejji

Homoud Alharbi

Leader Najm Al Hosiny

Sandu Golcea

Observer Abdulaziz Alharthi

Mahmoud Nagadi

Hind Aldossari

Laila Babsail

Visitor Aljoharah Almetrek

Abdulaziz Alrashed

sErbIaStudent Tamara Šumarac

Milan Kornjača

Milan Krstajić

Jovan Blanuša

Ilija Burić

Leader Aleksandar Krmpot

Mihailo Rabasović

44

sLOvakIaStudent Peter Kosec

Patrik Svancara

Patrik Turzak

Andrej Vlcek

Samuel Beznak

Leader Ivo Cap

Lubomir Mucha

Observer Lubomir Konrad

Visitor Klara Capova

sLOvEnIaStudent Domen Ipavec

Matevž Marinčič

Jan Šuntajs

Jurij Tratar

Miha Zgubič

Leader Ciril Dominko

Jurij Bajc

sOuTh aFrIcaStudent Thiolan Prevan Naidoo

Avthar Sewrathan

Xolela Jara

Lloyd Mahadeo

Shihal Menesher Sapry

Leader Mervlyn Moodley

Bagtyyar Jorayev

45

sPaInStudent Roberto Alegre

Francesc-Xavier Gispert Sánchez

David Trillo Fernández

Marc Rodà Llordés

Aitor Azemar

Leader Juan Leon

Esperanza García-Carpintero Romero

srI LankaStudent Chanaka Manoj Singhabahu

Dombagaha Gedara Prasad Randika Maithriepala

Liraj Harsha Prabath Kodithuwakku

M. Janidu Chandrashantha Gunarathna

Edurapotha Gamaralalage Inoka Amanthie Dharmasena

Leader Ramal Vernil Coorey

surInamEStudent Chaandnie Bandhoe

Raynesh Kanhai

Priya Kasimbeg

Suraj Kishoen Misier

Leader Tjien Bing Tan

Ignaas Jimidar

Visitor Chantal Hewitt

Lachman Jurgen

46

swEdEnStudent Johan Runeson

Andréas Sundström

Carl Smed

Viktor Djurberg

Simon Johansson

Leader Max Kesselberg

Bo Söderberg

Observer Anne-Sofie Mårtensson

Visitor Margareta Kesselberg

swITzErLandStudent Thanh Phong Lê

Dominic Schwarz

Sebastian Käser

Laura Gremion

Christoph Schildknecht

Leader Lionel Philippoz

Simon Birrer

Observer Johanna Nyffeler

syrIan arab rEPubLIc Student Ghadeer Shaaban

Osama Yaghi

Mohamad Nour Ahmad

Mohamed Alrazzouk

Leader Akil Salloum

Observer Farkad Alramadani

47

TaIwanStudent Kai-Chi Huang

Jun-Ting Hsieh

Wei-Jen Ko

Yu-Ting Liu

Chien-An Wang

Leader Chih-Ta Chia

Shang-Fang Tsai

Observer Chung-Yu Mou

Tzong-Jer Yang

Chon Saar Chu

Jiun-Huei Wu

Yen-Chen Yu

TaJIkIsTanStudent Abdukhomid Nurmatov

Rabboni Bafoev

Adhamzhon Shukurov

Shakhzodi Rustamdzhon

Isfandiyor Safarov

Leader Ilkhom Khotami

ThaILandStudent Pongsapuk Sawaddirak

Puthipong Worasaran

Paphop Sawasdee

Supanut Thanasilp

Nathanan Tantivasadakarn

Leader Sirapat Pratontep

Phichet Kittara

Observer Suwan Kusamran

Raksapol Thananuwong

48

TurkEy Student Atinc Cagan Sengul

Oguzhan Can

Abdurrahman Akkas

Mustafa Selman Akinci

Mehmet Said Onay

Leader Ibrahim Gunal

Onur Ozcan

TurkmEnIsTanStudent Mekan Toyjanov

Meylis Malikov

Kemal Babayev

Agajan Odayev

Övezmyrat Övezmyradow

Leader Halit Coşkun

Gylychmammet Orazov

ukraInEStudent Volodymyr Sivak

Vsevolod Bykov

Vladysslav Diachenko

Volodymyr Rozsokhovatskyi

Yevgen Cherniavskyi

Leader Boris Kreminskyi

Stanislav Vilchynskyi

Observer Bushtruk Artem

49

unITEd kIngdOmStudent Adam Brown

Richard Thorburn

Peter Budden

Eric Wieser

Frank Bloomfield

Leader Robin Hughes

Paul Nicholls

Observer Sian Owen

Visitor Muriel Irene Hughes

unITEd sTaTEs OF amErIcaStudent Allan Sadun

Eric Schneider

Jeffrey Cai

Jeffrey Yan

Kevin Zhou

Leader Paul Stanley

Andrew Lin

vIETnamStudent Xuan Hien Bui

Viet Thang Dinh

Phi Long Ngo

Ngoc Hai Dinh

Huy Quang Le

Leader The Khoi Nguyen

Minh Thi Tran

Observer Van Vinh Le

Van Pham Tran

Quang Tuan Ngo

Thai Hoc Bui

Van Vu Ha

Xuan Thanh Ha

50

51Photos by Andres Mihkeson, Siim Pille and Merily Salura

52

53Photos by Andres Mihkeson, Siim Pille and Merily Salura

54

OrganizerssTEErIng cOmmITTEE

ChairmanJanar Holm Estonian Ministry of Education and Research

SecretaryViire Sepp Gifted and Talented Development Centre

of University Tartu

MembersErgo Nõmmiste University of Tartu, Institute of Physics

Jaak Kikas University of Tartu

Jaan Kalda Tallinn University of Technology, Institute of Cybernetics

Jakob Kübarsepp Tallinn University of Technology

Kaido Reivelt Estonian Physical Society

Kristjan Haller University of Tartu

Marco Kirm University of Tartu, Institute of Physics

Peeter Saari Estonian Academy of Sciences

Rait Toompere Archimedes Foundation

Raivo Stern National Institute of Chemical Physics and Biophysics

Toomas Sõmera Estonian Information Technology Foundation

Ülle Kikas Estonian Ministry of Education and Research

55

acadEmIc cOmmITTEEJaak Kikas Head of the Academic Committee

Head of Experimental Examination

Jaan Kalda Head of Theoretical Examination

Alar Ainla

Eero Uustalu

Endel Soolo

Mihkel Heidelberg

Oleg Košik

Rünno Lõhmus

Siim Ainsaar

Stanislav Zavjalov

Taavi Adamberg

OrganIzIng cOmmITTEEEne Koitla Head of the Organizing Committee

Marily Hendrikson Project Manager

Annika Vihul Head of accounting,

transportation and accommodation

Eneli Sutt Head of information technology

Kerli Kusnets Head of media

Malle Tragon Head of events and catering

Anna Gureeva Heads of group leaders

Julia Šmakova Heads of group leaders

56

markErsHelle Kaasik Head of the Markers Team

Aigar Vaigu Estonia

Alar Ainla Estonia

Aleksandr Bitjukov Estonia

Aleksandr Morozenko Estonia

Aleksandr Pištšev Estonia

Andreas Valdmann Estonia

Andres Jaanson Estonia

Anna-Stiina Suur-Uski Finland

Ants Remm Estonia

Antti Karjalainen Finland

Arvo Mere Estonia

Bahar Mehmani Germany

Christian Laut Ebbesen Denmark

Eemeli Samuel Tomberg Finland

Eero Vaher Estonia

Endel Soolo Estonia

Erik Paemurru Estonia

Filip Studnička Czech Republic

Gleb Široki Estonia

Gyula Honyek Hungary

Hannu Jaakko Lauri Siikonen Finland

Heiki Niglas Estonia

Helle Kaasik Estonia

Henri Johannes Ylitie Finland

Herry Kwee Indonesia

Jaak Jaaniste Estonia

Jaakko Uusitalo Finland

Jaan Katus Estonia

Jaanus Sepp Estonia

Juho Kahala Finland

Kadi Liis Saar Estonia

Kert Pütsepp Estonia

Klára Baranyai Hungary

57

Kristian Kuppart Estonia

Madis Ollikainen Estonia

Maksim Säkki Estonia

Markko Paas Estonia

Mihkel Pajusalu Estonia

Mihkel Rähn Estonia

Mikko Ervasti Finland

Oleg Košik Estonia

Oleksii Chechkin Ukraine

Otso Olavi Ossian Huuska Finland

Ottb Rebane Estonia

Rauno Siinmaa Estonia

Reio Põder Estonia

Riho Taba Estonia

Roland Matt Estonia

Sami Kivistö Finland

Sanli Faez Germany

Shahabedin Chatraee Islamic Republic of Iran

Stanislav Zavjalov Estonia

Zainul Abidin Indonesia

Taavi Vaikjärv Estonia

Teemu Johannes Hynninen Finland

Tiit Sepp Estonia

Timo Olli Johannes Voipio Finland

Valter Kiisk Estonia

Vasja Susič Slovenia

Ville Suur-Uski Finland

58

vOLunTEErsAgnes Vask

Airike Jõesaar

Allan-Cristjan Puks

Anastassia Samovitš

Andreas Ragen Ayal

Andres Ainelo

Andres Allik

Andres Mihkelson

Anete Merilin Leetberg

Anete Sammler

Anete Viise

Anna Dunajeva

Anna Jazõkova

Anna Krajuškina

Annemari Sepp

Anni Müüripeal

Anni Sandra Varblane

Annika Lukner

Annika Pille

Ants Johanson

Anu Viks

Artur Panov

Arvo Ehrstein

Ats Kurvet

Auli Relve

Ave-Stina Udam

Bety Mehide

Brenda Rauniste

Diana Oidingu

Dmitri Lanevski

Donatas Braziulis

Egert Vinogradov

Elina Libek

Enna Elismäe

Eno Paenurk

Erik Ilbis

Eva Mõtshärg

Evelin Pihlap

Eveli Soo

Gerli Krjukov

Gerli Vaik

Gertrud Metsa

Grete Helena Roose

Grethe Aikevitšius

Hanna Britt Soots

Hanna Kadri Metsvaht

Hanna Kivila

Hanna Moor

Hanna-Loore Hansen

Hedvig Tamman

Helbe-Laura Nikitkina

Helena Ainsoo

Helena Talimaa

Heli Aomets

Heli Pärn

Henry Teigar

Ida Rahu

Inger Kangur

Ivo Kruusamägi

Jaanika Jensen

Jane Lihtmaa

Janne Disko

Jasper Kursk

Jelizaveta Dõljova

Jelizaveta Žatkina

Johanna-Maria Muuga

Joonas Jäme

Jorma Veiderpass

59

Juhani Almers

Julia Gavrilova

Kadi Ainsaar

Kadi Külasalu

Kadri Alumets

Kadri Ann Rebane

Kadri Eek

Kadri Tinn

Kaisa Jõgi

Karl Kütt

Karl Veskus

Karl-Mattias Tepp

Karolin Rõõm

Kaspar Märtens

Kati Randmäe

Katrin Tuude

Keidi Suursaar

Ken Riisalu

Kerli Kalk

Kerstin Kivila

Krista Kallavus

Kristi Kartus

Kristiina Štõkova

Kristin Ehala

Kristin Liiksaar

Kristine Diane Liive

Kristine Leetberg

Kristjan Kalve

Ksenia Kukuskina

Laura Liisa Lankei

Laura Soon

Liina Nõmm

Liis Kass

Liis Nurmis

Liis Talimaa

Liisa Hunt

Liisa Veerus

Liisi Liivalaid

Liisi Mõtshärg

Liisi Sünd

Liisu Miller

Lisett Kiudorv

Lona-Liisa Sutt

Ly Pärnaste

Maksim Ivanov

Maksim Mišin

Marek Järvik

Maria Krajuškina

Mari-Liis Jaansalu

Mariliis Maamägi

Mari-Liis Tamm

Maris Ertmann

Maris Palo

Marit Puusepp

Mark Gimbutas

Marren Tiivits

Mart Ernits

Marta Tanaga

Marta Vihtre

Mary-Ann Kubre

Merilin Kalavus

Merilin Vesingi

Merily Salura

Merle Lust

Merlin Russak

Mette-Triin Purde

Mihkel Lepik

Mihkel Tali

Minna-Triin Kohv

Mirjam Laurimäe

60

Mirjam Mikk

Morten Piibeleht

Natalia Nekrassova

Nele Kriisa

Olga Bulgakova

Oliver Grauberg

Ott Kekišev

Paap Koemets

Paul Liias

Pille-Riin Peet

Raimo Armus

Rando Porosk

Rasmus Kuusmann

Reile Juhanson

Rene Rünt

Riinu Ansper

Rudolf Bichele

Saile Mägi

Sander Benga

Sander Kütisaar

Sander Soo

Sander Udam

Sandhra-Mirella Valdma

Sergei Jakovlev

Siim Kaspar Uustalu

Siim Pille

Siiri Mägi

Sille Hausenberg

Simona Kalatšov

Sirje Kollom

Sten Aus

Stina Avvo

Teisi Timma

Terje Kapp

Tiina Pärtel

Tiina Turban

Triin Rebane

Triin Ärm

Triinu Hordo

Uku-Kaspar Uustalu

Ulla Meeri Liivamägi

Urmet Paloveer

Üllar Kivila

61Photos by Siim Pille, Andres Mihkeson and Henry Teigar

62

63

ProgramsStudentssunday, 15Th OF JuLyArrival and registration at Sokos Hotel Viru

17:00 – 17:30 Departure from the hotel,

transportation to Open Air Museum

18:00 – 21:00 Icebreaking – Estonian Open Air Museum

21:00 – 22:00 Arrival to the hotel

mOnday, 16Th OF JuLy07:00 – 08:00 Breakfast

08:00 – 09:00 Departure from the hotel, putting luggage onto buses

Mobile phones and laptops being collected by the organisers

09:15 – 09:45 Walk to Opening Ceremony

10:00 – 12:00 Opening Ceremony – NOKIA Concert Hall

The Ceremony will be broadcasted over the Internet

and videotaped.

12:00 – 13:30 Welcome Banquet – NOKIA Concert Hall

13:30 – 17:00 Transportation to Tartu hotels

Opening ceremony Photo by Andres Mihkeson

64

17:30 – 19:00 Free time and preparation for theory

19:00 – 21:00 Dinner – Restaurant Dorpat

TuEsday, 17Th OF JuLy06:00 – 07:30 Breakfast

08:00 Transportation to Theoretical Examination

09:00 – 14:00 Theoretical Examination –

Sports building of Estonian University of Life Sciences

14:00 – 16:00 Transportation and lunch – Tartu Adventure Park

15:00 – 19:00 Games & activities – Tartu Adventure Park

19:00 – 21:00 Dinner - Restaurant Dorpat

wEdnEsday, 18Th OF JuLy07:00 – 09:00 Breakfast

09:00 – 20:00 Excursion: Rakvere Castle

20:00 – 23:00 Free time and preparation for experiment

Thursday, 19Th OF JuLy06:00 – 06:45 Students group A: Breakfast

07:00 – 09:00 Students group B: Breakfast

07:00 Students group A: Departure to Experimental Examination

08:00 – 13:00 Students group A: Experimental Examination

09:00 – 12:30 Students group B: Excursion: AHHAA Science Centre

12:30 – 13:50 Students group B: Lunch – AHHAA Science Centre

13:00 – 14:30 Students group A: Lunch – Restaurant Dorpat

students in Tartu adventure Park Photo by Siim Pille

65

13:50 Students group B: Departure to Experimental Examination

14:30 – 17:30 Students group A: Excursion: AHHAA Science Centre

15:00 – 20:00 Students group B: Experimental Examination

17:30 – 19:00 Students group A: Free time

19:00 – 21:00 Students group A: Dinner – Restaurant Volga

20:00 – 22:00 Students group B: Dinner – Restaurant Atlantis

21:30 – 22:30 Students group A: Skype meeting with Leaders

22:30 – 23:30 Students group B: Skype meeting with Leaders

FrIday, 20Th OF JuLyTartu – the World Capital of Physics

08:00 – 10:00 Breakfast

10:00 – 17:00 Tartu – the World Capital of Physics –

public science activities

13:00 – 15:00 Lunch at Tartu restaurants

17:00 – 18:00 Lecture: Sir Harold Kroto

(The 1996 Nobel Prize in Chemistry) – Vanemuise Concert Hall

18:00 – 20:00 Reception by Mayor of Tartu – Vanemuise Concert Hall

saTurday, 21sT OF JuLy08:00 – 10:00 Breakfast

10:00 – 13:00 Transportation to Tallinn

13:00 – … Free time; lunch and dinner at Tallinn restaurants

Exam Photos by Siim Pille

66

sunday, 22nd OF JuLy07:00 – 09:00 Breakfast

09:00 – 13:00 Football tournament

13:00 – 15:00 Lunch – Football Stadium

14:00 – 18:00 Football tournament continued

18:00 – 23:00 Free time and dinner at Tallinn restaurants

mOnday, 23rd OF JuLy07.00 – 09.00 Breakfast

09.00 – 13.00 Free time

13.15 – 13.45 Walk to Closing Ceremony

14.00 – 17.00 Closing Ceremony – NOKIA Concert Hall


and videotaped.

17.00 – 17.30 Walk to the hotel

18.00 Transportation to the Farewall Party

18.30 – 01.00 Farewell Party – The Tallinn Song Festival Grounds

23.00 – … Round-the-clock transportation back to the hotel

TuEsday, 24Th OF JuLyDeparture

Football award. Football tournament Photos by Henry Teigar

67

Leaders and Observerssunday, 15Th OF JuLyArrival and Registration at Radisson Blu Hotel Olümpia

17:00 – 17:30 Departure from the hotel, transportation to Open Air Museum

18:00 – 21:00 Icebreaking - Estonian Open Air Museum

21:00 – 22:00 Arrival to the hotel

mOnday, 16Th OF JuLy07:00 – 09:00 Breakfast

09:15 – 09:45 Walk to Opening Ceremony

10:00 – 12:00 Opening Ceremony – NOKIA Concert Hall


and videotaped.

12.00 – 13.30 Welcome Banquet – NOKIA Concert Hall

13:30 – 14:00 Walk to the hotel

14:00 – 19:00 International Board Meeting: Discussion of theoretical

problems – Radisson Blu Hotel Olümpia Conference Centre

19:00 – 21:00 Dinner – Restaurant Senso

21:00 – … International Board Meeting:

Translation of theoretical problems

students in rakvere castle Photo by Siim Pille

68

TuEsday, 17Th OF JuLy06:00 – 07:00 Breakfast

07:00 Departure from the hotel

07:00 – 10:00 Excursion: trip to Saaremaa

10:00 – 13:00 Excursion: Kaali crater and Kuressaare

13:00 – 14:30 Lunch - Mändjala Camping

14:30 – 20:00 Excursion: Saaremaa and Muhu

20:00 – 21:30 Arrival to the hotel and dinner - Restaurant Senso

20:30 – 21:30 Distribution of theory papers from the IPhO office

wEdnEsday, 18Th OF JuLy07:00 – 09:00 Breakfast

09:00 – 12:00 Free time

11:30 – 13:00 Lunch - Restaurant Senso

13:00 – 19:00 International Board Meeting:

Discussion of experimental problems


21:00 – … International Board Meeting:

Translation of experimental problems

Thursday, 19Th OF JuLy07:00 – 09:00 Breakfast

09:00 – 12:00 Free time

12:00 – 13:00 Collection of marks from Leaders (theory) – Online


Estonian Open air museum. skype meeting with students Photos by Siim Pille, Henry Teigar

69

15:00 – 19:00 Free time

19:00 – 21:00 Leaders group A: Distribution of practical papers

from IPhO office


21:30 – 22:30 Leaders group A: Skype meeting with Students

22:30 – 23:30 Leaders group B: Skype meeting with Students

01:00 – 02:00 Leaders group B: Distribution of practical papers

from IPhO office

FrIday, 20Th OF JuLyTartu – the World Capital of Physics

07:00 – 09:00 Breakfast

07:00 – 08:00 Leaders group B: Distribution of practical papers

from IPhO office

09:00 – 11:30 Transportation to Tartu

12:00 – 17:00 Tartu – the World Capital of Physics –

Public science activities

13:00 – 15:00 Lunch at Tartu restaurants

17:00 – 18:00 Lecture: Sir Harold Kroto

(The 1996 Nobel Prize in Chemistry) - Vanemuise Concert Hall

18:00 – 20:00 Reception by Mayor of Tartu – Vanemuise Concert Hall

20:00 – 22:30 Transportation to Tallinn

22:30 – 23:30 Collection of marks from Leaders (experiment) – Online

visiting ahhaa science center Photo by Siim Pille

70

saTurday, 21sT OF JuLy07:00 – 09:00 Breakfast

10:00 – 12:00 International Board Meeting

11:00 Distribution of marks (theory) - Online



Moderation of theoretical papers

19:00 Distribution of marks (experiment) - Online


sunday, 22nd OF JuLy07:00 – 09:00 Breakfast


Moderation of experimental papers

12:00 – 14:00 Lunch – Restaurant Senso


Deciding final marks and medals

19:00 – … Free time and dinner at Tallinn restaurants

Farwell Party Photo by Merily Salura

71

mOnday, 23rd OF JuLy07:00 – 09:00 Breakfast

09:00 – 13:00 Free time

13:15 – 13:45 Walk to Closing Ceremony – NOKIA Concert Hall

14:00 – 17:00 Closing Ceremony


and videotaped.

17:00 – 17:30 Walk back to the hotel

18:00 Transportation to the Farewell Party

18:30 – 01:00 Farewell Party - The Tallinn Song Festival Grounds

23:00 – … Round-the-clock transportation back to the hotel

TuEsday, 24Th OF JuLyDeparture

closing ceremony Photo by Merily Salura

72

73

Problems and solutionsThe 43rd International Physics Olympiad — Theoretical CompetitionTartu, Estonia — Thursday, July 17th 2012The 43rd International Physics Olympiad — Theoretical Competition

Tartu, Estonia — Tuesday, July 17th 2012

• The examination lasts for 5 hours. There are 3 problems

worth in total 30 points. Please note that the point

values of the three theoretical problems are not

equal.

• You must not open the envelope with the prob-

lems before the sound signal of the beginning of

competition (three short signals).

• You are not allowed to leave your working place

without permission. If you need any assistance

(broken calculator, need to visit a restroom, etc), please

raise the corresponding flag (“help” or “toilet” with a

long handle at your seat) above your seat box walls and

keep it raised until an organizer arrives.

• Your answers must be expressed in terms of those

quantities, which are highlighted in the problem text,

and can contain also fundamental constants, if needed.

So, if it is written that “the box height is a and the

width — b” then a can be used in the answer, and b

cannot be used (unless it is highlighted somewhere else,

see below). Those quantities which are highlighted in

the text of a subquestion can be used only in the answer

to that subquestion; the quantities which are highlighted

in the introductory text of the Problem (or a Part of a

Problem), i.e. outside the scope of any subquestion, can

be used for all the answers of that Problem (or of that

Problem Part).

• Use only the front side of the sheets of paper.

• For each problem, there are dedicated Solution Sheets

(see header for the number and pictogramme). Write

your solutions onto the appropriate Solution Sheets. For

each Problem, the Solution Sheets are numbered; use

the sheets according to the enumeration. Always mark

which Problem Part and Question you are deal-

ing with. Copy the final answers into the appropriate

boxes of the Answer Sheets. There are also Draft pa-

pers; use these for writing things which you don’t want

to be graded. If you have written something what you

don’t want to be graded onto the Solution Sheets (such

as initial and incorrect solutions), cross these out.

• If you need more paper for a certain problem, please raise

the flag “help” and tell an organizer the problem num-

ber; you are given two Solution sheets (you can do this

more than once).

• You should use as little text as possible: try to ex-

plain your solution mainly with equations, numbers, sym-

bols and diagrams. When textual explanation is un-

avoidable, you are encouraged to provide English

translation alongside with the text in your native

language (if you mistranslate, or don’t translate

at all, your native language text will be used dur-

ing the Moderation).

• The first single sound signal tells you that there are 30

min of solving time left; the second double sound signal

means that 5 min is left; the third triple sound signal

marks the end of solving time. After the third sound

signal you must stop writing immediately. Put all

the papers into the envelope at your desk. You are not

allowed to take any sheet of paper out of the room.

If you have finished solving before the final sound signal,

please raise your flag.

— page 1 of 5 —

74

The 43rd International Physics Olympiad — Theoretical Competition





equal.






















Problem Part).

















more than once).


















— page 1 of 5 —






equal.






















Problem Part).

















more than once).


















— page 1 of 5 —

75






equal.






















Problem Part).

















more than once).


















— page 1 of 5 —

Problem T1. Focus on sketches (13 points)

Part A. Ballistics (4.5 points)

A ball thrown with an initial speed v0 moves in a homogeneous

gravitational field in x−z plane, where the x-axis is horizontal,

and z — vertical, antiparallel to the free fall acceleration g ;

neglect the air drag.

i. (0.8 pts) By adjusting the launching angle for a ball thrown

with a fixed initial speed v0 from the origin, targets can be

hit within the region given by

z ≤ z0 − kx2;

you can use this fact without proving it. Find the constants z0

and k.

ii. (1.2 pts) Now, the launching point can be

freely selected on the ground level z = 0, and

the launching angle can be adjusted as needed;

the aim is to hit the topmost point of a spherical

building of radius R (see fig.) with as small as

possible initial speed v0 (prior hitting the target, bouncing off

the roof is not allowed). Sketch qualitatively the shape of the

optimal trajectory of the ball (use the designated box on the

answer sheet). Note: the points are given only for the sketch.

iii. (2.5 pts) What is the minimal launching speed vmin needed

to hit the topmost point of a spherical building of radius R ?

La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part

B. Air flow around a wing (4 points)

For this Problem Part, the following information may be

useful. For a flow of liquid or gas in a tube, along a stream-

line p + ρgh + 1

2ρv2 = const., assuming that the velocity v

is much smaller than the sound speed. Here ρ is the density,

h — height, g — free fall acceleration, and p — hydrostatic

pressure. Streamlines are defined as the trajectories of fluid

particles (assuming that the flow pattern is stationary). Note

that the term 1

2ρv2 is called the dynamic pressure.

In the fig. below, a cross-section of an aircraft wing is de-

picted together with streamlines of the air flow around the

wing, as seen in the wing’s reference frame. Assume that

(a) the air flow is purely two-dimensional (i.e. that the velo-

city vectors of air lie in the figure plane); (b) the streamline

pattern is independent of the aircraft speed; (c) there is no

wind; (d) the dynamic pressure is much smaller than the at-

mospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to

take measurements from the fig. on the answer sheet .

i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s,

what is the speed of the air vP at the point P (marked in fig.)

with respect to the ground?

ii. (1.2 pts) In the case of high relative humidity, as the ground

speed of the aircraft increases over a critical value vcrit, a stream

of water droplets is created behind the wing. The droplets

emerge at a certain point Q. Mark the point Q in fig. on the

answer sheet. Explain qualitatively (using formulae and as few

text as possible) how you determined its position.

iii. (2.0 pts) Estimate the critical speed vcrit using the follow-

ing data: relative humidity of the air is r = 90% , specific heat

of air at constant pressure cp = 1.00 × 103 J/kg · K , pressure

of saturated water vapour: psa = 2.31 kPa at the temperat-

ure of the unperturbed air Ta = 293 K and psb = 2.46 kPa

at Tb = 294 K . Depending on your approximations you

may also need the specific heat of air at constant volume

cV = 0.717 × 103 J/kg · K . Note that the relative humidity is

defined as the ratio of the vapor pressure to the saturated vapor

pressure at the given temperature. Saturated vapor pressure is

defined as the vapor pressure by which vapor is in equilibrium

with the liquid.

— page 2 of 5 —










z ≤ z0 − kx2;


and k.
















line p + ρgh + 1






that the term 1































with the liquid.

— page 2 of 5 —

PrObLEm T1. FOcus On skETchEs (13 POInTs)

76










z ≤ z0 − kx2;


and k.
















line p + ρgh + 1






that the term 1































with the liquid.

— page 2 of 5 —










z ≤ z0 − kx2;


and k.
















line p + ρgh + 1






that the term 1































with the liquid.

— page 2 of 5 —










z ≤ z0 − kx2;


and k.
















line p + ρgh + 1






that the term 1































with the liquid.

— page 2 of 5 —Problem T1. Focus on sketches (13 points)









z ≤ z0 − kx2;


and k.
















line p + ρgh + 1






that the term 1































with the liquid.

— page 2 of 5 —










z ≤ z0 − kx2;


and k.
















line p + ρgh + 1






that the term 1































with the liquid.

— page 2 of 5 —

77










z ≤ z0 − kx2;


and k.
















line p + ρgh + 1






that the term 1































with the liquid.

— page 2 of 5 —

78

Part C. Magnetic straws (4.5 points)

Consider a cylindrical tube made of a super-

conducting material. The length of the tube

is l and the inner radius is r ; always l ≫ r.

The centre of the tube coincides with the

origin, and its axis coincides with the z-

axis. There is a magnetic flux Φ through

the central cross-section of the tube, z = 0,

x2 + y2 < r2.

Superconductor is a material which expels any magnetic

field (field is zero inside it).

i. (0.8 pts) Sketch five such magnetic field lines onto the des-

ignated box of the answer sheet which pass through the five

red dots marked on the axial cross-section of the tube.

ii. (1.2 pts) Find the z-directional tension force T in the

middle of the tube (i.e. the force by which two halves of the

tube, z > 0 and z < 0, interact with each other).

iii. (2.5 pts) Now there is another tube,

identical and parallel to the first one. The

second tube has opposite direction of the

magnetic field, and its centre is placed at

y = l , x = z = 0 (so that the tubes form

opposite sides of a square). Determine the

magnetic interaction force F between the two tubes.

— page 3 of 5 —









x2 + y2 < r2.
















— page 3 of 5 —









x2 + y2 < r2.
















— page 3 of 5 —

Problem T2. Kelvin water dropper (8 points)

The following facts about the surface tension may turn out

to be useful for this problem. For the molecules of a liquid, the

positions at the liquid-air interface are less favourable as com-

pared with the positions in the bulk of the liquid. Therefore,

this interface is ascribed the so-called surface energy U = σS,

where S is the surface area of the interface and σ — the surface

tension coefficient of the liquid. Further, two fragments of the

liquid surface pull each other with a force F = σl, where l is

the length of a straight line separating the fragments.

A long metallic pipe with internal diameter d

is pointing directly downwards; water is slowly

dripping from a nozzle at its lower end, see fig.

Water can be considered to be electrically con-

ducting; its surface tension is σ and density —

ρ . Always assume that d ≪ r. Here, r is the

radius of the droplet hanging below the nozzle,

which grows slowly in time until the droplet sep-

arates from the nozzle due to the free fall acceleration g .

Part A. Single pipe (4 points)

i. (1.2 pts) Find the radius rmax of a drop just before it sep-

arates from the nozzle.

ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s

electrostatic potential is ϕ . Find the charge Q of a drop when

its radius is r .

iii. (1.6 pts) For this question, assume that r is kept con-

stant and ϕ is slowly increased. The droplet becomes unstable

and breaks into two pieces if the hydrostatic pressure inside

the droplet becomes smaller than the atmospheric one. Find

the critical potential ϕmax at which this will happen.

Part B. Two pipes (4 points)

An apparatus called “Kelvin water dropper” consists of two

pipes (identical to the one described in Part A), connected

via a T-junction, see fig. The ends of both pipes are at the

centres of two cylindrical electrodes (with height L and dia-

meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n

droplets per unit time. Droplets fall from height H into con-

ductive bowls underneath the nozzles, cross-connected to the

electrodes as shown in Fig; the electrodes are connected via a

capacitance C . There is no net charge on the system of bowls

and electrodes. Note that the water container is grounded.

The first droplet to fall will have microscopic charge, which

will cause an imbalance between the two sides and a small

charge separation across the capacitor.

i. (1.2 pts) Express the modulus

of the charge Q0 of the drops separ-

ating at the instant when the capa-

citor’s charge is q in terms of rmax

(from Part A-i). Neglect the effect

described in Part A-iii.

ii. (1.5 pts) Find the dependence

of q on time t by approximating it

with a continuous function q(t) and

assuming that q(0) = q0 .

iii. (1.3 pts) The dropper’s functioning can be hindered by

the effect shown in Part A-iii. Additionally, a limit Umax to

the achievable voltage between the electrodes is set by the

electrostatic push between a droplet and the bowl beneath it;

find Umax.

— page 4 of 5 —

PrObLEm T2. kELvIn waTEr drOPPEr (8 POInTs)

79

























its radius is r .


































find Umax.

— page 4 of 5 —

























its radius is r .


































find Umax.

— page 4 of 5 —

























its radius is r .


































find Umax.

— page 4 of 5 —

























its radius is r .


































find Umax.

— page 4 of 5 —

























its radius is r .


































find Umax.

— page 4 of 5 —

80

























its radius is r .


































find Umax.

— page 4 of 5 —

PrObLEm T3. PrOTOsTar FOrmaTIOn (9 POInTs)Problem T3. Protostar formation (9 points)

Let us model the formation of a star as follows. A spherical

cloud of sparse interstellar gas, initially at rest, starts to col-

lapse due to its own gravity. The initial radius of the ball is

r0 and the mass — m . The temperature of the surroundings

(much sparser than the gas) and the initial temperature of the

gas is uniformly T0 . The gas may be assumed to be ideal.

The average molar mass of the gas is µ and its adiabatic in-

dex is γ > 4

3. Assume that Gmµ

r0

≫ RT0, where R is the gas

constant and G — the gravity constant.

i. (0.8 pts) During much of the collapse, the gas is so trans-

parent that any heat generated is immediately radiated away,

i.e. the ball stays in a thermodynamic equilibrium with its sur-

roundings. How many times (n) does the pressure increase

while the radius is halved ( r1 = 0.5r0 )? Assume that the gas

density stays uniform.

ii. (1 pt) Estimate the time t2 needed for the radius to shrink

from r0 to r2 = 0.95r0 . Neglect the change of the gravity field

at the position of a falling gas particle.

iii. (2.5 pts) Assuming that the pressure stays negligible, find

the time tr→0 needed for the ball to collapse from r0 down to

a much smaller radius using Kepler’s Laws for elliptical orbits.

iv. (1.7 pts) At some radius r3 ≪ r0, the gas becomes dense

enough to be opaque to the heat radiation. Calculate the

amount of heat Q radiated away during the collapse from the

radius r0 down to r3.

v. (1 pt) For radii smaller than r3 you may neglect heat ra-

diation. Determine how the temperature T of the ball depends

on its radius r < r3.

vi. (2 pts) Eventually we cannot neglect the effect of the pres-

sure on the dynamics of the gas and the collapse stops at

r = r4 (with r4 ≪ r3 ). However, the radiation can still be

neglected and the temperature is not yet high enough to ignite

nuclear fusion. The pressure of such a protostar is not uniform

anymore, but rough estimates with inaccurate numerical pre-

factors can still be done. Estimate the final radius r4 and the

respective temperature T4.

— page 5 of 5 —

81

Problem T3. Protostar formation (9 points)








dex is γ > 4

3. Assume that Gmµ

r0






























— page 5 of 5 —

Problem T3. Protostar formation (9 points)








dex is γ > 4

3. Assume that Gmµ

r0






























— page 5 of 5 —

82

Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)

i. (0.8 pts) When the stone is thrown vertically upwards, it

can reach the point x = 0, z = v20/2g (as it follows from the

energy conservation law). Comparing this with the inequality

z ≤ z0 − kx2 we conclude that

z0 = v20/2g. [0.3 pts]

Let us consider the asymptotics z → −∞; the trajectory of

the stone is a parabola, and at this limit, the horizontal dis-

placement (for the given z) is very sensitive with respect to the

curvature of the parabola: the flatter the parabola, the larger

the displacement. The parabola has the flattest shape when

the stone is thrown horizontally, x = v0t and z = −gt2/2, i.e.

its trajectory is given by z = −gx2/2v20. Now, let us recall

that z ≤ z0 − kx2, i.e. −gx2/2v20 ≤ z0 − kx2 ⇒ k ≤ g/2v2

0.

Note that k < g/2v20 would imply that there is a gap between

the parabolic region z ≤ z0 − kx2 and the given trajectory

z = −gx2/2v20 . This trajectory is supposed to be optimal for

hitting targets far below (z → −∞), so there should be no such

a gap, and hence, we can exclude the option k < g/2v20. This

leaves us with

k = g/2v20. [0.5 pts]

ii. (1.2 pts) Let us note that the

stone trajectory is reversible and due

to the energy conservation law, one

can equivalently ask, what is the min-

imal initial speed needed for a stone

to be thrown from the topmost point

of the spherical building down to the

ground without hitting the roof, and what is the respective tra-

jectory. It is easy to understand that the trajectory either needs

to touch the roof, or start horizontally from the topmost point

with the curvature radius equal to R. Indeed, if neither were

the case, it would be possible to keep the same throwing angle

and just reduce the speed a little bit — the stone would still

reach the ground without hitting the roof. Further, if it were

tangent at the topmost point, the trajectory wouldn’t touch

nor intersect the roof anywhere else, because the curvature of

the parabola has maximum at its topmost point. Then, it

would be possible to keep the initial speed constant, and in-

crease slightly the throwing angle (from horizontal to slightly

upwards): the new trajectory wouldn’t be neither tangent at

the top nor touch the roof at any other point; now we can re-

duce the initial speed as we argued previously. So we conclude

that the optimal trajectory needs to touch the roof somewhere,

as shown in Fig.

iii. (2.5 pts) The brute force approach would be writing down

the condition that the optimal trajectory intersects with the

building at two points and touches at one. This would be de-

scribed by a fourth order algebraic equation and therefore, it is

not realistic to accomplish such a solution within a reasonable

time frame.

Note that the interior of the building needs to lie inside the

region where the targets can be hit with a stone thrown from

the top with initial speed vmin. Indeed, if we can throw over

the building, we can hit anything inside by lowering the throw-

ing angle. On the other hand, the boundary of the targetable

region needs to touch the building. Indeed, if there were a

gap, it would be possible to hit a target just above the point

where the optimal trajectory touches the building; the traject-

ory through that target wouldn’t touch the building anywhere,

hence we arrive at a contradiction.

So, with v0 corresponding to the optimal trajectory, the tar-

getable region touches the building; due to symmetry, overall

there are two touching points (for smaller speeds, there would

be four, and for larger speeds, there would be none). With the

origin at the top of the building, the intersection points are

defined by the following system of equations:

x2 + z2 + 2zR = 0, z =v2

0

2g−

gx2

2v20

.

Upon eliminating z, this becomes a biquadratic equation for x:

x4

(

g

2v20

)2

+ x2

(

1

2−

gR

v20

)

+

(

v20

4g+ R

)

v20

g= 0.

Hence the speed by which the real-valued solutions disappear

can be found from the condition that the discriminant vanishes:(

1

2−

gR

v20

)2

=1

4+

gR

v20

=⇒gR

v20

= 2.

Bearing in mind that due to the energy conservation law, at

the ground level the squared speed is increased by 4gR. Thus

we finally obtain

vmin =√

v20 + 4gR = 3

√

gR

2.

Alternative solution using the fact that if a ball

is thrown from a point A to a point B (possibly at

different heights) with the minimal required launch-

ing speed, the initial and terminal velocities are equal.

This fact may be known to some of the contestants, but it

can be also easily derived. Indeed, suppose that when start-

ing with velocity v0 at point A, the ball will hit after time τ

the point B, and |v0| is the minimal speed by which hitting is

possible. Now, let us rotate the vector v0 slightly by adding

to it a perpendicular small vector u ⊥ v0 (|u| ≪ |v0|). With

the launching velocity v1 = v0 + u, the trajectory of the ball

will still go almost through point B: near the pont B, the

displacement of the trajectory cannot change linearly with |u|.

Indeed, a linear in |u| displacement would mean that with es-

sentially the same speed |v1| ≈ |v0|, it would be possible to

throw over point B, which is in a contradiction with the op-

timality of the original trajectory. Hence, the displacement

vector r0(τ) − r1(τ) needs to be parallel to the trajectory, i.e.

to the velocity vB of the ball at point B. Here, r0(t) [r1(t)]

is the radius vector of the ball as a function of time when it

was launched with velocity v0 [v1]. In a free-falling system of

— page 1 of 5 —






z0 = v20/2g. [0.3 pts]









0.






leaves us with

k = g/2v20. [0.5 pts]
























as shown in Fig.






time frame.

















x2 + z2 + 2zR = 0, z =v2

0

2g−

gx2

2v20

.


x4

(

g

2v20

)2

+ x2

(

1

2−

gR

v20

)

+

(

v20

4g+ R

)

v20

g= 0.



1

2−

gR

v20

)2

=1

4+

gR

v20

=⇒gR

v20

= 2.



we finally obtain

vmin =√

v20 + 4gR = 3

√

gR

2.






















— page 1 of 5 —

SolutionsPrObLEm T1. FOcus On skETchEs (13 POInTs)

83






z0 = v20/2g. [0.3 pts]









0.






leaves us with

k = g/2v20. [0.5 pts]
























as shown in Fig.






time frame.

















x2 + z2 + 2zR = 0, z =v2

0

2g−

gx2

2v20

.


x4

(

g

2v20

)2

+ x2

(

1

2−

gR

v20

)

+

(

v20

4g+ R

)

v20

g= 0.



1

2−

gR

v20

)2

=1

4+

gR

v20

=⇒gR

v20

= 2.



we finally obtain

vmin =√

v20 + 4gR = 3

√

gR

2.






















— page 1 of 5 —Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)





z0 = v20/2g. [0.3 pts]









0.






leaves us with

k = g/2v20. [0.5 pts]
























as shown in Fig.






time frame.

















x2 + z2 + 2zR = 0, z =v2

0

2g−

gx2

2v20

.


x4

(

g

2v20

)2

+ x2

(

1

2−

gR

v20

)

+

(

v20

4g+ R

)

v20

g= 0.



1

2−

gR

v20

)2

=1

4+

gR

v20

=⇒gR

v20

= 2.



we finally obtain

vmin =√

v20 + 4gR = 3

√

gR

2.






















— page 1 of 5 —

84






z0 = v20/2g. [0.3 pts]









0.






leaves us with

k = g/2v20. [0.5 pts]
























as shown in Fig.






time frame.

















x2 + z2 + 2zR = 0, z =v2

0

2g−

gx2

2v20

.


x4

(

g

2v20

)2

+ x2

(

1

2−

gR

v20

)

+

(

v20

4g+ R

)

v20

g= 0.



1

2−

gR

v20

)2

=1

4+

gR

v20

=⇒gR

v20

= 2.



we finally obtain

vmin =√

v20 + 4gR = 3

√

gR

2.






















— page 1 of 5 —reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,

(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.

We start again throwing a ball from the point O at the top

of the building, and notice that for the optimal trajctory, at

the point P , where it touches the building, the velocity is per-

pendicular both to the radius vector QP (where Q denotes the

building’s centre), and to the launching velocity. Hence, QP

and the launching velocity are parallel. Let O be the origin,

and let us denote ∠OQP = α. Then the trajectory is given by

z = x cot α −gx2

2v2 sin2 α.

Point P with coordinates x = R sin α and z = R(cos α − 1) be-

longs to this parabola, hence R = gR2

2v2 , from where we obtain

the previous result.

Part B. Mist (4 points)

i. (0.8 pts) In the plane’s reference frame, along the channel

between two streamlines the volume flux of air (volume flow

rate) is constant due to continuity. The volume flux is the

product of speed and channel’s cross-section area, which, due

to the two-dimensional geometry, is proportional to the channel

width and can be measured from the Fig. Due to the absence of

wind, the unperturbed air’s speed in the plane’s frame is just v0.

So, upon measuring the dimensions a = 10 mm and b = 13 mm

(see Fig), we can write v0a = ub and hence u = v0ab. Since at

point P , the streamlines are horizontal where all the velocities

are parallel, the vector addition is reduced to the scalar addi-

tion: the air’s ground speed vP = v0 − u = v0(1 −ab) = 23 m/s.

ii. (1.2 pts) Although the dynamic pressure 1

2ρv2 is relatively

small, it gives rise to some adiabatic expansion and compres-

sion. In expanding regions the temperature will drop and hence,

the pressure of saturated vapours will also drop. If the dew

point is reached, a stream of droplets will appear. This process

will start in a point where the adiabatic expansion is maximal,

i.e. where the hydrostatic pressure is minimal and consequently,

as it follows from the Bernoulli’s law p + 1

2ρv2 = const, the dy-

namic pressure is maximal: in the place where the air speed in

wing’s frame is maximal and the streamline distance minimal.

Such a point Q is marked in Fig.

iii. (2 pts) First we need to calculate the dew point for the air

of given water content (since the relative pressure change will

be small, we can ignore the dependence of the dew point on

pressure). The water vapour pressure is pw = psar = 2.08 kPa.

The relative change of the pressure of the saturated vapour is

small, so we can linearize its temperature dependence:

psa − pw

Ta − T=

psb − psa

Tb − Ta

=⇒ Ta − T = (Tb − Ta)(1 − r)psa

psb − psa

;

numerically T ≈ 291.5 K. Further we need to relate the air

speed to the temperature. To this end we need to use the en-

ergy conservation law. A convenient ready-to-use form of it is

provided by the Bernoulli’s law. Applying this law will give

a good approximation of the reality, but strictly speaking, it

needs to be modified to take into account the compressibility

of air and the associated expansion/contraction work. Con-

sider one mole of air, which has the mass µ and the volume

V = RT/p. Apparently the process is fast and the air par-

cels are large, so that heat transfer across the air parcels is

negligible. Additionally, the process is subsonic; all together

we can conclude that the process is adiabatic. Consider a seg-

ment of a tube formed by the streamlines. Let us denote the

physical quantities at its one end by index 1, and at the other

end — by index 2. Then, while one mole of gas flows into

the tube at one end, as much flows out at the other end. The

inflow carries in kinetic energy 1

2µv2

1 , and the outflow carries

out 1

2µv2

2. The inflowing gas receives work due to the pushing

gas equal to p1V1 = RT1, the outflowing gas performs work

p2V2 = RT2. Let’s define molar heat capacities CV = µcV and

Cp = µcp. The inflow carries in heat energy CV RT1, and the

outflow carries out CV RT2. All together, the energy balance

can be written as 1

2µv2 + CpT = const. From this we can

easily express ∆ v2

2= 1

Cv2

crit(a2

c2 − 1) = cp∆T , where c is the

streamline distance at the point Q, and further

vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s,

where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that

in reality, the required speed is probably somewhat higher, be-

cause for a fast condensation, a considerable over-saturation is

needed. However, within an order of magnitude, this estimate

remains valid.


i. (0.8 pts) Due to the superconduct-

ing walls, the magnetic field lines cannot

cross the walls, so the flux is constant

along the tube. For a closed contour in-

side the tube, there should be no circu-

lation of the magnetic field, hence the

field lines cannot be curved, and the field

needs to be homogeneous. The field lines

close from outside the tube, similarly to a solenoid.

ii. (1.2 pts) Let us consider the change of the magnetic energy

when the tube is stretched (virtually) by a small amount ∆l.

Note that the magnetic flux trough the tube is conserved: any

change of flux would imply a non-zero electromotive force dΦ

dt,

and for a zero resistivity, an infinite current. So, the induc-

tion B = Φ

πr2 . The energy density of the magnetic field is B2

2µ0

.

— page 2 of 5 —

85

reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,

(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.








z = x cot α −gx2

2v2 sin2 α.



















2ρv2 is relatively


















psa − pw

Ta − T=

psb − psa

Tb − Ta

=⇒ Ta − T = (Tb − Ta)(1 − r)psa

psb − psa

;


















2µv2


out 1

2µv2






can be written as 1



2= 1

Cv2

crit(a2



vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s,





remains valid.















dt,


tion B = Φ


2µ0

.

— page 2 of 5 —


(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.








z = x cot α −gx2

2v2 sin2 α.



















2ρv2 is relatively


















psa − pw

Ta − T=

psb − psa

Tb − Ta

=⇒ Ta − T = (Tb − Ta)(1 − r)psa

psb − psa

;


















2µv2


out 1

2µv2






can be written as 1



2= 1

Cv2

crit(a2



vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s,





remains valid.















dt,


tion B = Φ


2µ0

.

— page 2 of 5 —reference one can easily see that r0(t) − r1(t) = (v0 − v1)t. So,

(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.








z = x cot α −gx2

2v2 sin2 α.



















2ρv2 is relatively


















psa − pw

Ta − T=

psb − psa

Tb − Ta

=⇒ Ta − T = (Tb − Ta)(1 − r)psa

psb − psa

;


















2µv2


out 1

2µv2






can be written as 1



2= 1

Cv2

crit(a2



vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s,





remains valid.















dt,


tion B = Φ


2µ0

.

— page 2 of 5 —

86


(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.








z = x cot α −gx2

2v2 sin2 α.



















2ρv2 is relatively


















psa − pw

Ta − T=

psb − psa

Tb − Ta

=⇒ Ta − T = (Tb − Ta)(1 − r)psa

psb − psa

;


















2µv2


out 1

2µv2






can be written as 1



2= 1

Cv2

crit(a2



vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s,





remains valid.















dt,


tion B = Φ


2µ0

.

— page 2 of 5 —

87


(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.








z = x cot α −gx2

2v2 sin2 α.



















2ρv2 is relatively


















psa − pw

Ta − T=

psb − psa

Tb − Ta

=⇒ Ta − T = (Tb − Ta)(1 − r)psa

psb − psa

;


















2µv2


out 1

2µv2






can be written as 1



2= 1

Cv2

crit(a2



vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s,





remains valid.















dt,


tion B = Φ


2µ0

.

— page 2 of 5 —


(v0 − v1)τ = uτ vB ⇒ u vB ⇒ v0 ⊥ vB.








z = x cot α −gx2

2v2 sin2 α.



















2ρv2 is relatively


















psa − pw

Ta − T=

psb − psa

Tb − Ta

=⇒ Ta − T = (Tb − Ta)(1 − r)psa

psb − psa

;


















2µv2


out 1

2µv2






can be written as 1



2= 1

Cv2

crit(a2



vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s,





remains valid.















dt,


tion B = Φ


2µ0

.

— page 2 of 5 —Thus, the change of the magnetic energy is calculated as

∆W =B2

2µ0

πr2∆l =Φ2

2µ0πr2∆l.

This energy increase is achieved owing to the work done by the

stretching force, ∆W = T ∆l. Hence, the force

T =Φ2

2µ0πr2.

iii. (2.5 pts) Let us analyse, what would be the change of

the magnetic energy when one of the straws is displaced to a

small distance. The magnetic field inside the tubes will remain

constant due to the conservation of magnetic flux, but outside,

the magnetic field will be changed. The magnetic field out-

side the straws is defined by the following condition: there is

no circulation of B (because there are no currents outside the

straws); there are no sources of the field lines, other than the

endpoints of the straws; each of the endpoints of the straws is

a source of streamlines with a fixed magnetic flux ±Φ. These

are exactly the same condition as those which define the elec-

tric field of four charges ±Q. We know that if the distance

between charges is much larger than the geometrical size of

a charge, the charges can be considered as point charges (the

electric field near the charges remains almost constant, so that

the respective contribution to the change of the overall electric

field energy is negligible). Therefore we can conclude that the

endpoints of the straws can be considered as magnetic point

charges. In order to calculate the force between two magnetic

charges (magnetic monopoles), we need to establish the corres-

pondence between magnetic and electric quantities.

For two electric charges Q separated by a distance a, the

force is F = 1

4πε0

Q2

a2 , and at the position of one charge, the elec-

tric field of the other charge has energy density w = 1

32π2ε0

Q2

a4 ;

hence we can write F = 8πwa2. This is a universal expression

for the force (for the case when the field lines have the same

shape as in the case of two opposite and equal by modulus elec-

tric charges) relying only on the energy density, and not related

to the nature of the field; so we can apply it to the magnetic

field. Indeed, the force can be calculated as a derivative of

the full field energy with respect to a virtual displacement of

a field line source (electric or magnetic charge); if the energy

densities of two fields are respectively equal at one point, they

are equal everywhere, and so are equal the full field energies.

As it follows from the Gauss law, for a point source of a fixed

magnetic flux Φ at a distance a, the induction B = 1

4πΦ

a2 . So,

the energy density w = B2

2µ0

= 1

32π2µ0

Φ2

a4 , hence

F =1

4πµ0

Φ2

a2.

For the two straws, we have four magnetic charges. The lon-

gitudinal (along a straw axis) forces cancel out (the diagonally

positioned pairs of same-sign-charges push in opposite direc-

tions). The normal force is a superposition of the attraction

due to the two pairs of opposite charges, F1 = 1

4πµ0

Φ2

l2 , and

the repulsive forces of diagonal pairs, F2 =√

2

8πµ0

Φ2

2l2 . The net

attractive force will be

F = 2(F1 − F2) =4 −

√2

8πµ0

Φ2

l2.

Alternative solution based on dipole energy calcu-

lation. It is known that the axial component of the magnetic

induction created by a solenoidal current of surface density j

is proportional to the solid angle Ω under which the current is

seen from the given point:

B = µ0jΩ/4π;

this can be easily derived from the Biot-Savart law. Let the

distance between the tubes be a (we’ll take derivative over a),

and let us consider a first tube’s point which has a coordinate

x (with 0 ≤ x ≤ l) from where the direction to the one end-

point of the other tube forms an angle α = arctan a/x with

the tube’s axis. From that point, the open circular face of the

other tube forms a solid angle Ω = πr2 sin2 α cos α/a2, so that

its contribution to the axial magnetic field at the point x

B =µ0jr2 sin2 α cos α

4a2=

Φ sin2 α cos α

4πa2.

The solenoidal current at that point forms a magnetic dipole

dm = πr2jdx = Φdx/µ0,

which has potential energy

dU = Bdm = sin2 α cos αΦ2dx/4πµ0a2.

When integrating over x, α varies from arctan a/l to 0, so that

U1 =

∫

dU =

∫

sin2 α cos αΦ2dx

4πµ0a2=

∫

cos αΦ2dα

4πµ0a.

Bearing in mind that the other end-circle of the other tube con-

tributes the same amount to the magnetic interaction energy,

we find

U = 2U1 =Φ2

2πµ0

(

1

a−

1√

a2 + l2

)

.

Upon taking derivative over a and using a = l we obtain the

same result for F as previously.

— page 3 of 5 —

88

Thus, the change of the magnetic energy is calculated as

∆W =B2

2µ0

πr2∆l =Φ2

2µ0πr2∆l.



T =Φ2

2µ0πr2.























force is F = 1

4πε0

Q2



32π2ε0

Q2

a4 ;













4πΦ

a2 . So,


2µ0

= 1

32π2µ0

Φ2

a4 , hence

F =1

4πµ0

Φ2

a2.






4πµ0

Φ2

l2 , and


2

8πµ0

Φ2

2l2 . The net


F = 2(F1 − F2) =4 −

√2

8πµ0

Φ2

l2.






B = µ0jΩ/4π;










4a2=

Φ sin2 α cos α

4πa2.






U1 =

∫

dU =

∫

sin2 α cos αΦ2dx

4πµ0a2=

∫

cos αΦ2dα

4πµ0a.



we find

U = 2U1 =Φ2

2πµ0

(

1

a−

1√

a2 + l2

)

.



— page 3 of 5 —


∆W =B2

2µ0

πr2∆l =Φ2

2µ0πr2∆l.



T =Φ2

2µ0πr2.























force is F = 1

4πε0

Q2



32π2ε0

Q2

a4 ;













4πΦ

a2 . So,


2µ0

= 1

32π2µ0

Φ2

a4 , hence

F =1

4πµ0

Φ2

a2.






4πµ0

Φ2

l2 , and


2

8πµ0

Φ2

2l2 . The net


F = 2(F1 − F2) =4 −

√2

8πµ0

Φ2

l2.






B = µ0jΩ/4π;










4a2=

Φ sin2 α cos α

4πa2.






U1 =

∫

dU =

∫

sin2 α cos αΦ2dx

4πµ0a2=

∫

cos αΦ2dα

4πµ0a.



we find

U = 2U1 =Φ2

2πµ0

(

1

a−

1√

a2 + l2

)

.



— page 3 of 5 —

89



∆W =B2

2µ0

πr2∆l =Φ2

2µ0πr2∆l.



T =Φ2

2µ0πr2.























force is F = 1

4πε0

Q2



32π2ε0

Q2

a4 ;













4πΦ

a2 . So,


2µ0

= 1

32π2µ0

Φ2

a4 , hence

F =1

4πµ0

Φ2

a2.






4πµ0

Φ2

l2 , and


2

8πµ0

Φ2

2l2 . The net


F = 2(F1 − F2) =4 −

√2

8πµ0

Φ2

l2.






B = µ0jΩ/4π;










4a2=

Φ sin2 α cos α

4πa2.






U1 =

∫

dU =

∫

sin2 α cos αΦ2dx

4πµ0a2=

∫

cos αΦ2dα

4πµ0a.



we find

U = 2U1 =Φ2

2πµ0

(

1

a−

1√

a2 + l2

)

.



— page 3 of 5 —Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)

i. (1.2 pts) Let us write the force balance for the droplet.

Since d ≪ r, we can neglect the force π4

∆pd2 due to the excess

pressure ∆p inside the tube. So, the gravity force 4

3πr3

maxρg

is balanced by the capillary force. When the droplet separates

from the tube, the water surface forms in the vicinity of the

nozzle a “neck”, which has vertical tangent. In the horizontal

cross-section of that “neck”, the capillary force is vertical and

can be calculated as πσd. So,

rmax = 3

√

3σd

4ρg.

ii. (1.2 pts) Since d ≪ r, we can neglect the change of the

droplet’s capacitance due to the tube. On the one hand, the

droplet’s potential is ϕ; on the other hand, it is 1

4πε0

Q

r. So,

Q = 4πε0ϕr.

iii. (1.6 pts) Excess pressure inside the droplet is caused by

the capillary pressure 2σ/r (increases the inside pressure), and

by the electrostatic pressure 1

2ε0E2 = 1

2ε0ϕ2/r2 (decreases the

pressure). So, the sign of the excess pressure will change, if1

2ε0ϕ2

max/r2 = 2σ/r, hence

ϕmax = 2√

σr/ε0.

The expression for the electrostatic pressure used above can

be derived as follows. The electrostatic force acting on a surface

charge of density σ and surface area S is given by F = σS · E,

where E is the field at the site without the field created by the

surface charge element itself. Note that this force is perpen-

dicular to the surface, so F/S can be interpreted as a pressure.

The surface charge gives rise to a field drop on the surface equal

to ∆E = σ/ε0 (which follows from the Gauss law); inside the

droplet, there is no field due to the conductivity of the droplet:

E − 1

2∆E = 0; outside the droplet, there is field E = E + 1

2∆E,

therefore E = 1

2E = 1

2∆E. Bringing everything together, we

obtain the expression used above.

Note that alternatively, this expression can be derived by

considering a virtual displacement of a capacitor’s surface and

comparing the pressure work p∆V with the change of the elec-

trostatic field energy 1

2ε0E2∆V .

Finally, the answer to the question can be also derived from

the requirement that the mechanical work dA done for an in-

finitesimal droplet inflation needs to be zero. From the en-

ergy conservation law, dW + dWel

= σ d(4πr2) + 1

2ϕ2

max dCd,

where the droplet’s capacitance Cd = 4πε0r; the electrical work

dWel

= ϕmaxdq = 4πε0ϕ2maxdr. Putting dW = 0 we obtain an

equation for ϕmax, which recovers the earlier result.


i. (1.2 pts) This is basically the same as Part A-ii, except

that the surroundings’ potential is that of the surrounding

electrode, −U/2 (where U = q/C is the capacitor’s voltage)

and droplet has the ground potential (0). As it is not defined

which electrode is the positive one, opposite sign of the po-

tential may be chosen, if done consistently. Note that since

the cylindrical electrode is long, it shields effectively the en-

vironment’s (ground, wall, etc) potential. So, relative to its

surroundings, the droplet’s potential is U/2. Using the result

of Part A we obtain

Q = 2πε0Urmax = 2πε0qrmax/C.

ii. (1.5 pts) The sign of the droplet’s charge is the same as

that of the capacitor’s opposite plate (which is connected to

the farther electrode). So, when the droplet falls into the bowl,

it will increase the capacitor’s charge by Q:

dq = 2πε0UrmaxdN = 2πε0rmaxndtq

C,

where dN = ndt is the number of droplets which fall during

the time dt This is a simple linear differential equation which

is solved easily to obtain

q = q0eγt, γ =2πε0rmaxn

C=

πε0n

C3

√

6σd

ρg.

iii. (1.3 pts) The droplets can reach the bowls if their mech-

anical energy mgH (where m is the droplet’s mass) is large

enough to overcome the electrostatic push: The droplet starts

at the point where the electric potential is 0, which is the sum of

the potential U/2, due to the electrode, and of its self-generated

potential −U/2. Its motion is not affected by the self-generated

field, so it needs to fall from the potential U/2 down to the po-

tential −U/2, resulting in the change of the electrostatic energy

equal to UQ ≤ mgH , where Q = 2πε0Urmax (see above). So,

Umax =mgH

2πε0Umaxrmax

,

∴ Umax =

√

Hσd

2ε0rmax

= 6

√

H3gσ2ρd2

6ε30

.

— page 4 of 5 —

90

Problem T2. Kelvin water dropper (8 points)Part A. Single pipe (4 points)





3πr3

maxρg






rmax = 3

√

3σd

4ρg.




4πε0

Q

r. So,

Q = 4πε0ϕr.




2ε0E2 = 1



2ε0ϕ2


ϕmax = 2√

σr/ε0.










E − 1


2∆E,

therefore E = 1

2E = 1







2ε0E2∆V .





= σ d(4πr2) + 1

2ϕ2

max dCd,


dWel













of Part A we obtain







C,





C=

πε0n

C3

√

6σd

ρg.










Umax =mgH

2πε0Umaxrmax

,

∴ Umax =

√

Hσd

2ε0rmax

= 6

√

H3gσ2ρd2

6ε30

.

— page 4 of 5 —






3πr3

maxρg






rmax = 3

√

3σd

4ρg.




4πε0

Q

r. So,

Q = 4πε0ϕr.




2ε0E2 = 1



2ε0ϕ2


ϕmax = 2√

σr/ε0.










E − 1


2∆E,

therefore E = 1

2E = 1







2ε0E2∆V .





= σ d(4πr2) + 1

2ϕ2

max dCd,


dWel













of Part A we obtain







C,





C=

πε0n

C3

√

6σd

ρg.










Umax =mgH

2πε0Umaxrmax

,

∴ Umax =

√

Hσd

2ε0rmax

= 6

√

H3gσ2ρd2

6ε30

.

— page 4 of 5 —

91






3πr3

maxρg






rmax = 3

√

3σd

4ρg.




4πε0

Q

r. So,

Q = 4πε0ϕr.




2ε0E2 = 1



2ε0ϕ2


ϕmax = 2√

σr/ε0.










E − 1


2∆E,

therefore E = 1

2E = 1







2ε0E2∆V .





= σ d(4πr2) + 1

2ϕ2

max dCd,


dWel













of Part A we obtain







C,





C=

πε0n

C3

√

6σd

ρg.










Umax =mgH

2πε0Umaxrmax

,

∴ Umax =

√

Hσd

2ε0rmax

= 6

√

H3gσ2ρd2

6ε30

.

— page 4 of 5 —

92

Problem T3. Protostar formation (9 points)i. (0.8 pts)

T = const =⇒ pV = const

V ∝ r3

∴ p ∝ r−3 =⇒p(r1)

p(r0)= 23 = 8.

ii. (1 pt) During the period considered the pressure is negli-

gible. Therefore the gas is in free fall. By Gauss’ theorem and

symmetry, the gravitational field at any point in the ball is

equivalent to the one generated when all the mass closer to the

center is compressed into the center. Moreover, while the ball

has not yet shrunk much, the field strength on its surface does

not change much either. The acceleration of the outermost

layer stays approximately constant. Thus,

t ≈

√

2(r0 − r2)

g

where

g ≈Gm

r20

,

∴ t ≈

√

2r20(r0 − r2)

Gm=

√

0.1r30

Gm.

iii. (2.5 pts) Gravitationally the outer layer of the ball is in-

fluenced by the rest just as the rest were compressed into a

point mass. Therefore we have Keplerian motion: the fall of

any part of the outer layer consists in a halfperiod of an ultra-

elliptical orbit. The ellipse is degenerate into a line; its foci are

at the ends of the line; one focus is at the center of the ball (by

Kepler’s 1st law) and the other one is at r0, see figure (instead

of a degenerate ellipse, a strongly elliptical ellipse is depicted).

The period of the orbit is determined by the longer semiaxis of

the ellipse (by Kepler’s 3rd law). The longer semiaxis is r0/2

and we are interested in half a period. Thus, the answer is

equal to the halfperiod of a circular orbit of radius r0/2:

(

2π

2tr→0

)2r0

2=

Gm

(r0/2)2=⇒ tr→0 = π

√

r30

8Gm.

Alternatively, one may write the energy conservation lawr2

2−

Gmr

= E (that in turn is obtainable from Newton’s

II law r = − Gmr2 ) with E = − Gm

r0

, separate the variables

(drdt

= −

√

2E + 2Gmr

) and write the integral t = −∫

dr√2E+ 2Gm

r

.

This integral is probably not calculable during the limitted

time given during the Olympiad, but a possible approach can

be sketched as follows. Substituting√

2E + 2Gmr

= ξ and√

2E = υ, one gets

t∞

4Gm=

∫

∞

0

dξ

(υ2 − ξ2)2

=1

4υ3

∫

∞

0

[

υ

(υ − ξ)2+

υ

(υ + ξ)2+

1

υ − ξ+

1

υ + ξ

]

dξ.

Here (after shifting the variable) one can use∫

dξξ

= ln ξ and∫

dξ

ξ2 = − 1

ξ, finally getting the same answer as by Kepler’s laws.

iv. (1.7 pts) By Clapeyron–Mendeleyev law,

p =mRT0

µV.

Work done by gravity to compress the ball is

W = −

∫

p dV = −mRT0

µ

∫ 4

3πr3

3

4

3πr3

0

dV

V=

3mRT0

µln

r0

r3

.

The temperature stays constant, so the internal energy does not

change; hence, according to the 1st law of thermodynamics, the

compression work W is the heat radiated.

v. (1 pt) The collapse continues adiabatically.

pV γ = const =⇒ T V γ−1 = const.

∴ T ∝ V 1−γ∝ r3−3γ

∴ T = T0

(r3

r

)3γ−3

.

vi. (2 pts) During the collapse, the gravitational energy is con-

verted into heat. Since r3 ≫ r4, The released gravitational en-

ergy can be estimated as ∆Π = −Gm2(r−14 −r−1

3 ) ≈ −Gm2/r4

(exact calculation by integration adds a prefactor 3

5); the ter-

minal heat energy is estimated as ∆Q = cVmµ

(T4 − T0) ≈

cVmµ

T4 (the approximation T4 ≫ T0 follows from the result

of the previous question, when combined with r3 ≫ r4). So,

∆Q = Rγ−1

mµ

T4 ≈mµ

RT4. For the temperature T4, we can use

the result of the previous question, T4 = T0

(

r3

r4

)3γ−3

. Since

initial full energy was approximately zero, ∆Q + ∆Π ≈ 0, we

obtain

Gm2

r4

≈m

µRT0

(

r3

r4

)3γ−3

=⇒ r4 ≈ r3

(

RT0r3

µmG

)1

3γ−4

.

Therefore,

T4 ≈ T0

(

RT0r3

µmG

)

3γ−3

4−3γ

.

Alternatively, one can obtain the result by approximately

equating the hydrostatic pressure ρr4Gmr2

4

to the gas pressure

p4 = ρ

µRT4; the result will be exactly the same as given above.

— page 5 of 5 —

PrObLEm T3. PrOTOsTar FOrmaTIOn (9 POInTs)

93



V ∝ r3

∴ p ∝ r−3 =⇒p(r1)

p(r0)= 23 = 8.









t ≈

√

2(r0 − r2)

g

where

g ≈Gm

r20

,

∴ t ≈

√

2r20(r0 − r2)

Gm=

√

0.1r30

Gm.













(

2π

2tr→0

)2r0

2=

Gm

(r0/2)2=⇒ tr→0 = π

√

r30

8Gm.


2−

Gmr



r0


(drdt

= −

√

2E + 2Gmr


dr√2E+ 2Gm

r

.




2E + 2Gmr

= ξ and√

2E = υ, one gets

t∞

4Gm=

∫

∞

0

dξ

(υ2 − ξ2)2

=1

4υ3

∫

∞

0

[

υ

(υ − ξ)2+

υ

(υ + ξ)2+

1

υ − ξ+

1

υ + ξ

]

dξ.


dξξ

= ln ξ and∫

dξ

ξ2 = − 1



p =mRT0

µV.


W = −

∫

p dV = −mRT0

µ

∫ 4

3πr3

3

4

3πr3

0

dV

V=

3mRT0

µln

r0

r3

.






∴ T ∝ V 1−γ∝ r3−3γ

∴ T = T0

(r3

r

)3γ−3

.




3 ) ≈ −Gm2/r4


5); the ter-


(T4 − T0) ≈

cVmµ



∆Q = Rγ−1

mµ

T4 ≈mµ



(

r3

r4

)3γ−3

. Since


obtain

Gm2

r4

≈m

µRT0

(

r3

r4

)3γ−3

=⇒ r4 ≈ r3

(

RT0r3

µmG

)1

3γ−4

.

Therefore,

T4 ≈ T0

(

RT0r3

µmG

)

3γ−3

4−3γ

.



4

to the gas pressure

p4 = ρ


— page 5 of 5 —



V ∝ r3

∴ p ∝ r−3 =⇒p(r1)

p(r0)= 23 = 8.









t ≈

√

2(r0 − r2)

g

where

g ≈Gm

r20

,

∴ t ≈

√

2r20(r0 − r2)

Gm=

√

0.1r30

Gm.













(

2π

2tr→0

)2r0

2=

Gm

(r0/2)2=⇒ tr→0 = π

√

r30

8Gm.


2−

Gmr



r0


(drdt

= −

√

2E + 2Gmr


dr√2E+ 2Gm

r

.




2E + 2Gmr

= ξ and√

2E = υ, one gets

t∞

4Gm=

∫

∞

0

dξ

(υ2 − ξ2)2

=1

4υ3

∫

∞

0

[

υ

(υ − ξ)2+

υ

(υ + ξ)2+

1

υ − ξ+

1

υ + ξ

]

dξ.


dξξ

= ln ξ and∫

dξ

ξ2 = − 1



p =mRT0

µV.


W = −

∫

p dV = −mRT0

µ

∫ 4

3πr3

3

4

3πr3

0

dV

V=

3mRT0

µln

r0

r3

.






∴ T ∝ V 1−γ∝ r3−3γ

∴ T = T0

(r3

r

)3γ−3

.




3 ) ≈ −Gm2/r4


5); the ter-


(T4 − T0) ≈

cVmµ



∆Q = Rγ−1

mµ

T4 ≈mµ



(

r3

r4

)3γ−3

. Since


obtain

Gm2

r4

≈m

µRT0

(

r3

r4

)3γ−3

=⇒ r4 ≈ r3

(

RT0r3

µmG

)1

3γ−4

.

Therefore,

T4 ≈ T0

(

RT0r3

µmG

)

3γ−3

4−3γ

.



4

to the gas pressure

p4 = ρ


— page 5 of 5 —

94



V ∝ r3

∴ p ∝ r−3 =⇒p(r1)

p(r0)= 23 = 8.









t ≈

√

2(r0 − r2)

g

where

g ≈Gm

r20

,

∴ t ≈

√

2r20(r0 − r2)

Gm=

√

0.1r30

Gm.













(

2π

2tr→0

)2r0

2=

Gm

(r0/2)2=⇒ tr→0 = π

√

r30

8Gm.


2−

Gmr



r0


(drdt

= −

√

2E + 2Gmr


dr√2E+ 2Gm

r

.




2E + 2Gmr

= ξ and√

2E = υ, one gets

t∞

4Gm=

∫

∞

0

dξ

(υ2 − ξ2)2

=1

4υ3

∫

∞

0

[

υ

(υ − ξ)2+

υ

(υ + ξ)2+

1

υ − ξ+

1

υ + ξ

]

dξ.


dξξ

= ln ξ and∫

dξ

ξ2 = − 1



p =mRT0

µV.


W = −

∫

p dV = −mRT0

µ

∫ 4

3πr3

3

4

3πr3

0

dV

V=

3mRT0

µln

r0

r3

.






∴ T ∝ V 1−γ∝ r3−3γ

∴ T = T0

(r3

r

)3γ−3

.




3 ) ≈ −Gm2/r4


5); the ter-


(T4 − T0) ≈

cVmµ



∆Q = Rγ−1

mµ

T4 ≈mµ



(

r3

r4

)3γ−3

. Since


obtain

Gm2

r4

≈m

µRT0

(

r3

r4

)3γ−3

=⇒ r4 ≈ r3

(

RT0r3

µmG

)1

3γ−4

.

Therefore,

T4 ≈ T0

(

RT0r3

µmG

)

3γ−3

4−3γ

.



4

to the gas pressure

p4 = ρ


— page 5 of 5 —

95

Grading scheme: Theory The 43rd International Physics Olympiad — July 2012

Grading scheme: Theory

General rules This grading scheme describes the number of

points allotted for each term entering a useful formula. These

terms don’t need to be separately described: if a formula is

written correctly, full marks (the sum of the marks of all the

terms of that formula) are given. If a formula is not written

explicitly, but it is clear that individual terms are written bear-

ing the equation in mind (eg. indicated on a diagram), marks

for these terms will be given. Some points are allotted for

mathematical calculations.

If a certain term of a useful formula is written incor-

rectly, 0.1 is subtracted for a minor mistake (eg. missing non-

dimensional factor); no mark is given if the mistake is major

(with non-matching dimensionality). The same rule is applied

to the marking of mathematical calculations: each minor mis-

take leads to a subtraction of 0.1 pts (as long as the remaining

score for that particular calculation remains positive), and no

marks are given in the case of dimensional mistakes.

No penalty is applied in these cases when a mistake is clearly

just a rewriting typo (i.e. when there is no mistake in the draft).

If formula is written without deriving: if it is simple

enough to be derived in head, full marks, otherwise zero

marks.

If there two solutions on Solution sheets, one correct and

another incorrect: only the one wich corresponds to the An-

swer Sheets is taken into account. What is crossed out is

never considered.

No penalty is applied for propagating errors unless the cal-

culations are significantly simplified (in which case mathemat-

ical calculations are credited partially, according to the degree

of simplification, with marking granularity of 0.1 pts).

— page 1 of 5 —

The 43rd International Physics Olympiad — July 2012

Grading scheme: Theory


points allotted for each term entering a useful formula. These

terms don’t need to be separately described: if a formula is

written correctly, full marks (the sum of the marks of all the

terms of that formula) are given. If a formula is not written

explicitly, but it is clear that individual terms are written bear-

ing the equation in mind (eg. indicated on a diagram), marks

for these terms will be given. Some points are allotted for

mathematical calculations.

If a certain term of a useful formula is written incor-

rectly, 0.1 is subtracted for a minor mistake (eg. missing non-

dimensional factor); no mark is given if the mistake is major

(with non-matching dimensionality). The same rule is applied

to the marking of mathematical calculations: each minor mis-

take leads to a subtraction of 0.1 pts (as long as the remaining

score for that particular calculation remains positive), and no

marks are given in the case of dimensional mistakes.



If formula is written without deriving: if it is simple

enough to be derived in head, full marks, otherwise zero

marks.

If there two solutions on Solution sheets, one correct and

another incorrect: only the one wich corresponds to the An-

swer Sheets is taken into account. What is crossed out is

never considered.





— page 1 of 5 —

96


i. (0.8 pts) If the parameters are derived using the particular

cases of throwing up and throwing horizontally:

for throwing up, zg = v20/2 — 0.2 pts;

from where z0 = v20/2g — 0.1 pts;

noticing that for horizontally thrown ball,

the trajectory has the same shape as z = −kx2 — 0.2 pts;

finding this trajectory, z = −gx2/2v20 — 0.2 pts;

Concluding k = g/2v20 — 0.1 pts;

If, instead of studying the trajectory of a horizontally

thrown ball (for which 0.5 pts were allocated), a trajectory

of a ball thrown at 45 degrees is studied:

distance is max. when the angle is 45 — 0.2

Finding this maximal distance v2/g — 0.2

Obtaining k — 0.1

If the parameters are derived using condition that the quad-

ratic equation for the tangent of the throwing angle has exactly

one real solution:

Requiring x = v cos αt — 0.1 pts;

Requiring z = v sin αt − gt2/2 — 0.2 pts;

Eliminating t: z = x tan α − gx2/v2 cos2 α — 0.1 pts;

Obtaining z − x tan α + gx2/v2(1 + tan2 α) = 0 — 0.1 pts;

Requiring that the discriminant is 0 — 0.2 pts;

Obtaining z0 = v20/2g, k = gx2/2v2

0 — 0.1 pts;

(If one of the two is incorrect — 0 pts for the last line)

ii. (1.2 pts)

Trajectory hits the sphere when descending — 0.7 pts(if the top of the parabola is higher than 5

2R — 0.5 pts);

Trajectory touches the sphere when ascending — 0.5 pts.

Trajectory touches the sphere at its top oris clearly non-parabolic or starts inside the sphere

or intersects the sphere: total — 0 pts.

iii. (2.5 pts) If the analysis is based on a trajectory which is

wrong in principle, 0 pts.

For minimal speed, z = z0 − kx2 touches the sphere — 1.5 pts

Shifting the origin to the sphere’s top for simplicity — 0.1 pts

Then z =v2

0

2g−

gx2

2v20

— 0.2 pts

and x2 + z2 + 2zR = 0 — 0.2 pts

⇒ x4

(

g

2v2

0

)2

+ x2

(

1

2−

gR

v2

0

)

+(

v2

0

4g+ R

)

v2

0

g= 0 — 0.1 pts

(full 0.6 pts if equivalent eq. is obtained for non-shifted origin)

Discriminant equals 0 — 0.2 pts

from where v20 = 0.5gR — 0.1 pts

hence vmin =√

4.5gR — 0.1 pts

If alternative solution is followed:


Noting that the touching point vel. is ⊥ to launching vel.— 0.6 pts

Angle from centre to touching pt P = launching angle — 0.6 pts

Cond. that P belongs to the trajectory — 0.6 pts

From this cond., final answer obtained — 0.6 pts

For a brute force approach:

Obtaining 4th order equationfor intersection points x (or y) — 0.5 pts

which is reduced to cubic(divided by x) — 0.1 pts

Mentioning that it has exactly one pos. root — 0.2 pts(equivalently, an extrememum coincides with a root or there

are exactly two distinct real roots.)

Obtaining quadratic eq. for x-coord. of extrema — 0.2 pts

Finding the roots of it — 0.2 pts

Selecting the larger root x∗ of it — 0.2 pts

Requiring that x∗ is the root of the cubic eq — 0.2 pts

Obtaining the speed as a function of α — 0.2 pts

Finding the minimum of that function — 0.7 pts

Part B. Air flow around a wing (4 points)

i. (0.8 pts)

Using the wing’s frame of reference — 0.1 pts

taking streamline distance measurment at P — 0.2 pts(if off by 1 mm or more, 0.1 pts)

measuring unperturbed streamline distance — 0.1 pts(if off by 1 mm or more, 0 pts)

at P , streamlines are horiz. ⇒ scalar adding — 0.1 pts

Writing continuity condition — 0.2 pts

Finding final answer — 0.1 pts

(if the speed in wing’s frame is given as final anser, points

are lost for 4th and 6th line)

ii. (1.2 pts)

Writing down continuity condition — 0.2 pts(or stating: smaller streamline distance ⇒ larger speed)

Writing Bernoulli’s law — 0.4 pts0 pts out of 0.4 if the Bernoulli law includes ρgh

(or stating that larger speed ⇒ smaller pressure)

Writing adiabatic law — 0.4 pts

— page 2 of 5 —

PrObLEm T1. FOcus On skETchEs (13 POInTs)

97















Obtaining k — 0.1



one real solution:







0 — 0.1 pts;


ii. (1.2 pts)


2R — 0.5 pts);








Then z =v2

0

2g−

gx2

2v20

— 0.2 pts

and x2 + z2 + 2zR = 0 — 0.2 pts

⇒ x4

(

g

2v2

0

)2

+ x2

(

1

2−

gR

v2

0

)

+(

v2

0

4g+ R

)

v2

0

g= 0 — 0.1 pts




hence vmin =√

4.5gR — 0.1 pts



















i. (0.8 pts)









ii. (1.2 pts)





— page 2 of 5 —Problem T1. Focus on sketches (13 points)Part A. Ballistics (4.5 points)














Obtaining k — 0.1



one real solution:







0 — 0.1 pts;


ii. (1.2 pts)


2R — 0.5 pts);








Then z =v2

0

2g−

gx2

2v20

— 0.2 pts

and x2 + z2 + 2zR = 0 — 0.2 pts

⇒ x4

(

g

2v2

0

)2

+ x2

(

1

2−

gR

v2

0

)

+(

v2

0

4g+ R

)

v2

0

g= 0 — 0.1 pts




hence vmin =√

4.5gR — 0.1 pts



















i. (0.8 pts)









ii. (1.2 pts)





— page 2 of 5 —

98















Obtaining k — 0.1



one real solution:







0 — 0.1 pts;


ii. (1.2 pts)


2R — 0.5 pts);








Then z =v2

0

2g−

gx2

2v20

— 0.2 pts

and x2 + z2 + 2zR = 0 — 0.2 pts

⇒ x4

(

g

2v2

0

)2

+ x2

(

1

2−

gR

v2

0

)

+(

v2

0

4g+ R

)

v2

0

g= 0 — 0.1 pts




hence vmin =√

4.5gR — 0.1 pts



















i. (0.8 pts)









ii. (1.2 pts)





— page 2 of 5 —(or stating that smaller pressure ⇒ lower temperature)

Finding Q as in the figure in the solutions — 0.2 pts(if Q marked below the wing’s tip — 0.1 pts)

iii. (2 pts)

Finding the dew point: idea of linearization — 0.2 pts

Expression and/or numerical value for dew point — 0.2 pts

Deriving 1

2µv2 + cpT = const:

1 mole of gas carries kin. en. 1

2µv2 — 0.2 pts

1 mole of gas carries heat en. CV T — 0.2 pts

work done on 1 mole of gas: p1V1 − p2V2 — 0.4 pts

en.: 1

2µ(v2

2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts

using ideal gas law obtaining 1

2v2 + cpT = const — 0.2 pts

Alternative approximate approach:

Bernoulli’s law 1

2ρv2 + p = const — 0.3 pts

0 pts out of 0.3 if the Bernoulli law includes ρgh

Adiabatic law pV γ = const — 0.3 pts

⇒ p1−γT γ = const — 0.2 pts

Approximation ∆p = γγ−1

pT

∆T — 0.2 pts

Leading to 1

2v2 + R

µ

cp

cp−cVT = const — 0.1 pts

Leading to 1


And further (for either approach):

Bringing it to ∆v2

2= 1

2v2

crit(a2

c2 − 1) = cp∆T — 0.1 pts

Measuring a and c — 0.2 pts

Obtaining vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s — 0.1 pts


i. (0.8 pts)

Lines straight and parallel inside — 0.6 pts

(if not drawn over the entire length, subtract 0.2)

(can be slightly curved close the tube’s end)

Lines curve outwards slightly after the exit — 0.2 pts

(if so curved that more than one closed loop on both sides is

depicted in Fig, 0 pts out of 0.2)

ii. (1.2 pts)

Expressing induction as B = Φ/πr2 — 0.2 pts

Stating that w = B2

2µ0

— 0.2 pts

Idea of using virtual lengthening — 0.2 pts

(Introducing a lengthening ∆l is enough)

Expressing ∆W = B2

2µ0

πr2∆l — 0.2 pts

(Same marks if W expressed for entire tube)

Equating ∆W = T ∆l — 0.2 pts

Expressing T = Φ2

2µ0πr2 . — 0.2 pts

iii. (2.5 pts)

Idea of using analogy with el. charges — 1 pt

(Sketch with a quadrupole conf. of el. charges is enough)

Finding the force between two magn. charges via matching el.

and magn. quantities is worth 1 pts in total, split down as

follows:

Expressing el. stat. force via en. density — 0.5 pts

Using the obtained Eq. to obtain F = 1

4πµ0

Φ2

a2 — 0.5 pts

Any other matching scheme is graded analogously; e.g. find-

ing such a Q which has the same en. density as Φ — 0.5 pts;

expressing the force between magn. charges (Φ) as the force

between the matching el. charges (Q) — 0.5 pts. Declaring

matching pairs Q ↔ Φ and 1

4πε0

↔1

4πµ0

without energy-based-

motivation gives only 0.5 pts.

Noting that to tubes comp. of force = 0 — 0.2 pts

When expressing F = 2(F1 − F2): for factor “2” — 0.1 pts

and for finding F2 — 0.2 pts

(if wrong sign for F2, subtract 0.1)

If alternative solution is followed

Plan to express inter. energy as a function of a

intending to find F as a derivative — 0.2 pts

Calculating B(x) — 0.8 pts

If estimated without dependance on x, 0.4

considering a tube as an array of dipoles — 0.3 pts

expressing dm = Sjdx — 0.3 pts

relating j to Φ — 0.2 pts

Expressing U =∫

B(x)dm — 0.4 pts

If estimated as BSjl, 0.2

Finding F = dU/da — 0.2 pts

Taking into account the factor “ 2”— 0.1 pts

— page 3 of 5 —

99

(or stating that smaller pressure ⇒ lower temperature)


iii. (2 pts)



Deriving 1



2µv2 — 0.2 pts



en.: 1

2µ(v2

2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts




Bernoulli’s law 1






pT

∆T — 0.2 pts

Leading to 1

2v2 + R

µ

cp


Leading to 1




2= 1

2v2

crit(a2

c2 − 1) = cp∆T — 0.1 pts


Obtaining vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s — 0.1 pts


i. (0.8 pts)







ii. (1.2 pts)


Stating that w = B2

2µ0

— 0.2 pts




2µ0




Expressing T = Φ2

2µ0πr2 . — 0.2 pts

iii. (2.5 pts)





follows:



4πµ0

Φ2

a2 — 0.5 pts






4πε0

↔1

4πµ0















Expressing U =∫

B(x)dm — 0.4 pts




— page 3 of 5 —

100



iii. (2 pts)



Deriving 1



2µv2 — 0.2 pts



en.: 1

2µ(v2

2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts




Bernoulli’s law 1






pT

∆T — 0.2 pts

Leading to 1

2v2 + R

µ

cp


Leading to 1




2= 1

2v2

crit(a2

c2 − 1) = cp∆T — 0.1 pts


Obtaining vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s — 0.1 pts


i. (0.8 pts)







ii. (1.2 pts)


Stating that w = B2

2µ0

— 0.2 pts




2µ0




Expressing T = Φ2

2µ0πr2 . — 0.2 pts

iii. (2.5 pts)





follows:



4πµ0

Φ2

a2 — 0.5 pts






4πε0

↔1

4πµ0















Expressing U =∫

B(x)dm — 0.4 pts




— page 3 of 5 —



iii. (2 pts)



Deriving 1



2µv2 — 0.2 pts



en.: 1

2µ(v2

2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts




Bernoulli’s law 1






pT

∆T — 0.2 pts

Leading to 1

2v2 + R

µ

cp


Leading to 1




2= 1

2v2

crit(a2

c2 − 1) = cp∆T — 0.1 pts


Obtaining vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s — 0.1 pts


i. (0.8 pts)







ii. (1.2 pts)


Stating that w = B2

2µ0

— 0.2 pts




2µ0




Expressing T = Φ2

2µ0πr2 . — 0.2 pts

iii. (2.5 pts)





follows:



4πµ0

Φ2

a2 — 0.5 pts






4πε0

↔1

4πµ0















Expressing U =∫

B(x)dm — 0.4 pts




— page 3 of 5 —

101



iii. (2 pts)



Deriving 1



2µv2 — 0.2 pts



en.: 1

2µ(v2

2 − v21) + CV (T2 − T1) = p1V1 − p2V2 — 0.2 pts




Bernoulli’s law 1






pT

∆T — 0.2 pts

Leading to 1

2v2 + R

µ

cp


Leading to 1




2= 1

2v2

crit(a2

c2 − 1) = cp∆T — 0.1 pts


Obtaining vcrit = c

√

2cp∆T

a2 − c2≈ 23 m/s — 0.1 pts


i. (0.8 pts)







ii. (1.2 pts)


Stating that w = B2

2µ0

— 0.2 pts




2µ0




Expressing T = Φ2

2µ0πr2 . — 0.2 pts

iii. (2.5 pts)





follows:



4πµ0

Φ2

a2 — 0.5 pts






4πε0

↔1

4πµ0















Expressing U =∫

B(x)dm — 0.4 pts




— page 3 of 5 —


i. (1.2 pts) For the terms entering the force balance of a

droplet immediately before separation from the nozzle, the

points are given as follows:

mg — 0.2 pts;

m = ρV — 0.1 pts;

V =4

3πr3

max — 0.1 pts;

πσd — 0.4 pts.

(if geometrically obtained cos α is included, 0.2 pts)

Force balance equation including these terms — 0.2 pts;

For expressing rmax from the equation — 0.2 pts.

ii. (1.2 pts)

Stating that the droplet’s potential is ϕ — 0.2 pts

(if used correctly, this does not need to be explicitly stated).

Expressing the droplet’s potential as1

4πε0

Q

r— 0.8 pts

(without correct sign — 0.6 pts).

Expressing Q from the obtained equation — 0.2 pts.

iii. (1.6 pts) The components of the excess pressure are graded

as follows.

2σ/r — 0.5 pts;

1

2ε0E2 — 0.4 pts;

bringing it to the form1

2ε0ϕ2/r2 — 0.2 pts.

Noticing that the two effects have opposite sign — 0.2 pts

(if used correctly, this does not need to be stated separately).

Equation stating that the excess pressure is 0 — 0.1 pts;

Expressing ϕmax from the obtained equation — 0.2 pts.

In the case of energy-balance-based solution, the distribu-

tion of marks is as follows.

surface energy change as 4πσd(r2) = 8πσrdr — 0.5 pts;

electrostatic energy change as 2πε0ϕ2maxdr — 0.5 pts;

electrostatic work as dAel

= 4πε0ϕ2maxdr — 0.3 pts;

equation stating that en. change equals to work — 0.1 pts;

expressing ϕmax from the obtained equation — 0.2 pts.

0 pts if energy are equated (without virtual displacement).

If factor 1

2missing before the electrostat. force (but other-

wise correct), -0.2 pts).


i. (1.2 pts)

Stating that the surroundings’ potential is − U/2 — 0.4 pts;

(if not stated but used correctly — full marks; opposite signs

are allowed to be chosen consistently)

stating that the droplet’s potential is 0 — 0.4 pts

(if not stated but used correctly — full marks)

Applying formula U = q/C — 0.2 pts

Using the result of Part A with

ϕ = U/2 to obtain the final result — 0.2 pts.

the solutions with U instead of U/2 will qualify for 0.4 pts

for the first two lines (i.e. in total up to 0.8)

ii. (1.5 pts)

Stating that a droplet will increase the capacitor’s

charge by its own charge Q = 2πε0qrmax/C — 0.4 pts;

(0 pts if wrong sign)

(0.2 pts if redundant factor “2”)

Expressing dq = Q dN — 0.2 pts;

Substituting dN = n dt — 0.2 pts;

solving the obtained differential equation — 0.5 pts;

determining the integration constant from

the initial condition — 0.1 pts;

substituting rmax from above — 0.1 pts.

iii. (1.3 pts) Equation for the energy balance of a droplet:

expressing the droplet’s electrostatic energy change during

the fall as UQ — 0.6 pts

(0.3 for UQ/2)

expressing the droplet’s gravitational energy change as mgH

— 0.2 pts

noticing that at the limit voltage, droplet’s terminal kinetic

energy is zero — 0.3 pts

expressing energy conservation law equation — 0.1 pts

expressing the final answer — 0.1 pts.

(if bowl considered as a point charge, only the mgh line

and the kin. en. lines are applicable)

If instead of the energy balance, the force balance is used

(which is incorrect), partial credit for the terms entering the

equation is given as follows.

gravity force gm — 0.2 pts

electric field estimated as E ≈ U/(H − L/2) — 0.2 pts

(0.1, if not realized that this is an approximation)

electric force F = EQ — 0.1 pts

writing down force balance equation — 0.1 pts.


(if bowl considered as a point charge, only the first line and

the force balance line are applicable)

— page 4 of 5 —


102





mg — 0.2 pts;

m = ρV — 0.1 pts;

V =4

3πr3

max — 0.1 pts;

πσd — 0.4 pts.




ii. (1.2 pts)




4πε0

Q

r— 0.8 pts




as follows.

2σ/r — 0.5 pts;

1

2ε0E2 — 0.4 pts;


2ε0ϕ2/r2 — 0.2 pts.














If factor 1




i. (1.2 pts)











ii. (1.5 pts)














(0.3 for UQ/2)


— 0.2 pts


















— page 4 of 5 —





mg — 0.2 pts;

m = ρV — 0.1 pts;

V =4

3πr3

max — 0.1 pts;

πσd — 0.4 pts.




ii. (1.2 pts)




4πε0

Q

r— 0.8 pts




as follows.

2σ/r — 0.5 pts;

1

2ε0E2 — 0.4 pts;


2ε0ϕ2/r2 — 0.2 pts.














If factor 1




i. (1.2 pts)











ii. (1.5 pts)














(0.3 for UQ/2)


— 0.2 pts


















— page 4 of 5 —





mg — 0.2 pts;

m = ρV — 0.1 pts;

V =4

3πr3

max — 0.1 pts;

πσd — 0.4 pts.




ii. (1.2 pts)




4πε0

Q

r— 0.8 pts




as follows.

2σ/r — 0.5 pts;

1

2ε0E2 — 0.4 pts;


2ε0ϕ2/r2 — 0.2 pts.














If factor 1




i. (1.2 pts)











ii. (1.5 pts)














(0.3 for UQ/2)


— 0.2 pts


















— page 4 of 5 —

103





mg — 0.2 pts;

m = ρV — 0.1 pts;

V =4

3πr3

max — 0.1 pts;

πσd — 0.4 pts.




ii. (1.2 pts)




4πε0

Q

r— 0.8 pts




as follows.

2σ/r — 0.5 pts;

1

2ε0E2 — 0.4 pts;


2ε0ϕ2/r2 — 0.2 pts.














If factor 1




i. (1.2 pts)











ii. (1.5 pts)














(0.3 for UQ/2)


— 0.2 pts


















— page 4 of 5 —

104


In thermodynamic equilibrium T = const — 0.2 pts.(if not stated but used correctly — full marks)

pV = const — 0.3 pts.

V ∝ r3 — 0.1 pts.

p ∝ r−3 — 0.1 pts.

p(r1)

p(r0)= 8 — 0.1 pts.

ii. (1 pt)

t ≈

√

2(r0 − r2)

g— 0.4 pts.

g ≈Gm

r20

— 0.4 pts.

t ≈

√

0.1r30

Gm— 0.2 pts.

iii. (2.5 pts)

First solution:

Understanding that we have effectivelyinteraction of two point masses — 0.5 pts.

(equivalently one may mention the ∝ r−2 force coming from

Gauss’ law; if not stated, but used correctly — full marks)

Idea of the motion as an ultraelliptical orbit — 1 pt.

Period of the elliptical orbit is equal to the periodof the circular orbit of the same longer semiaxis — 0.4 pts.


The longer semiaxis is r0/2 — 0.1 pts.

Equation(s) for the period — 0.3 pts.

We need half a period — 0.1 pts.

Final answer tr→0 = π

√

r30

8Gm— 0.1 pts.

Alternative solution:




Energy conservation as a differential equation — 0.3 pts.

(if only expressed through v (like mv2

2− GMm

r= E) — 0.1 pts.;

if the differential equation of Newton’s 2nd law (r = −Gmr2 ) is

given instead — 0.2 pts.)

t =

∫

dr√

2E + 2Gmr

(whichever the sign is) — 0.4 pts.

Integration and final answer — 1.3 pts.

iv. (1.7 pts)

Radiated heat equals the compression work — 1 pt.

[for mentioning or using the 1st law of thermodynamics (pos-

sibly in a wrong way) — 0.5 pts.]

W = −

∫

p dV or W = −∑

p ∆V — 0.3 pts.

(with either “+” or “−” — give full marks)

p =mRT0

µV— 0.2 pts.

Calculating the integral, W =3mRT0

µln

r0

r3

— 0.2 pts.

v. (1 pt)

The collapse continues adiabatically.(If used, but not written down, give full marks.) — 0.3 pts.

pV γ = const — 0.3 pts.

T ∝ V 1−γ — 0.2 pts.

T = T0

(r3

r

)3γ−3

— 0.2 pts.

vi. (2 pts)

In case of using energies (marks must not be subtracted for

fewer approximations, even in the final answer):

T4 = T0

(

r3

r4

)3γ−3

— 0.1 pts.

∆Q + ∆Π ≈ 0 — 0.4 pts.

∆Π ≈ −Gm2/r4 — 0.6 pts.

∆Q = mcV T4 — 0.4 pts.

cV ≈R

µ— 0.3 pts.

Final answer r4 ≈ r3

(

RT0r3

µmG

)1

3γ−4

— 0.1 pts.

Final answer T4 ≈ T0

(

RT0r3

µmG

)

3γ−3

4−3γ

— 0.1 pts.

In case of using pressures (again, fewer approximations are per-

mitted):

T4 = T0

(

r3

r4

)3γ−3

— 0.1 pts.

p4 = phydrostatic — 0.4 pts.

p4 =ρ

µRT4 — 0.5 pts.

phydrostatic ≈ ρgr4 — 0.4 pts.

g ≈Gm

r24

— 0.4 pts.


(

RT0r3

µmG

)1

3γ−4

— 0.1 pts.


(

RT0r3

µmG

)

3γ−3

4−3γ

— 0.1 pts.

— page 5 of 5 —

PrObLEm T3. PrOTOsTar FOrmaTIOn (9 POInTs)

105




V ∝ r3 — 0.1 pts.

p ∝ r−3 — 0.1 pts.

p(r1)

p(r0)= 8 — 0.1 pts.

ii. (1 pt)

t ≈

√

2(r0 − r2)

g— 0.4 pts.

g ≈Gm

r20

— 0.4 pts.

t ≈

√

0.1r30

Gm— 0.2 pts.

iii. (2.5 pts)

First solution:











√

r30

8Gm— 0.1 pts.







2− GMm

r= E) — 0.1 pts.;



t =

∫

dr√

2E + 2Gmr



iv. (1.7 pts)




W = −

∫

p dV or W = −∑

p ∆V — 0.3 pts.


p =mRT0

µV— 0.2 pts.


µln

r0

r3

— 0.2 pts.

v. (1 pt)



T ∝ V 1−γ — 0.2 pts.

T = T0

(r3

r

)3γ−3

— 0.2 pts.

vi. (2 pts)



T4 = T0

(

r3

r4

)3γ−3

— 0.1 pts.

∆Q + ∆Π ≈ 0 — 0.4 pts.

∆Π ≈ −Gm2/r4 — 0.6 pts.

∆Q = mcV T4 — 0.4 pts.

cV ≈R

µ— 0.3 pts.


(

RT0r3

µmG

)1

3γ−4

— 0.1 pts.


(

RT0r3

µmG

)

3γ−3

4−3γ

— 0.1 pts.


mitted):

T4 = T0

(

r3

r4

)3γ−3

— 0.1 pts.


p4 =ρ

µRT4 — 0.5 pts.


g ≈Gm

r24

— 0.4 pts.


(

RT0r3

µmG

)1

3γ−4

— 0.1 pts.


(

RT0r3

µmG

)

3γ−3

4−3γ

— 0.1 pts.

— page 5 of 5 —Problem T3. Protostar formation (9 points)i. (0.8 pts)



V ∝ r3 — 0.1 pts.

p ∝ r−3 — 0.1 pts.

p(r1)

p(r0)= 8 — 0.1 pts.

ii. (1 pt)

t ≈

√

2(r0 − r2)

g— 0.4 pts.

g ≈Gm

r20

— 0.4 pts.

t ≈

√

0.1r30

Gm— 0.2 pts.

iii. (2.5 pts)

First solution:











√

r30

8Gm— 0.1 pts.







2− GMm

r= E) — 0.1 pts.;



t =

∫

dr√

2E + 2Gmr



iv. (1.7 pts)




W = −

∫

p dV or W = −∑

p ∆V — 0.3 pts.


p =mRT0

µV— 0.2 pts.


µln

r0

r3

— 0.2 pts.

v. (1 pt)



T ∝ V 1−γ — 0.2 pts.

T = T0

(r3

r

)3γ−3

— 0.2 pts.

vi. (2 pts)



T4 = T0

(

r3

r4

)3γ−3

— 0.1 pts.

∆Q + ∆Π ≈ 0 — 0.4 pts.

∆Π ≈ −Gm2/r4 — 0.6 pts.

∆Q = mcV T4 — 0.4 pts.

cV ≈R

µ— 0.3 pts.


(

RT0r3

µmG

)1

3γ−4

— 0.1 pts.


(

RT0r3

µmG

)

3γ−3

4−3γ

— 0.1 pts.


mitted):

T4 = T0

(

r3

r4

)3γ−3

— 0.1 pts.


p4 =ρ

µRT4 — 0.5 pts.


g ≈Gm

r24

— 0.4 pts.


(

RT0r3

µmG

)1

3γ−4

— 0.1 pts.


(

RT0r3

µmG

)

3γ−3

4−3γ

— 0.1 pts.

— page 5 of 5 —

106




V ∝ r3 — 0.1 pts.

p ∝ r−3 — 0.1 pts.

p(r1)

p(r0)= 8 — 0.1 pts.

ii. (1 pt)

t ≈

√

2(r0 − r2)

g— 0.4 pts.

g ≈Gm

r20

— 0.4 pts.

t ≈

√

0.1r30

Gm— 0.2 pts.

iii. (2.5 pts)

First solution:











√

r30

8Gm— 0.1 pts.







2− GMm

r= E) — 0.1 pts.;



t =

∫

dr√

2E + 2Gmr



iv. (1.7 pts)




W = −

∫

p dV or W = −∑

p ∆V — 0.3 pts.


p =mRT0

µV— 0.2 pts.


µln

r0

r3

— 0.2 pts.

v. (1 pt)



T ∝ V 1−γ — 0.2 pts.

T = T0

(r3

r

)3γ−3

— 0.2 pts.

vi. (2 pts)



T4 = T0

(

r3

r4

)3γ−3

— 0.1 pts.

∆Q + ∆Π ≈ 0 — 0.4 pts.

∆Π ≈ −Gm2/r4 — 0.6 pts.

∆Q = mcV T4 — 0.4 pts.

cV ≈R

µ— 0.3 pts.


(

RT0r3

µmG

)1

3γ−4

— 0.1 pts.


(

RT0r3

µmG

)

3γ−3

4−3γ

— 0.1 pts.


mitted):

T4 = T0

(

r3

r4

)3γ−3

— 0.1 pts.


p4 =ρ

µRT4 — 0.5 pts.


g ≈Gm

r24

— 0.4 pts.


(

RT0r3

µmG

)1

3γ−4

— 0.1 pts.


(

RT0r3

µmG

)

3γ−3

4−3γ

— 0.1 pts.

— page 5 of 5 —

107

The 43rd International Physics Olympiad — Experimental Competition

Tartu, Estonia — Thursday, July 19th 2012


worth in total 20 points. There are two tables in your

disposal (in two neighbouring cubicles), the apparatus of

Problem E1 is on one table and the apparatus of Prob-

lem E2 is on the other table; you can move freely between

these tables. However, you are not allowed to move

any piece of experimental setup from one table to

the other.

• Initially the experimental equipment on one table is

covered and on the other table is boxed. You must

neither remove the cover nor open the box nor

open the envelope with the problems before the

sound signal of the beginning of competition

(three short signals).


without permission. If you need any assistance (mal-

functioning equipment, broken calculator, need to

visit a restroom, etc), please raise the corresponding flag

(“help” or “toilet” with a long handle at your seat)

above your seat box walls and keep it raised until an or-

ganizer arrives.

















more than once).

• You should use as little text as possible: try to

explain your solution mainly with equations, numbers,

tables, symbols and diagrams. When textual explana-

tion is unavoidable, you are encouraged to provide Eng-

lish translation alongside with the text in your native

language (if you mistranslate, or don’t translate at all,

your native language text will be used during the Moder-

ation).

• Avoid unnecessary movements during the experimental

examination and do not shake the walls of your box: the

laser experiment requires stability.

• Do not look into the laser beam or its reflections! It may

permanently damage your eyes.










— page 1 of 4 —

The 43rd International Physics Olympiad — Experimental CompetitionTartu, Estonia — Thursday, July 19th 2012

108










the other.













ganizer arrives.

















more than once).








ation).















— page 1 of 4 —










the other.













ganizer arrives.

















more than once).








ation).















— page 1 of 4 —










the other.













ganizer arrives.

















more than once).








ation).















— page 1 of 4 —

109

Problem E1. The magnetic permeability of water(10 points)

The effect of a magnetic field on most of the substancesbut ferromagnetics is rather weak. This is because the energydensity of the magetic field in substances of relative magneticpermeability µ is given by formula w = 1

2µµ0

B2, and typically,

µ is very close to 1. Still, with suitable experimental techniquessuch effects are firmly observable. In this problem we study theeffect of a magnetic field, created by a permanent neodymiummagnet, on water and use the results to calculate the magneticpermeability of water. You are not asked to estimate anyuncertainties throughout this problem and you do notneed to take into account the effects of the surface ten-sion.The setup comprises of 1 a stand (the highlighted numbers

correspond to the numbers in the fig.), 3 a digital caliper, 4

a laser pointer, 5 a dish with water and 7 a cylindricalpermanent magnet in it (the magnet is axially magnetised).The dish is fixed to the base of the stand by the magnet’s pull.The laser is fixed to the caliper, the base of which is fastenedto the stand; the caliper allows horizontal displacement of thelaser. The on-off button of the laser can be kept down withthe help of 13 the white conical tube. The depth of the wa-ter above the magnet should be reasonably close to 1 mm (ifshallower, the water surface becomes so curved that it will bedifficult to take readings from the screen). 15 A cup of water

and 16 a syringe can be used for the water level adjustment

(to raise the level by 1 mm, add 13 ml of water). 2 A sheet ofgraph paper (the “screen”) is to be fixed to the vertical plate

with 14 small magnetic tablets. If the laser spot on the screenbecomes smeared, check for a dust on the water surface (andblow away).

The remaining legend for the figure is as follows: 6 the

point where the laser beam hits the screen; 11 the LCD screen

of the caliper, 10 the button which switches the caliper units

between millimeters and inches; 8 on-off switch; 9 buttonfor setting the origin of the caliper reading. Beneath the laserpointer, there is one more button of the caliper, which tempor-arily re-sets the origin (if you pushed it inadvertently, push itonce again to return to the normal measuring mode).

Numerical values for your calculations: horizontal distancebetween the magnet’s centre and the screen L0 = 490 mm.Check (and adjust, if needed) the alignment of the centre ofthe magnet in two perpendicular directions. The vertical axisof the magnet must intersect with the laser beam, and it mustalso intersect with 12 the black line on the support plate.Magnetic induction at the magnet’s axis, at the height of1 mm from the flat surface B0 = 0.50 T; Density of waterρw = 1000 kg/m3; free-fall acceleration g = 9.8 m/s2; va-cuum permeability µ0 = 4π × 10−7 H/m.

WARNINGS:

⋄ The laser orientation is pre-adjusted, do not move it!

⋄ Do not look into the laser beam or its reflections!

⋄ Do not try to remove the strong neodymium magnet!

⋄ Do not put magnetic materials close to the magnet!

⋄ Turn off the laser when not used, batteries drain in 1 h!

Part A. Qualitative shape of the water surface (1 points)When a cylindrical magnet is placed below water surface, thelatter becomes curved. Observe, what is the shape of the watersurface above the magnet; according to this observation, decideif the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).

Write the letter corresponding to the correct option into theAnswer Sheet, together with an inequality µ > 1 or µ < 1. Forthis part, you don’t need to motivate your answer.

Part B. Exact shape of the water surface (7 points)

Curving of the water surface can be checked with high sensitiv-ity by measuring the reflection of the laser beam by the surface.We use this effect to calculate the dependence of the depth ofthe water on the horizontal position above the magnet.

i. ( 1.6 pts) Measure the dependence of the height y of thelaser spot on the screen on the caliper reading x (c.f. figure).You should cover all the usable range of caliper displacements.Fill the results into the table in the Answer Sheet.

ii. (0.7 pts) Draw the graph of the obtained dependence.

iii. ( 0.7 pts) Using the obtained graph, determine the angleα0 between the beam and the horizontal area of the watersurface.

iv. ( 1.4 pts) Please note that the slope (tan β) of the watersurface can be expressed as follows:

tan β ≈ β ≈cos2 α0

2·

y − y0 − (x − x0) tan α0

L0 + x − x0

,

where y0 is the height of the laser spot on the screen whenthe beam is reflected from the water surface at the axis of themagnet, and x0 — the respective position of the caliper.

Calculate the values of the water surface slope and fill theminto the Table on the Answer Sheet. Please note that it maybe possible to simplify your calculations if you substitute somecombination of terms in the expression from the previous taskwith a reading from the last graph.

v. ( 1.6 pts) Calculate the height of the water surface relativeto the surface far from the magnet as a function of x and fill itinto the Table on the Answer Sheet.

vi. (1 pt) Draw the graph of the latter dependence. Indicateon it the region where the beam hits the water surface directlyabove the magnet.

Part C. Magnetic permeability (2 points)

Using the results of Part B, calculate the value of µ − 1 (theso-called magnetic susceptibility), where µ is the relative mag-netic permeability of the water. Write your final formula andthe numerical result into the Answer Sheet.

— page 2 of 4 —



2µµ0

B2, and typically,












WARNINGS:















2·

y − y0 − (x − x0) tan α0

L0 + x − x0

,







— page 2 of 4 —

PrObLEm E1. ThE magnETIc PErmEabILITy OF waTEr (10 POInTs)

110



2µµ0

B2, and typically,












WARNINGS:















2·

y − y0 − (x − x0) tan α0

L0 + x − x0

,







— page 2 of 4 —



2µµ0

B2, and typically,












WARNINGS:















2·

y − y0 − (x − x0) tan α0

L0 + x − x0

,







— page 2 of 4 —



2µµ0

B2, and typically,












WARNINGS:















2·

y − y0 − (x − x0) tan α0

L0 + x − x0

,







— page 2 of 4 —



2µµ0

B2, and typically,












WARNINGS:















2·

y − y0 − (x − x0) tan α0

L0 + x − x0

,







— page 2 of 4 —

111



2µµ0

B2, and typically,












WARNINGS:















2·

y − y0 − (x − x0) tan α0

L0 + x − x0

,







— page 2 of 4 —

112

Problem E2. Nonlinear Black Box (10 points)In simple problems, electrical circuits are assumed to consistof linear elements, for which electrical characteristics are dir-ectly proportional to each other. Examples include resistance(V = RI), capacitance (Q = CV ) and inductance (V = LI),where R, C and L are constants. In this problem, however, weexamine a circuit containing nonlinear elements, enclosed intoa black box, for which the assumption of proportionality nolonger holds.The setup comprises of a multimeter (labelled “IPhO-measure”), a current source, a black box containing non-linear elements, and four test leads with stackable connectorsfor wiring. Be careful not to break the seal on the black box.

The multimeter can measure current and voltage simultan-eously. You can record with it up to 2000 data points, eachconsisting of: voltage V , current I, power P = IV , resistanceR = V/I, voltage time-derivative V , current time-derivative Iand time t. See its manual for details. If you go beyond 2000data points, the oldest data will be overwritten.

OUTIN

Multimeter

GND

A V

The constant current source supplies stable current as longas the voltage across its terminals stays between −0.6125Vand 0.6125V. When switched off, the constant current sourcebehaves as a large (essentially infinite) resistance.

Current source

+

-

I=6mA

U=-612.5mV...612.5mV

The black box contains an electric double layer capacitor(which is a slightly nonlinear high capacitance capacitor), anunknown nonlinear element and an inductor L = 10µH of negli-gible resistance, switchable as indicated on the circuit diagram.The nonlinear element can be considered as a resistance with anonlinear dependence between the voltage and the current [ Iis a continuous function of V with I(0) = 0]. Likewise, for thecapacitor, the differential capacitance C(V ) = dQ/dV is notexactly constant. We say that the voltage on the blackbox is positive when the potential on its red terminal ishigher than the potential on the black terminal. Pos-

itive voltage will be acquired when the terminals ofmatching colours on the black box and the currentsource are connected (you are allowed to use negativevoltages).

Black box

Nonlinear

device

+

-C(V)

Here it is safe to discharge the capacitor in the black box byshorting its inputs either by itself or through the IN and OUTterminals on multimeter: the internal resistance of this capa-citor is enough to keep the current from damaging anything.

You are not asked to estimate any uncertaintiesthroughout this problem.

Part A. Circuit without inductance (7 points)In this part, keep the switch on the black box closed (push “I”down), so that the inductance is shorted.

Please note that some measurements may take considerabletime, therefore it is recommended to read through all the tasksof part A to avoid unnecessary work.i. (1 pt) Confirm that the output current of the current sourceis approximately 6 mA, and determine the range within whichit varies for voltages between 0 and +480 mV. Document thecircuit diagram.ii. (1.2 pts) Show that the differential capacitance C(V ) usedin the black box is approximately 2 F by measuring its valuefor a single voltage of your choice C(V0) = C0. Document thecircuit diagram.iii. (2.2 pts) Neglecting the nonlinearity of the capacitance[C(V ) ≈ C0], determine the current–voltage characteristic ofthe nonlinear element used in the black box. Plot the I(V )curve for obtainable positive voltages on the black box ontothe answer sheet. Document the circuit diagram.iv. (2.6 pts) Using measurements taken from the whole rangeof obtainable voltages, calculate and plot the C(V ) curve forobtainable positive voltages on the black box to the answersheet. Write down the minimal and maximal values of differ-ential capacitance Cmin, Cmax. Document the circuit diagram.

Part B. Circuit with inductance (3 points)Enable the inductance by opening the switch on the black box(push “0” down). Using the same method as in pt. A-iii, meas-ure and plot the current–voltage characteristic of the nonlin-ear element. Describe any significant differences between thecurves of parts A and B and suggest a reason using qualitativearguments.

Here you need to know that the nonlinear element has actu-ally also a capacitance (ca 1 nF) which is connected in parallelto the nonlinear resistance.

— page 3 of 4 —



OUTIN

Multimeter

GND

A V


Current source

+

-

I=6mA

U=-612.5mV...612.5mV



Black box

Nonlinear

device

+

-C(V)







— page 3 of 4 —

PrObLEm E2. nOnLInEar bLack bOx (10 POInTs)

OUTIN

Multimeter

GND

A V

113



OUTIN

Multimeter

GND

A V


Current source

+

-

I=6mA

U=-612.5mV...612.5mV



Black box

Nonlinear

device

+

-C(V)







— page 3 of 4 —



OUTIN

Multimeter

GND

A V


Current source

+

-

I=6mA

U=-612.5mV...612.5mV



Black box

Nonlinear

device

+

-C(V)







— page 3 of 4 —



OUTIN

Multimeter

GND

A V


Current source

+

-

I=6mA

U=-612.5mV...612.5mV



Black box

Nonlinear

device

+

-C(V)







— page 3 of 4 —


114



OUTIN

Multimeter

GND

A V


Current source

+

-

I=6mA

U=-612.5mV...612.5mV



Black box

Nonlinear

device

+

-C(V)







— page 3 of 4 —

IPhO-measure: short manualIPhO-measure is a multimeter capable of measuring voltage Vand current I simultaneously. It also records their time derivat-ives V and I, their product P = V I, ratio R = V/I, and timet of the sample. Stored measurements are organized into sep-arate sets; every stored sample is numbered by the set numbers and a counter n inside the set. All saved samples are writtento an internal flash memory and can later be retrieved.

Electrical behaviourThe device behaves as an ammeter and a voltmeter connectedas follows. OUTIN

Multimeter

GND

A V

InternalRange resistance

Voltmeter 0 . . . 2 V 1 MΩVoltmeter 2 . . . 10 V 57 kΩAmmeter 0 . . . 1 A 1 Ω

Basic usage

• Push “Power” to switch the IPhO-measure on. Thedevice is not yet measuring; to start measuring, push“Start”. Alternatively, you can now start browsing yourstored data, see below.

• To browse previously saved samples (through all sets),press “Previous” or “Next”. Hold them down longerto jump directly between sets.

• While not measuring, push “Start” to start measuringa new set.

• While measuring, push “Sample” to store a datasample (with the current readings).

• While measuring, you can also browse other samples ofthe current set, using “Previous” and “Next”.

• Press “Stop” to end a set and stop measuring. Thedevice is still on, you are ready to start a new measuringsession, or browsing stored data.

• Pushing “Power” turns the device off. The device willshow the text “my mind is going . . . ”; don’t worry, all thedata measurements will be stored and you will be able tobrowse them after you switch the device on, again. Savedsamples will not be erased.

Display

A displayed sample consists of nine variables:

1. index n of the sample in the set;2. index s of the set;3. time t since starting the set;4. voltmeter output V ;5. rate of change of V (the time derivative V ); if derivative

cannot be reliably taken due to fluctuations, “+nan/s” isshown;

6. ammeter output I;7. rate of change of I (the time derivative I); if derivative


8. product P = V I;9. ratio R = V/I.

If any of the variables is out of its allowed range, its displayshows “+inf” or “-inf”.

— page 4 of 4 —



Multimeter

GND

A V



Basic usage








Display








— page 4 of 4 —

115



Multimeter

GND

A V



Basic usage








Display








— page 4 of 4 —



Multimeter

GND

A V



Basic usage








Display








— page 4 of 4 —

116

Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)

Observing reflections from the water surface (in particular,

those of straight lines, such as the edge of a sheet of paper),

it is easy to see that the profile has one minimum and has a

relatively flat bottom, ie. the correct answer is “Option D” (full

marks are given also for Option B). This profile implies that

water is pushed away from the magnet, which means µ < 1

(recall that ferromagnets with µ > 1 are pulled).


i. (1.6 pts) The height of the spot on the screen y is tabulated

below as a function of the horizontal position x of the caliper.

Note that the values of y in millimetres can be rounded to in-

tegers (this series of measurements aimed as high as possible

precision).

x (mm) 10 15 20 25 30 32 34 36

y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5

x (mm) 38 40 42 44 46 48 50 52

y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4

x (mm) 54 56 58 60 62 64 66 68

y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6

x (mm) 70 72 74 76 78 80 85 90

y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9

ii. (0.7 pts)

On this graph, the data of to two different water levels are

depicted; blue curve corresponds to a water depth of ca 2 mm

(data given in the table above); the violet one — to 1 mm.

iii. (0.5 pts) If the water surface were flat, the dependence of x

on y would be linear, and the tangent of the angle α0 would be

given by tan α0 = ∆y

∆x, where ∆x is a horizontal displacement of

the pointer, and ∆y — the respective displacement of the spot

height. For the extreme positions of the pointer, the beam hits

the water surface so far from the magnet that there, the surface

is essentially unperturbed; connecting the respective points on

the graph, we obtain a line corresponding to a flat water sur-

face — the red line. Using these two extreme data points we

can also easily calculate the angle α0 = arctan 74.9−11.590−10

≈ 38.

iv. (1.4 pts) For faster calculations, y − y0 − (x − x0) tan α0

(appearing in the formula given) can be read from the previous

graph as the distance between red and blue line; the red line

is given by equation yr = y0 + (x − x0) tan α0. One can also

precalculate 1

2cos2 α0 ≈ 0.31. The calculations lead to the fol-

lowing table (with z = tan β · 105 ; as mentioned above, during

the competition, lesser precision with two significant numbers

is sufficient).

x (mm) 10 15 20 25 30 32 34 36

z 0 10 27 66 204 303 473 591

x (mm) 38 40 42 44 46 48 50 52

z 597 428 239 128 53 26 0 -26

x (mm) 54 56 58 60 62 64 66 68

z -72 -145 -278 -449 -606 -536 -388 -254

x (mm) 70 72 74 76 78 80 85 90

z -154 -74 -40 -20 -6 2 -2 0

— page 1 of 4 —














precision).

x (mm) 10 15 20 25 30 32 34 36

y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5

x (mm) 38 40 42 44 46 48 50 52

y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4

x (mm) 54 56 58 60 62 64 66 68

y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6

x (mm) 70 72 74 76 78 80 85 90

y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9

ii. (0.7 pts)















≈ 38.





precalculate 1




is sufficient).

x (mm) 10 15 20 25 30 32 34 36

z 0 10 27 66 204 303 473 591

x (mm) 38 40 42 44 46 48 50 52

z 597 428 239 128 53 26 0 -26

x (mm) 54 56 58 60 62 64 66 68

z -72 -145 -278 -449 -606 -536 -388 -254

x (mm) 70 72 74 76 78 80 85 90

z -154 -74 -40 -20 -6 2 -2 0

— page 1 of 4 —


Solutions

117














precision).

x (mm) 10 15 20 25 30 32 34 36

y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5

x (mm) 38 40 42 44 46 48 50 52

y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4

x (mm) 54 56 58 60 62 64 66 68

y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6

x (mm) 70 72 74 76 78 80 85 90

y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9

ii. (0.7 pts)















≈ 38.





precalculate 1




is sufficient).

x (mm) 10 15 20 25 30 32 34 36

z 0 10 27 66 204 303 473 591

x (mm) 38 40 42 44 46 48 50 52

z 597 428 239 128 53 26 0 -26

x (mm) 54 56 58 60 62 64 66 68

z -72 -145 -278 -449 -606 -536 -388 -254

x (mm) 70 72 74 76 78 80 85 90

z -154 -74 -40 -20 -6 2 -2 0

— page 1 of 4 —Problem E1. The magnetic permeability of water(10 points)Part A. Qualitative shape of the water surface (1 points)













precision).

x (mm) 10 15 20 25 30 32 34 36

y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5

x (mm) 38 40 42 44 46 48 50 52

y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4

x (mm) 54 56 58 60 62 64 66 68

y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6

x (mm) 70 72 74 76 78 80 85 90

y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9

ii. (0.7 pts)















≈ 38.





precalculate 1




is sufficient).

x (mm) 10 15 20 25 30 32 34 36

z 0 10 27 66 204 303 473 591

x (mm) 38 40 42 44 46 48 50 52

z 597 428 239 128 53 26 0 -26

x (mm) 54 56 58 60 62 64 66 68

z -72 -145 -278 -449 -606 -536 -388 -254

x (mm) 70 72 74 76 78 80 85 90

z -154 -74 -40 -20 -6 2 -2 0

— page 1 of 4 —














precision).

x (mm) 10 15 20 25 30 32 34 36

y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5

x (mm) 38 40 42 44 46 48 50 52

y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4

x (mm) 54 56 58 60 62 64 66 68

y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6

x (mm) 70 72 74 76 78 80 85 90

y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9

ii. (0.7 pts)















≈ 38.





precalculate 1




is sufficient).

x (mm) 10 15 20 25 30 32 34 36

z 0 10 27 66 204 303 473 591

x (mm) 38 40 42 44 46 48 50 52

z 597 428 239 128 53 26 0 -26

x (mm) 54 56 58 60 62 64 66 68

z -72 -145 -278 -449 -606 -536 -388 -254

x (mm) 70 72 74 76 78 80 85 90

z -154 -74 -40 -20 -6 2 -2 0

— page 1 of 4 —














precision).

x (mm) 10 15 20 25 30 32 34 36

y (mm) 11.5 15.6 19.8 24.3 30.2 33.2 37.2 40.5

x (mm) 38 40 42 44 46 48 50 52

y (mm) 42.2 41.4 40.3 40.3 40.8 42 43.2 44.4

x (mm) 54 56 58 60 62 64 66 68

y (mm) 45.3 45.8 45.4 44.4 43.6 46.2 50 53.6

x (mm) 70 72 74 76 78 80 85 90

y (mm) 56.7 59.5 61.6 63.5 65.3 67 70.9 74.9

ii. (0.7 pts)















≈ 38.





precalculate 1




is sufficient).

x (mm) 10 15 20 25 30 32 34 36

z 0 10 27 66 204 303 473 591

x (mm) 38 40 42 44 46 48 50 52

z 597 428 239 128 53 26 0 -26

x (mm) 54 56 58 60 62 64 66 68

z -72 -145 -278 -449 -606 -536 -388 -254

x (mm) 70 72 74 76 78 80 85 90

z -154 -74 -40 -20 -6 2 -2 0

— page 1 of 4 —


v. (1.6 pts) The water height can be obtained as the integral

h =∫

tan βdx. Thus, we calculate the water height row-by-

row, by adding to the height in the previous row the product

of the horizontal displacement xi+1 −xi with the average slope1

2(tan βi+1 + tan βi).

x (mm) 10 15 20 25 30 32 34 36

−h (µm) 0 0 1 4 10 15 23 34

x (mm) 38 40 42 44 46 48 50 52

−h (µm) 46 56 63 66 68 69 69 69

x (mm) 54 56 58 60 62 64 66 68

−h (µm) 68 66 61 54 44 32 23 17

x (mm) 70 72 74 76 78 80 85 90

−h (µm) 12 10 9 8 8 8 8 8

Note that the water level height at the end of the table should

be also 0 (this corresponds also to an unperturbed region); the

non-zero result is explained by the measurement uncertainties.

One can improve the result by subtracting from h a linear trend

8 µm ·x−10 mm

80 mm.

If the water level above the magnet is 1 mm, the water level

descends below its unperturbed level at the axis of the magnet

by ca 120 µm.

vi. (1 pt)

Similarly to the previous figure, blue curve corresponds to

a water depth of ca 2 mm, (data given in the table above), and

the violet one — to 1 mm.

The position of the magnet can be found by measuring the

caliper (find the positions when the laser beam hits the edges of

the magnet and determine the distance between these positions

— the result is ca 24 mm), and using the symmetry: magnet

is placed symmetrically with respect to the surface elevation

curve.


Water surface takes an equipotential shape; for a unit volume of

water, the potential energy associated with the magnetic inter-

action is B2

2µ0

(µ−1−1) ≈ B2 1−µ

2µ0

; the potential energy associated

with the Earth’s gravity is ρgh. At the water surface, the sum

of those two needs to be constant; for a point at unperturbed

surface, this expression equals to zero, so B2 µ−1

2µ0

+ ρgh = 0

and hence, µ − 1 = 2µ0ρgh/B2. Here, h = 120 µm stands for

the depth of the water surface at the axis of the magnet; note

that we have compensated the cumulative error as described at

the end of the previous task and obtained h as the difference

between the depth at the magnet’s axis (121 µm) and the half-

depth at the right-hand-side of the graph (1 µm). Putting in

the numbers, we obtain µ − 1 = −1.2 × 10−5 .

— page 2 of 4 —

118


h =∫





x (mm) 10 15 20 25 30 32 34 36

−h (µm) 0 0 1 4 10 15 23 34

x (mm) 38 40 42 44 46 48 50 52

−h (µm) 46 56 63 66 68 69 69 69

x (mm) 54 56 58 60 62 64 66 68

−h (µm) 68 66 61 54 44 32 23 17

x (mm) 70 72 74 76 78 80 85 90

−h (µm) 12 10 9 8 8 8 8 8





8 µm ·x−10 mm

80 mm.



by ca 120 µm.

vi. (1 pt)









curve.




action is B2

2µ0

(µ−1−1) ≈ B2 1−µ

2µ0





2µ0

+ ρgh = 0








— page 2 of 4 —


h =∫





x (mm) 10 15 20 25 30 32 34 36

−h (µm) 0 0 1 4 10 15 23 34

x (mm) 38 40 42 44 46 48 50 52

−h (µm) 46 56 63 66 68 69 69 69

x (mm) 54 56 58 60 62 64 66 68

−h (µm) 68 66 61 54 44 32 23 17

x (mm) 70 72 74 76 78 80 85 90

−h (µm) 12 10 9 8 8 8 8 8





8 µm ·x−10 mm

80 mm.



by ca 120 µm.

vi. (1 pt)









curve.




action is B2

2µ0

(µ−1−1) ≈ B2 1−µ

2µ0





2µ0

+ ρgh = 0








— page 2 of 4 —

119


h =∫





x (mm) 10 15 20 25 30 32 34 36

−h (µm) 0 0 1 4 10 15 23 34

x (mm) 38 40 42 44 46 48 50 52

−h (µm) 46 56 63 66 68 69 69 69

x (mm) 54 56 58 60 62 64 66 68

−h (µm) 68 66 61 54 44 32 23 17

x (mm) 70 72 74 76 78 80 85 90

−h (µm) 12 10 9 8 8 8 8 8





8 µm ·x−10 mm

80 mm.



by ca 120 µm.

vi. (1 pt)









curve.




action is B2

2µ0

(µ−1−1) ≈ B2 1−µ

2µ0





2µ0

+ ρgh = 0








— page 2 of 4 —

PrObLEm E2. nOnLInEar bLack bOx (10 POInTs)Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)

It is possible to make all the measurements needed for this

problem with a single circuit as shown in the figure. While the

current source is switched on, we are charging the capacitor in

the black box, until the current I(Vmax) through the nonlinear

element equals to the output current I0 of the current source.

Vmax = 540±40mVs varies from one experimental setup to an-

other. When the current source is switched off or disconnected,

the capacitor will discharge through the nonlinear element.

Multimeter

Current source

IN OUT GND

+−Switch

O

I

Black boxSwitch

O

I

i. (1 pt) During charging of the capacitor from V = 0 to

V = Vmax we note that the output of the current source is con-

stant (I0 = 6.0 mA) close to the precision of the multimeter.

ii. (1.2 pts) Using the definition of differential capacitance,

we can calculate the current through the capacitor in the black

box from the time derivative of the voltage on the black box.

Ic =dQ

dt=

dQ

dV

dV

dt= C(V )V

There are several ways to determine the capacitance used in

the black box based on chosen voltage.

• When the voltage on the black box is close to zero, the

current through the nonlinear element is also close to

zero, because I(V = 0) = 0. After switching the current

source on, most of the input current I0 will at first go

through the capacitor.

C0 = I0/V↑(V = 0)

This can be measured more precisely after first reversing

the polarity of the current source and charging the capa-

citor backwards, because the multimeter does not display

derivatives when they change sharply (as in few moments

after switching the current source on).

Example measurements taken this way follow.

V↑(0) (mV/s) 3.51 3.32 3.55

C0 (F) 1.71 1.81 1.69

C0 = 1.74 F

• When the voltage on the black box is Vmax, the current

through the nonlinear element is I0. Switching the cur-

rent source off, we will have the capacitor discharging

with the same current.

C0 = −I0/V↓(V = Vmax)

• We can also measure the capacitance for any intermediate

voltage as in A-iv.

iii. (2.2 pts) If we neglect the nonlinearity of the capacitor,

there are (at least) two ways to obtain the current–voltage char-

acteristic of the nonlinear element in the black box.

• Applying Kirchhoff’s I law to the charging capacitor,

I(V ) = Ic − C0V↑(V ).

An I(V ) characteristic obtained by charging the capacitor

is shown on the following figure.

• Applying Kirchhoff I law to the discharging capacitor,

I(V ) = −C0V↓(V ).

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

V (V)

I (m

A)

Part APart B

iv. (2.6 pts) In order to obtain the differential capacitance,

we solve a system of linear equations by eliminating I(V ):

I0 = V↑C(V ) + I(V )

I(V ) = −V↓C(V );=⇒ C(V ) =

I0

V↑ − V↓

.

Therefore we need to take measurements during both charging

and discharging the capacitor in the black box at the same

voltages. A graph of measurement results follows.

— page 3 of 4 —

Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)









Multimeter

Current source

IN OUT GND

+−Switch

O

I

Black boxSwitch

O

I







Ic =dQ

dt=

dQ

dV

dV

dt= C(V )V








C0 = I0/V↑(V = 0)







V↑(0) (mV/s) 3.51 3.32 3.55

C0 (F) 1.71 1.81 1.69

C0 = 1.74 F





C0 = −I0/V↓(V = Vmax)


voltage as in A-iv.





I(V ) = Ic − C0V↑(V ).




I(V ) = −C0V↓(V ).

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

V (V)

I (m

A)

Part APart B



I0 = V↑C(V ) + I(V )

I(V ) = −V↓C(V );=⇒ C(V ) =

I0

V↑ − V↓

.




— page 3 of 4 —

120










Multimeter

Current source

IN OUT GND

+−Switch

O

I

Black boxSwitch

O

I







Ic =dQ

dt=

dQ

dV

dV

dt= C(V )V








C0 = I0/V↑(V = 0)







V↑(0) (mV/s) 3.51 3.32 3.55

C0 (F) 1.71 1.81 1.69

C0 = 1.74 F





C0 = −I0/V↓(V = Vmax)


voltage as in A-iv.





I(V ) = Ic − C0V↑(V ).




I(V ) = −C0V↓(V ).

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

V (V)

I (m

A)

Part APart B



I0 = V↑C(V ) + I(V )

I(V ) = −V↓C(V );=⇒ C(V ) =

I0

V↑ − V↓

.




— page 3 of 4 —Problem E2. Nonlinear Black Box (10 points)Part A. Circuit without inductance (7 points)









Multimeter

Current source

IN OUT GND

+−Switch

O

I

Black boxSwitch

O

I







Ic =dQ

dt=

dQ

dV

dV

dt= C(V )V








C0 = I0/V↑(V = 0)







V↑(0) (mV/s) 3.51 3.32 3.55

C0 (F) 1.71 1.81 1.69

C0 = 1.74 F





C0 = −I0/V↓(V = Vmax)


voltage as in A-iv.





I(V ) = Ic − C0V↑(V ).




I(V ) = −C0V↓(V ).

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

V (V)

I (m

A)

Part APart B



I0 = V↑C(V ) + I(V )

I(V ) = −V↓C(V );=⇒ C(V ) =

I0

V↑ − V↓

.




— page 3 of 4 —

121










Multimeter

Current source

IN OUT GND

+−Switch

O

I

Black boxSwitch

O

I







Ic =dQ

dt=

dQ

dV

dV

dt= C(V )V








C0 = I0/V↑(V = 0)







V↑(0) (mV/s) 3.51 3.32 3.55

C0 (F) 1.71 1.81 1.69

C0 = 1.74 F





C0 = −I0/V↓(V = Vmax)


voltage as in A-iv.





I(V ) = Ic − C0V↑(V ).




I(V ) = −C0V↓(V ).

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

V (V)

I (m

A)

Part APart B



I0 = V↑C(V ) + I(V )

I(V ) = −V↓C(V );=⇒ C(V ) =

I0

V↑ − V↓

.




— page 3 of 4 —

122

0 0.1 0.2 0.3 0.4 0.51.7

1.75

1.8

1.85

1.9

1.95

2

2.05

2.1

V (V)

C (F

)

Part B. Circuit with inductance (3 points)

Measuring and plotting the current–voltage characteristic of

the nonlinear element in the same way as in part A-iii, we

obtain a graph that differs only in the negative differential res-

istance (I ′(V ) < 0) region, in our case 70 mV < V < 330 mV.

This is the region where, when we look at small-signal oscil-

lations, the nonlinear element behaves as a negative-valued

Ohmic resistance. After enabling the inductance we have a

LC circuit whose oscillations are amplified (instead of being

dampened) by the negative differential resistance. Because the

resonant frequency ω =√

1

LCp

∼ 30 MHz (with Cp being the

capacitance of the nonlinear element) is high, we actually meas-

ure the average current through the nonlinear element, while

the real current oscillates all over the region of negative differ-

ential resistance.

— page 4 of 4 —

0 0.1 0.2 0.3 0.4 0.51.7

1.75

1.8

1.85

1.9

1.95

2

2.05

2.1

V (V)

C (F

)


Measuring and plotting the current–voltage characteristic of

the nonlinear element in the same way as in part A-iii, we

obtain a graph that differs only in the negative differential res-

istance (I ′(V ) < 0) region, in our case 70 mV < V < 330 mV.

This is the region where, when we look at small-signal oscil-

lations, the nonlinear element behaves as a negative-valued

Ohmic resistance. After enabling the inductance we have a

LC circuit whose oscillations are amplified (instead of being

dampened) by the negative differential resistance. Because the

resonant frequency ω =√

1

LCp

∼ 30 MHz (with Cp being the

capacitance of the nonlinear element) is high, we actually meas-

ure the average current through the nonlinear element, while

the real current oscillates all over the region of negative differ-

ential resistance.

— page 4 of 4 —










Multimeter

Current source

IN OUT GND

+−Switch

O

I

Black boxSwitch

O

I







Ic =dQ

dt=

dQ

dV

dV

dt= C(V )V








C0 = I0/V↑(V = 0)







V↑(0) (mV/s) 3.51 3.32 3.55

C0 (F) 1.71 1.81 1.69

C0 = 1.74 F





C0 = −I0/V↓(V = Vmax)


voltage as in A-iv.





I(V ) = Ic − C0V↑(V ).




I(V ) = −C0V↓(V ).

0 0.1 0.2 0.3 0.4 0.50

1

2

3

4

5

6

V (V)

I (m

A)

Part APart B



I0 = V↑C(V ) + I(V )

I(V ) = −V↓C(V );=⇒ C(V ) =

I0

V↑ − V↓

.




— page 3 of 4 —

123


Grading scheme: Experiment


points allotted for the design of the experiments, measurements,

plotting, and formulae used for calculations. In the case of a for-

mula, points are allotted for each term entering it. If a certain

term of a useful formula is written incorrectly, 0.1 is subtrac-

ted for a minor mistake (eg. missing non-dimensional factor);

no mark is given if the mistake is major (with non-matching

dimensionality). Points for the data measurements and calcu-

lations are not given automatically: the data which are clearly

wrong are not credited.

If the numerical data miss units, but the units can be

guessed, 25% of points will be subtracted for the corresponding

line in this grading scheme (rounded to one decimal place). The

same rule applies if there is a typo with units with a missing

or redundant prefactor (millli-, micro-, etc); no mark is given

if the mistake is dimensional (e.g Ampere instead of Volt).







— page 1 of 4 —




points allotted for the design of the experiments, measurements,

plotting, and formulae used for calculations. In the case of a for-

mula, points are allotted for each term entering it. If a certain

term of a useful formula is written incorrectly, 0.1 is subtrac-

ted for a minor mistake (eg. missing non-dimensional factor);

no mark is given if the mistake is major (with non-matching

dimensionality). Points for the data measurements and calcu-

lations are not given automatically: the data which are clearly

wrong are not credited.

If the numerical data miss units, but the units can be

guessed, 25% of points will be subtracted for the corresponding

line in this grading scheme (rounded to one decimal place). The

same rule applies if there is a typo with units with a missing

or redundant prefactor (millli-, micro-, etc); no mark is given

if the mistake is dimensional (e.g Ampere instead of Volt).







— page 1 of 4 —


124


Correct choice (B or D) — 0.5 pts;

Correct sign (µ < 1) — 0.5 pts;


i. (1.6 pts) Data

Sufficient number of reasonably accurate ±2 mm

data points: (n − 9)/10 pts but ≥ 0 and ≤ 0.9 — ≤ 0.9 pts;

(-0.1 if there is a sign error throughout the whole measure-

ments)

Sufficient range of horizontal displacements: (x − 45 mm)/50

rounded to one decimal but ≥ 0 and ≤ 0.7 — ≤ 0.7 pts.

ii. (0.7 pts) Graph

Axes supplied with units — 0.1 pts;

Data points correctly plotted — 0.4 pts;

(each clearly wrong point on graph: −0.1 pt down to total 0)

More densely spaced data points

in the regions of fast change — 0.2 pts.

iii. (0.7 pts) Angle (≈ 38 )

Idea: using the flat regions

far from the magnet — 0.5 pts

if the central part of the magnet is used: 0.1

Correct value: ±5 / ±10 / > 10 — 0.2/0.1/0 pts;

iv. (1.4 pts) Calculated table

According to the number of correctly calculated data points:

0.1n up to n = 10;

1 + 0.05(n − 10) rounded down one decimal place for n > 10

(but ≤ 1.4);

v. (1.6 pts) Height profile

Idea of calculation: integration of surface slope — 0.7 pts

(if not stated but used correctly: full marks)

∆x multiplied by mean slope at that interval — 0.3 pts

For n correctly calculated data points: — n/30 pts

(but no more than 0.6; rounded down to one decimal place).

vi. (1 pt) Graph

Units on axes — 0.1 pts

Data points correctly plotted — 0.4 pts


Correct qualitative shape: height difference

in wings less than the 20% of maximum variation) — 0.1 pts

flat central region: the water level height variation in the

region spanning ±10 mm around the magnet’s axis

is less than 20 µm) — 0.2 pts

Magnet indicated correctly: (width 24 mm ± 2 mm) — 0.1 pts

symmetrically positioned — 0.1 pts


Concept of equipotentiality — 0.8 pts

Formula correctly includes magnetic energy — 0.4 pts

Formula correctly includes gravitational energy — 0.2 pts

height at the middle corrected for the integration error — 0.2 pts

Value calculated correctly from the existing data — 0.1 pts

Value: magnitude correct within 50% — 0.2 pts

(no marks if not obtained from the experimental data)

correct sign — 0.1 pts

— page 2 of 4 —





i. (1.6 pts) Data




ments)



ii. (0.7 pts) Graph






iii. (0.7 pts) Angle (≈ 38 )







0.1n up to n = 10;


(but ≤ 1.4);







vi. (1 pt) Graph




















— page 2 of 4 —





i. (1.6 pts) Data




ments)



ii. (0.7 pts) Graph






iii. (0.7 pts) Angle (≈ 38 )







0.1n up to n = 10;


(but ≤ 1.4);







vi. (1 pt) Graph




















— page 2 of 4 —


125





i. (1.6 pts) Data




ments)



ii. (0.7 pts) Graph






iii. (0.7 pts) Angle (≈ 38 )







0.1n up to n = 10;


(but ≤ 1.4);







vi. (1 pt) Graph




















— page 2 of 4 —


126



Typical I(V ) and C(V ) curves can be found in solutions

file and sample filled-in answer sheets. Because I(V ) and C(V )

curves shape, typical capacitance, I0 and Vmax varies a bit from

one setup to another, reference curves for particular setup can

be acquired for grading when deemed necessary.

For each question, if the position of the switches on the

circuit diagram is not indicated, take -0.1 pts from the marks

for the circuit.

i. (1 pt)

Correct circuit — 0.3 pts

Measurements that cover 0 V to 480 mV — 0.3 pts

Correct value Imin (±0.4 mA) — 0.2 pts

Correct value Imax (±0.4 mA) — 0.2 pts

(unless only one data point)

In case of (single) measurement without the black box:

Circuit diagram — 0.1 pts

Measurement with the correct result — 0.1 pts

If only Imin, Imax together with a correct scheme docu-

mented, 0.7 pts in total; if only Imin, Imax, 0 pts in total; if a

scheme without the black box and a single I measurement,

0.2 pts in total.

ii. (1.2 pts) Measuring C0 at V = 0


(-0.1 for measuring voltage on ammeter + black box)

(-0.1 for wrong polarity of the black box)

Realising that Ic = C0V — 0.2 pts

Using the fact that Ic = I0 when V = 0 — 0.2 pts

Correct result — 0.2 pts

-0.1 for error between ±30% and ±50%

linear extrapolation to obtain V at V = 0 — 0.4 pts

Alternatively instead of the last line

Three or more measurements — 0.2 pts

Precharging the capacitor to a negative voltage — 0.2 pts

Alternate solution C0 at Vmax




Using the fact that Ic = −I0 when V = Vmax — 0.2 pts


linear extrapolation to obtain V at V = Vmax — 0.2 pts


iii. (2.2 pts) For method:

If charging the capacitor:




Realising that I(V ) = I0 − C0V↑ — 0.2 pts

If discharging the capacitor:

Correct circuit (there are several) — 0.2 pts


Realising that I(V ) = −C0V↓— 0.2 pts

For measurements:

Total # of data correct data points

10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts

Additionaly for correct data points:

In range 0mV - 80mV at least 4/5 data pts — 0.1/0.2 pts


In range 200mV - 400mV at least 3 data pts — 0.1 pts


For plotting:




Correct qualitative shape — 0.3 pts

(single maximum, single minimum with flat bottom, followed

by fast rise)

— page 3 of 4 —

127










for the circuit.

i. (1 pt)












0.2 pts in total.































For measurements:


10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts






For plotting:






by fast rise)

— page 3 of 4 —Problem E2. Nonlinear Black Box (10 points)

Part A. Circuit without inductance (7 points)








for the circuit.

i. (1 pt)












0.2 pts in total.































For measurements:


10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts






For plotting:






by fast rise)

— page 3 of 4 —









for the circuit.

i. (1 pt)












0.2 pts in total.































For measurements:


10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts






For plotting:






by fast rise)

— page 3 of 4 —

128

iv. (2.6 pts)




Idea to use the reverse cylce to pt iii— 0.4 pts

Writing the lin. eq for finding C(V )— 0.4 pts

(No pts if only one Eq.)

Expressing from there C(V )— 0.1 pts

Idea to use the same voltages for both cycles— 0.4 pts

(0.3 pts if intermediate values read from graph)

For plotting:




For correct data points:





For finding Cmax and Cmin:

Finding reasonable Cmax — 0.1 pts

Finding reasonable Cmin — 0.1 pts







For plotting:





(two sharp falls, plateau in between)

For detecting differences:

Correct range for V :— 0.2 pts

Correct cond. for I(V ) — 0.5 pts

Noting that we have now LC-circuit— 0.2 pts

Noting that neg. resist. → instability— 0.4 pts

(or something equivalent)

Mentioning emergence of oscillations— 0.2 pts

Noting that I(V ) is the average current— 0.3 pts

— page 4 of 4 —









for the circuit.

i. (1 pt)












0.2 pts in total.































For measurements:


10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 — 0.1/0.2/0.3 pts






For plotting:






by fast rise)

— page 3 of 4 —

129

iv. (2.6 pts)




Idea to use the reverse cylce to pt iii— 0.4 pts

Writing the lin. eq for finding C(V )— 0.4 pts

(No pts if only one Eq.)

Expressing from there C(V )— 0.1 pts

Idea to use the same voltages for both cycles— 0.4 pts

(0.3 pts if intermediate values read from graph)

For plotting:









For finding Cmax and Cmin:

Finding reasonable Cmax — 0.1 pts

Finding reasonable Cmin — 0.1 pts







For plotting:





(two sharp falls, plateau in between)

For detecting differences:

Correct range for V :— 0.2 pts

Correct cond. for I(V ) — 0.5 pts

Noting that we have now LC-circuit— 0.2 pts

Noting that neg. resist. → instability— 0.4 pts

(or something equivalent)

Mentioning emergence of oscillations— 0.2 pts

Noting that I(V ) is the average current— 0.3 pts

— page 4 of 4 —

130

131

Results

Gold MedalistsAttila Szabó, Hungary

Hengyun Zhou, People’s Republic of China

Yijun Jiang, People’s Republic of China

Chien-An Wang, Taiwan

Kai-Chi Huang, Taiwan

Paphop Sawasdee, Thailand

Eric Schneider, United States of America

Jun-Ting Hsieh, Taiwan

Siyuan Wei, People’s Republic of China

Chi Shu, People’s Republic of China

Wonseok Lee, Republic of Korea

Wenzhuo Huang, People’s Republic of China

Yu-Ting Liu, Taiwan

Wei-Jen Ko, Taiwan

Rahul Trivedi, India

Lev Ginzburg, Russia

Phot

o by

Hen

ry T

eiga

r

132

Tudor Giurgică-Tiron, Romania

Allan Sadun, United States of America

Kazumi Kasaura, Japan

Nikita Sopenko, Russia

Ding Yue, Republic of Singapore

Lam Ho Tat, Hong Kong

Sooshin Kim, Republic of Korea

Jaemo Lim, Republic of Korea

Lorenz Eberhardt, Germany

Ivan Ivashkovskiy, Russia

Michal Pacholski, Poland

Qiao Gu, Germany

Adrian Nugraha Utama, Indonesia

Albert Samoilenka, Belarus

Puthipong Worasaran, Thailand

Jaan Toots, Estonia

Nathanan Tantivasadakarn, Thailand

Phi Long Ngo, Vietnam

Yuichi Enoki, Japan

Ihar Lobach, Belarus

Veselin Karadzhov, Bulgaria

Ivan Tadeu Ferreira Antunes Filho, Brazil

Ilya Vilkoviskiy, Kazakhstan

Kevin Zhou, United States of America

Ang Yu Jian, Republic of Singapore

Fung Tsz Chai, Hong Kong

Kuan Jun Jie, Joseph, Republic of Singapore

Ngoc Hai Dinh, Vietnam

Huan Yan Qi, Republic of Singapore

gold medalists Photo by Henry Teigar

133

Silver MedalistsAlexandra Vasilyeva, Russia

Bijoy Singh Kochar, India

Jeevana Priya Inala, India

Zoltán Laczkó, Hungary

Dan-Cristian Andronic, Romania

Ramtin Yazdanian, Islamic Republic of Iran

Vladysslav Diachenko, Ukraine

Sebastian Linß, Germany

Kunal Singhal, India

Kohei Kawabata, Japan

Volodymyr Sivak, Ukraine

David Frenklakh, Russia

Vsevolod Bykov, Ukraine

Theodor Misiakiewicz, France

Woojin Kweon, Republic of Korea

Kaisarbek Omirzakhov, Kazakhstan

Huy Quang Le, Vietnam

Ondřej Bartoš, Czech Republic

Jakub Vošmera, Czech Republic

Stanislav Fořt, Czech Republic

Pongsapuk Sawaddirak, Thailand

Suyeon Choi, Republic of Korea

Yordan Yordanov, Bulgaria

Hiromasa Nakatsuka, Japan

Aliaksey Khatskevich, Belarus

Peter Kosec, Slovakia

Chan Cheuk Lun, Hong Kong

Itay Knaan Harpaz, Israel

Tasuku Omori, Japan

Vadzim Reut, Belarus

Filip Ficek, Poland

Daumantas Kavolis, Lithuania

Tanel Kiis, Estonia

Katerina Naydenova, Bulgaria

Jeffrey Yan, United States of America

Kristjan Kongas, Estonia

Phot

o by

Hen

ry T

eiga

r

134

Georg Krause, Germany

Yevgen Cherniavskyi, Ukraine

Jonathan Dong, France

Kaur Aare Saar, Estonia

Abdurrahman Akkas, Turkey

Jeffrey Cai, United States of America

Georgijs Trenins, Latvia

Henry Honglei Wu, Canada

Patrik Svancara, Slovakia

Oliver Edtmair, Austria

Amir Yousefi, Islamic Republic of Iran

Lubomír Grund, Czech Republic

Patrik Turzak, Slovakia

Milan Krstajić, Serbia

Atinc Cagan Sengul, Turkey

Ittai Rubinstein, Israel

Mehmet Said Onay, Turkey

Jean Douçot, France

Matias Mannerkoski, Finland

Soo Wah Ming, Wayne, Republic of Singapore

Mussa Rajamov, Kazakhstan

Lai Kwun Hang, Hong Kong

Tudor Ciobanu, Romania

Roland Papp, Hungary

Eric Wieser, United Kingdom

Nurzhas Aidynov, Kazakhstan

Adam Brown, United Kingdom

Žygimantas Stražnickas, Lithuania

Paul Kirchner, France

Atli Thor Sveinbjarnarson, Iceland

Sebastian Florin Dumitru, Romania

Péter Juhász, Hungary

Christoph Schildknecht, Switzerland

Wai Hong Lei, Macao, China

Andres Erbsen, Estonia

silver medalists Photo by Karl Veskus

135

Bronze MedalistsMilan Kornjača, Serbia

Petar Tadic, Montenegro

Cristian Zanoci, Moldova

Nicholas Salmon, Australia

Oguzhan Can, Turkey

Mustafa Selman Akinci, Turkey

Volodymyr Rozsokhovatskyi, Ukraine

Richard Thorburn, United Kingdom

Andrej Vlcek, Slovakia

Xuan Hien Bui, Vietnam

Ivan Senyushkin, Kazakhstan

Chen Solomon, Israel

Martijn van Kuppeveld, Netherlands

Jonathan Lay, Australia

Mohamad Ansarifard, Islamic Republic of Iran

Supanut Thanasilp, Thailand

Jan Rydzewski, Poland

Johan Runeson, Sweden

Iiro Lehto, Finland

Mehrdad Malak Mohammadi,

Islamic Republic of Iran

Hólmfríður Hannesdóttir, Iceland

Simon Pirmet, France

Simon Blouin, Canada

Roberto Albesiano, Italy

Viet Thang Dinh, Vietnam

Pulkit Tandon, India

Sajad Khodadadian, Islamic Republic of Iran

Christoph Weis, Austria

Kacper Oreszczuk, Poland

Yigal Zegelman, Israel

Frank Bloomfield, United Kingdom

Tamara Šumarac, Serbia

Vu Phan Thanh, Germany

Federica Maria Surace, Italy

Christopher Whittle, Australia

Francisco Machado, Portugal

Konstantin Gundev, Bulgaria

Martin Stadler, Austria

Martin Raszyk, Czech Republic

Thanh Phong Lê, Switzerland

Aliaksandr Yankouski, Belarus

Sepehr Ebadi, Canada

Domen Ipavec, Slovenia

Phot

o by

Hen

ry T

eiga

r

136

Ruben Doornenbal, Netherlands

Yun Jia (Melody) Guan, Canada

Ion Toloaca, Moldova

Wai Pan Si, Macao, China

Eden Segal, Israel

Jose Luciano De Morais Neto, Brazil

Tobias Karg, Austria

Sergi Chalalauri, Georgia

Meylis Malikov, Turkmenistan

Virab Gevorgyan, Armenia

Tristan Downing, Canada

Luqman Fathurrohim, Indonesia

Aitor Azemar, Spain

Dinis Cheian, Moldova

Lo Hei Chun, Hong Kong

Eric Huang, Australia

Jorge Torres-Ramos, Mexico

Roberta Răileanu, Romania

Jovan Blanuša, Serbia

Michele Fava, Italy

Selver Pepic, Bosnia and Herzegovina

Vardan Avetisyan, Armenia

Maximilian Ruep, Austria

Saba Kharabadze, Georgia

Sebastian Käser, Switzerland

Giorgi Kobakhidze, Georgia

Troy Figiel, Netherlands

Áron Dániel Kovács, Hungary

Ilija Burić, Serbia

Nicoleta Colibaba, Moldova

Abdullah Alsalloum, Saudi Arabia

Haji Piriyev, Azerbaijan

Javier Mendez-Ovalle, Mexico

Ka Fai Chan, Macao, China

Guilherme Renato Martins Unzer, Brazil

Muhammad Taimoor Iftikhar, Pakistan

Emmanouil Vourliotis, Greece

Jemal Shengelia, Georgia

Lara Timbo Araujo, Brazil

Ilie Popanu, Moldova

Francesc-Xavier Gispert Sánchez, Spain

Battsooj Bayarsaikhan, Mongolia

Samuel Bosch, Croatia

Koen Dwarshuis, Netherlands

Shinjini Saha, Bangladesh

Leandro Salemi, Belgium

Chan Lon Wu, Macao, China

Adhamzhon Shukurov, Tajikistan

Peter Budden, United Kingdom

bronze medalists Photo by Henry Teigar

137

Honorable MentionsJyri Maanpää, Finland

Sandro Maludze, Georgia

Siobhan Tobin, Australia

Simão João, Portugal

Tomas Čerškus, Lithuania

Osama Yaghi, Syrian Arab Republic

Koay Hui Wen, Malaysia

Kasper Tolborg, Denmark

Ramadhiansyah Ramadhiansyah, Indonesia

Ooi Chun Yeang, Malaysia

Dominic Schwarz, Switzerland

Bruno Buljan, Croatia

Bartlomiej Zawalski, Poland

Shovon Biswas, Bangladesh

Homoud Alharbi, Saudi Arabia

Bilguun Batjargal, Mongolia

Mantas Abazorius, Lithuania

Nudzeim Selimovic, Bosnia and Herzegovina

Ahmed Maksud, Bangladesh

Arttu Yli-Sorvari, Finland

Farid Mammadov, Azerbaijan

Karlo Sepetanc, Croatia

Romain Falla, Belgium

Fotios Vogias, Greece

Miha Zgubič, Slovenia

Tsogt Baigalmaa, Mongolia

Mohamed Alrazzouk, Syrian Arab Republic

Battushig Myanganbayar, Mongolia

Aram Mkrtchyan, Armenia

Munkhtsetseg Battulga, Mongolia

Michalis Halkiopoulos, Greece

Nikolaj Theodor Thams, Denmark

Federico Re, Italy

Stefan Alexis Sigurðsson, Iceland

Phot

o by

Hen

ry T

eiga

r

138

Martin Vlashi, Italy

Mathias Stichelbaut, Belgium

Tapio Hautamäki, Finland

Jurij Tratar, Slovenia

Wai Hei Hoi, Macao, China

Ali Alhulaymi, Saudi Arabia

Kaloyan Darmonev, Bulgaria

Pedro Paredes, Portugal

Laura Gremion, Switzerland

Yeoh Chin Vern, Malaysia

Grgur Simunic, Croatia

Matheus Marreiros, Portugal

Thijs van der Gugten, Netherlands

Salizhan Kylychbekov, Kyrgyzstan

Molte Emil Strange Andersen, Denmark

Dale Alexander Hughes, Ireland

Luka Ivanovskis, Latvia

I Made G.N. Kumara, Indonesia

Mohammad Alhejji, Saudi Arabia

David Trillo Fernández, Spain

Aleksandr Petrosyan, Armenia

Vladimir Pejovic, Montenegro

Andres Rios Tascon, Colombia

Eduardo Acosta-Reynoso, Mexico

Werdi Wedana Gunawan, Indonesia

Ghadeer Shaaban, Syrian Arab Republic

Maris Serzans, Latvia

Pétur Rafn Bryde, Iceland

Kemal Babayev, Turkmenistan

honorable mentioned students Photo by Henry Teigar

139

Special PrizesIPhO 2012 absOLuTE wInnErAttila Szabó, Hungary

sPEcIaL PrIzEs FOr ThE bEsT sOLuTIOn OF ThE ExPErImEnTaLChristoph Schildknecht, Switzerland

Ivan Ivashkovskiy, Russia

Chi Shu, People’s Republic of China

sPEcIaL PrIzEs FOr ThE bEsT sOLuTIOn OF ThE ThEOrETIcaLEric Schneider, United States of America

bEsT In ExPErImEnTKai-Chi Huang, Taiwan

bEsT In ThEOryAttila Szabó, Hungary

PrIzEs OF assOcIaTIOn OF asIa PacIFIc PhysIcaL sOcIETIEsJeevana Priya Inala, India

Hengyun Zhou, People’s Republic of China

PrIzE OF EurOPEan PhysIcaL sOcIETyAttila Szabó, Hungary

sPEcIaL PrIcE FOr ThE bEsT gIrL IPhO 2012Alexandra Vassiljeva, Russia

sPEcIaL PrIcE FOr ThE bEsT EsTOnIan sTudEnTJaan Toots, Estonia

sPEcIaL PrIcE FOr ThE bEsT InnOvaTIvE sOLuTIOnLev Ginzburg, Russia

Absolute winner Attila Szabó with Jaan Kalda

Photo by Henry Teigar

140

statistics of the competition resultsJaan Kalda

Academic Committee of IPhO 2012

Let us analyze the overall difficulty of the problem set using the score-rank rela-

tionship presented in Fig 1. For an ideal well-balanced problem set, the net scores

of the contestants should be equi-spaced: the point-rank graph should be a nearly

straight line connecting the winner with a maximal score and ending at the last-

place-owner with 0 points. Bearing in mind that the performance of the contest-

ants is not fully predictable, such a problem set is hardly achievable. One can also

argue that in order to determine the absolute winners and gold medalists more

reliably, it is better to have larger point differences at the small ranks.

Fig 1. Point-rank graph for the total score; blue line - before moderation;

black line - after moderation.

The graph in Fig 1 tells us that the simple questions might have been slightly too

simple: the linear trend at the middle of the graph breaks at the right-hand-side

of the graph, where the curves turn steeply down. Meanwhile, the difficult ques-

tions were difficult, indeed, and provided a good separation between the very best

contestants; this is evidenced by another steeper segment at the left-hand-side.

141

It can also be of interest to compare the curves before and after the moderation.

This year, the typical distance between the two curves is less than 1 pt; without

going into details, one can say that this is quite a reasonable result. Ideally, a

smaller distance between the two curves should imply a better initial grading.

However, another factor here is the “flexibility” of the graders during the mod-

eration. For a partially solved problem, the decision of how many points should

be awarded is always slightly subjective. While the markers try to settle at the

middle of the uncertainty interval (to provide as fair a grading as possible), the

leaders tend to ask as many points as possible. (This is why the graders had been

instructed to grade “generously”: if doubting between two options, opt for more

points.) However, being too generous, i.e. giving too many marks for a partially

solved problem, will be unfair in respect to those who have solved the problem

flawlessly. Thus, regardless of how good the initial grading was, there will always

be some room for negotiation during the moderations, and a score shift is some-

times explained by the willingness of the graders to accept the arguments of the

leaders. As an example, this year there were two “compromises” due to which

the graders went through all the examination papers a second time. For all the

experimental tasks, marks were added for a correct plotting of wrong data points,

and for the Problem T3-iv and T3-v, partial credit for an incorrectly written first

law of thermodynamics was increased.

Next, let us have a look at the distribution of the theoretical and experi-

mental marks separately.

142

Fig 2. Point-rank graph for the scores of the experiment and for the theory

For the theoretical examination, the linear trend extends down to the right-

most corner of the graph. Therefore, none of the questions were too simple! As for

the experiment, the graph is qualitatively very similar to the graph of the total

scores.

Let us dwell even deeper into detail and have a look at the distribution of

points for all the theoretical problems.

143

Fig 3. Point-rank graphs for the scores of the Problem T1, T2 and T3.

The simple questions of problem T1 were, indeed, simple: the total cost of

these was 3x0.8=2.4 pts and ca 40% of the contestants got at least this much.

However, the supposedly medium difficulty questions, each worth 1.2 and total-

ling in 3.6 pts, were actually quite difficult: answering all the simple and medium

questions would have resulted in 6 pts, and only 6% of contestants got 6 points or

more.

The contestants with top scores are as follows (all gold medalists).

12.1: Attila Szabó (HUN);

10.9: Eric Schneider (USA);

10.2: Hengyun Zhou (CHN);

10.1: Yijun Jiang (CHN);

9.3: Ilya Vilkoviskiy (KAZ);

9.1: Paphop Sawasdee (THA), Wenzhuo Huang (CHN);

8.7: Chien-An Wang (TWN);

8.2: Wonseok Lee (KOR);

7.7: Jun-Ting Hsieh (TWN);

7.6: Ding Yue (SGP);

144

7.5: Kuan Jun Jie, Joseph (SGP);

7.2: Sooshin Kim (KOR), Siyuan Wei (CHN);

7.0: Phi Long Ngo (VNM).

Next, about Problem T2. As you can see, this is a problem with a perfect balance

between simple and difficult questions: there is almost a linear line connecting

the upper left corner with the lower right corner. The contestants with top scores

are as follows (all gold medalists, unless otherwise noted).

8.0 pts: Chien-An Wang (TWN), Yijun Jiang (CHN);

7.9 pts: Jun-Ting Hsieh (TWN), Tudor Giurgică-Tiron (ROU);

7.8 pts: Hengyun Zhou (CHN), Chi Shu (CHN), Rahul Trivedi (IND), David

Frenklakh (RUS, Silver), Kacper Oreszczuk (POL, Bronze),

7.7 pts: Wenzhuo Huang (CHN), Siyuan Wei (CHN), Jaemo Lim (KOR), Tanel

Kiis (EST, Silver).

Finally, Problem T3. A slight score saturation can be observed for this problem:

the curve “hits the roof” (i.e. the maximal value of 9.0 pts) at the upper left cor-

ner. There would have been probably a better balance between difficult and easy

questions if the hint about Kepler’s laws were not given in the text of the problem.

However, including the hint was the wish of the International Board, and the

problem set as a whole was difficult enough even with the hint included ...

The contestants with a full score (9.0 pts; all gold medalists): Attila Szabó

(HUN), Paphop Sawasdee (THA), Chien-An Wang (TWN), Siyuan Wei (CHN),

Yuichi Enoki (JPN), Rahul Trivedi (IND), Puthipong Worasaran (THA), Tudor

Giurgică-Tiron (ROU), Ngoc Hai Dinh (VNM), Alexandra Vasilyeva (RUS, Silver),

Volodymyr Sivak (UKR, Silver), Bijoy Singh Kochar (IND, Silver), Nurzhas Aidynov

(KAZ, Silver), Cristian Zanoci (MDA, Bronze).

Next, let us have a look at the experimental problems.

145

Fig 4. Point-rank graphs for the scores of the Problems E1 and E2.

Problem E1 has a nice distribution at the upper left corner, but too steep a

fall-off at the right edge – the simplest tasks of this problem were perhaps too

simple. The contestants with top scores are as follows (all gold medalists, unless

otherwise noted).

10: Jaan Toots (EST);

9.9: Kai-Chi Huang (TWN), Ivan Ivashkovskiy (RUS);

9.8: Wei-Jen Ko (TWN);

9.7: Attila Szabó (HUN), Siyuan Wei (CHN);

9.6: Allan Sadun (USA);

9.5: Hengyun Zhou (CHN), Chien-An Wang (TWN);

9.4: Wonseok Lee (KOR);

9.3: Jun-Ting Hsieh (TWN);

9.2: Eric Schneider (USA), Wenzhuo Huang (CHN);

9.1: Ngoc Hai Dinh (VNM), Alexandra Vasilyeva (RUS, Silver), Chi Shu (CHN).

Meanwhile, Problem E2 was intended to be a difficult problem, aimed at

finding the winner of the best experimentalist’s prize. And difficult it was: it

had actually no easy tasks, as evidenced by a concave shape of the curve. The con-

testants with top scores are as follows (all gold medalists, unless otherwise noted).

8.8: Chi Shu (CHN);

8.5: Kai-Chi Huang (TWN);

146

8.3: Christoph Schildknecht (CHE, Silver);

8.1: Ivan Ivashkovskiy (RUS);

7.7: Attila Szabó (HUN);

7.5: Hengyun Zhou (CHN), Huan Yan Qi (SGP);

7.4: Lev Ginzburg (RUS);

7.2: Abdurrahman Akkas (TUR, Silver);

7: Kristjan Kongas (EST, Silver);

6.9: Yu-Ting Liu (TWN), Adam Brown (GBR, Silver);

6.7: Kevin Zhou (USA), Frank Bloomfield (GBR, Bronze).

And now, it is time to have a look at the most difficult questions (tasks). Let us

start with the three parts of Problem 1.

Fig 5. Point-rank graphs for the scores of Tasks T1A , T1B, and T1C.

I was quite sure that question iii of Part A would be very difficult for the con-

testants, and question iii of Part C would be extremely difficult, and I was not

mistaken. However, I did believe that question iii of Part B was not that difficult

(just difficult, not “very” or “extremely”), and my colleagues from the Academic

Committee agreed. However, here we were mistaken: Part B turned out to be the

most difficult part!

Part 1A: only ca 20% of students were able to figure out the correct shape

of the trajectory. Meanwhile, there was also a considerable number of those

147

who got everything correctly done, including q. iii! This is an interesting case

because in order to be able to solve this problem only a moderate physical educa-

tion is needed. This is evidenced by the fact that among the best solvers of Part

1A, there are several students whose overall results were not so good. One can

only hypothesize that had they passed a full course of physics covering all the

Syllabus of the IPhO, they would have been able to get gold medals. The contest-

ants with top scores are as follows (all gold medalists, unless otherwise noted).

4.5 pts: Attila Szabó (HUN), Hengyun Zhou (CHN), Eric Schneider (USA), Wenzhuo

Huang (CHN), Yijun Jiang (CHN), Rahul Trivedi (IND), Ilya Vilkoviskiy (KAZ),

Kuan Jun Jie (SGP), Joseph Ramadhiansyah Ramadhiansyah (IDN, Honourable

Mention);

4.4 pts: Paphop Sawasdee (THA), Jeffrey Cai (USA, Silver), Puthipong Worasaran

(THA), Nathanan Tantivasadakarn (THA);

4.2 pts: Michele Fava (ITA, Bronze);

4.1 pts: Ding Yue (SGP);

4 pts: Hakon Tásken (NOR, Participation Certificate).

Part 1B: the list of top-solvers is shorter than before because all the others just

did not get enough marks for q. iii to be qualified as someone who really solved

this problem. As usual, everyone below got a gold medal, unless otherwise noted.

3.9 pts: Jun-Ting Hsieh (TWN);

3.8 pts: Attila Szabó (HUN);

3.6 pts: Sooshin Kim (KOR);

3.5 pts: Yijun Jiang (CHN);

3.4 pts: Siyuan Wei (CHN), Kai-Chi Huang (TWN), Wonseok Lee (KOR);

3.3 pts: Ihar Lobach (BLR);

3.1 pts: Wenzhuo Huang (CHN);

3 pts: Georgijs Trenins (LVA, Silver), Karlo Sepetanc (HRV, Honourable Mention).

Finally, Part 1C: note that about half of those who performed very well here

(listed below) lost 0.2 in q. i for drawing too curved field lines. There were only

four contestants who got the idea of magnetic charges and realized it flawlessly

(these are the first three in the list below, and Kunal Singhal). However, owing

to the fact that there is also another way of calculating the force (via integrating

over dipoles), several contestants got a correct estimate of the force, and, thereby,

148

collected enough marks to be listed below

4.5 pts: Ilya Vilkoviskiy (KAZ);

4.3 pts: Chien-An Wang (TWN);

4.2 pts: Paphop Sawasdee (THA);

4 pts: Wonseok Lee (KOR), Wei-Jen Ko (TWN);

3.9 pts: Eric Schneider (USA);

3.8 pts: Attila Szabó (HUN), Yuichi Enoki (JPN), Hengyun Zhou (CHN);

3.7 pts: Phi Long Ngo (VNM);

3.6 pts: Kazumi Kasaura (JPN);

3.5 pts: Kunal Singhal (IND, Silver);

3.4 pts: Yu-Ting Liu (TWN);

3 pts: Jun-Ting Hsieh (TWN), Siyuan Wei (CHN), Ivan Tadeu Ferreira Antunes

Filho (BRA).

The rest of the theoretical test was not that difficult (q iii of Problem T3 would

have been quite difficult, but with the hint inserted by the International Board it

no longer was). So we won’t dwell more on the theoretical results, and we’ll switch

to the really tricky experimental tasks: A-iv and B of Problem E2.

Fig 6. Point-rank graphs for the scores of the Tasks E2B and E2A-iv.

In the case of Task A-iv, the number of those who really got the correct idea

how to measure C(V) was really small - essentially only those who are listed below.

149

2.6 pts: Kuan Jun Jie, Joseph (SGP), Kai-Chi Huang (TWN), Lev Ginzburg (RUS),

Ivan Ivashkovskiy (RUS);

2.5 pts: Chi Shu (CHN), Qiao Gu (DEU), Kevin Zhou (USA), Allan Sadun (USA)

2.4 pts: Yu-Ting Liu (TWN), Kristjan Kongas (EST, Silver), Adrian Nugraha

Utama (IDN), Tudor Giurgică-Tiron (ROU), Sebastian Linß (DEU, Silver).

Finally, Task B. There was a surprisingly small number of those contestants

who noticed that the difference in the graphs of Part A and Part B is localized to the

region of negative differential resistance. As for the explanation of the phenome-

non (which consists of three key elements), none of the contestants managed to

list all the key elements flawlessly, and the only one to get it almost done (with

some omissions in the average current part) was Christoph Schildknecht; the

next two in the list below mentioned one key element. And so, the best results

for Task 2B:

2.9 pts: Christoph Schildknecht (CHE, Silver);

2.3 pts: Attila Szabó (HUN);

2.1 pts: Chi Shu (CHN);

2 pts: Kai-Chi Huang (TWN), Luka Ivanovskis (LVA, Honourable Mention).

For those who want to go beyond this statistical analysis, there is also an Excel

file http://www.ioc.ee/~kalda/ipho/Results_for_web.xlsx (the names of those who got less

than 12.4 pts have been stripped).

The problems of the 43rd IPhO have been thought to be difficult, and it has

even been stated that the problem set was the most difficult one during the last

20 years. In order to make a comparative study about how difficult the problems

actually were several types of data are needed, which are not freely available for all

the Olympiads. Still, I managed to get more or less what is needed (overall number

of participants, number of medals, medal boundaries in points, the scores of the

absolute winners) for the period covering 1994–2012. The graph is shown below.

(The last two digits of the year are shown alongside the curve – except for some

curves in the central densely populated region; note that the curves which are

based only on the number of medals and on the medal boundaries are interpolated

and smooth.)

150

Fig 7. Point-rank graphs for the overall scores of the last nineteen IPhO-s

(for those years for which the complete data were unavailable, the curves are

interpolated between the datapoints corresponding to the medal boundaries).

Even with these data, the IPhOs apart as much as 19 years are not fully compa-

rable. I have a feeling that, in average, the preparation level of the leading group of

contestants has risen significantly. So, the graph here does not allow comparison

of the absolute difficulties of the problems, but only relative ones – relative to the

preparation level of the students. One should also bear in mind that the scores of

absolute winners have a high intrinsic variability (there is essentially no statistical

averaging); c.f. this year: the first and second places were separated by a huge

margin of 3 pts.

Therefore, the conclusion is that the claim, about this being the most difficult

problem set in the last 20 years, was slightly exaggerated. The problems in Beijing

in 1994 were even more difficult, at least in relative terms, and at least when

leaving out the contestants with ranks from 2 to 6.

151

International boardMinutesmInuTEs OF ThE mEETIngs OF ThE InTErnaTIOnaL bOard durIng ThE 43rd InTErnaTIOnaL PhysIcs OLymPIad In TaLLInn, EsTOnIa, JuLy 15Th – 24Th, 2012

1. A total number of 378 contestants from the following 80 countries were

present at the 43rd International Physics Olympiad:

Albania, Armenia, Australia, Austria, Azerbaijan, Bangladesh, Belarus,

Belgium, Bolivia, Bosnia & Herzegovina, Brazil, Bulgaria, Canada, China,

Colombia, Croatia, Cyprus, Czech Republic, Denmark, El Salvador, Estonia,

Finland, France, Georgia, Germany, Greece, Hong Kong, Hungary, Iceland, India,

Indonesia, Iran, Ireland, Israel, Italy, Japan, Kazakhstan, Kuwait, Kyrgyzstan,

Latvia, Liechtenstein, Lithuania, Macau, Macedonia, Malaysia, Mexico,

Moldova, Mongolia, Montenegro, Netherlands, Nigeria, Norway, Pakistan,

Poland, Portugal, Puerto Rico, Republic of Korea, Romania, Russia, Saudi Arabia,

Serbia, Singapore, Slovakia, Slovenia, South Africa*, Spain, Sri Lanka, Suriname,

Sweden, Switzerland, Syria, Taiwan, Tajikistan, Thailand, Turkey, Turkmenistan,

Ukraine, United Kingdom, USA, Vietnam.

* : new country invited by the Organizing Committee to the IPhO this year.

2. Results of marking the papers by the organizers were presented.

The best score (45.8 points) was achieved by Mr. Attila Szabó from Hungary (the

overall winner of the 43rd IPhO). The following limits for awarding the medals and

the honorable mention were established according to the Statutes:

Gold Medal 31.0 points

152

Silver Medal 23.9 points

Bronze Medal 17.2 points

Honorable Mention 12.4 points

According to the above limits, 45 Gold Medals, 71 Silver Medals, 92 Bronze

Medals, and 63 Honorable Mentions were awarded. The grade lists of the awardees

were distributed to all the delegation leaders in print.

3. In addition to the regular prizes, the following special prizes were awarded:

The Overall Winner

Mr. Attila Szabó, Hungary

Best in Theory


Best in Experiment

Mr. Kai-Chi Huang, Taiwan

Special Prize for the Best Solution of the Theoretical

Mr. Eric Schneider, United States of America

Special Prizes for the Best Solution of the Experimental

Mr. Christoph Schildknecht, Switzerland

Mr. Ivan Ivashkovskiy, Russia

Mr. Chi Shu, People`s Republic of China

Special Prize for the best innovative solution

Mr. Lev Ginzburg, Russia

Special Prize for the best Estonian student

Mr. Jaan Toots, Estonia

Special Prize for the Best Girl IPhO 2012

Ms. Alexandra Vassiljeva, Russia

Prize of European Physical Society


Prizes of Association of Asia Pacific Physical Societies

Ms. Jeevana Priya Inala, India

Mr. Hengyun Zhou, People`s Republic of China

4. The following three leaders were designated by the International Board

to serve as consultants to the local Academic Committee for grading the

examination papers :

153

Prof. Niels Hartling (Denmark), Prof. Suwan Kusamran (Thailand), and Prof.

Helmuth Mayr (Austria)

5. The International Board discussed a proposal, presented by the President of

the IPhO, Dr. Hans Jordens, of establishing a legal body of a fund-raising

bank account, named “the Foundation of International Physics Olympiad”,

which would be affiliated with the IPhO. The objective of the non-profit

Foundation is to raise funds for helping economy-weak countries to send

students to participate in the IPhO and for providing financial support to

future hosting countries in case of need. The Statutes of the Foundation

which were examined and revised by the Advisory Committee had been

disseminated to all leaders three months prior to the IB meeting. The

proposal was accepted by the International Board with an overwhelming

majority in the affirmative. The Board of the Foundation comprises at least

3 members. In addition to the IPhO President and Secretary, a Treasurer

will be nominated and elected in the next IPhO.

6. The term of the current President of the IPhO will expire in 2013. Since

no specific regulation on the election of presidency is stated in the IPhO

Statutes, the International Board unanimously agreed to adopt the same

procedures as for the electing of the IPhO Secretary, as specified in the

Regulations to the Article #8, for the election of a new president for the

term 2013–2018.

7. The International Board discussed a proposal submitted by Icelandic

leader, Dr. Martin Swift, which asked to modify the regulations on using

“non-programmable” pocket calculators as stated in the Article #5 of the

IPhO Statutes. The International Board approved to form a specific commit-

tee to deliberate on revising the concerned regulations. The IPhO Secretariat

has invited the following three leaders to form the calculator-committee:

Dr. Martin Swift (Iceland), Dr. Eli Raz (Israel), and Dr. Matthew Verdon

(Australia).

8. The IPhO Secretariat received an official letter from the Korean Physical

Society which stated that the Republic of Korea would prefer to host the

154

58th IPhO in 2027 instead of the 63th IPhO in 2032 as previously proposed.

The request was accepted by the International Board.

9. The President of the IPhOs, Prof. Hans Jordens, acting on behalf of all the

participants, expressed his gratitude to the Minister of Education and

Research, Prof. Dr. Jaak Aaviksoo, and deep thanks to Mrs Ene Koitla (Head

of the Organizing Committee), Prof. Jaak Kikas (Head of the Academic

Committee), Dr. Jaan Kalda (Co-head of the Academic Committee), Dr.

Viire Sepp (Academic Secretary of the Steering Committee) and all other

members involved in organizing the IPhO2012 for excellent preparation

and conduction of the 43rd International Physics Olympiad. Deep thanks

were also conveyed to Tallinn University of Technology, the University of

Tartu, the Estonian Physical Society, and the Ministry of Education and

Research of Estonia, all the sponsors, graders, guides and other people who

contributed to the success of the Olympiad.

10. The Danish delegates disseminated printed materials about the 44th IPhO

in 2013 to all the delegations and described the present state of the prepara-

tory works to ensure smooth organization of the next Olympiad.

11. At the Closing Ceremony of the Olympiad, on behalf of the organizers of

the next International Physics Olympiad, Prof. Niels Hartling announced

that the 44th International Physics Olympiad would be organized in

Copenhagen, Denmark from July 7th to 15th, 2013 and cordially invited

all the participating countries to attend the competition.

(signature)

......................

Prof. Hans Jordens

President of the IPhO

(signature)

......................

Prof. Ming-Juey Lin

Secretary of the IPhO

(signature)

......................

Prof. Jaak Kikas

Head of the Academic

Committee of IPhO2012

Tallinn, Estonia

July 22nd, 2012

155Photos by Merily Salura, Henry Teigar and Karl Veskus

156

157

StatutessTaTuTEs OF ThE InTErnaTIOnaL PhysIcs OLymPIads

Version accepted in 1999 in Padova (Italy)

Revised: 2000 - Leicester (Great Britain)

2001 - Antalya (Turkey)

2002 - Bali (Indonesia)

2004 - Pohang (Korea)

2006 - Singapore

2008 - Hanoi (Vietnam)

§1

In recognition of the growing significance of physics in all fields of science and

technology, and in the general education of young people, and with the aim of

enhancing the development of international contacts in the field of school educa-

tion in physics, an annual physics competition has been organised for secondary

school students. The competition is called the International Physics Olympiad and

is a competition between individuals.

§2

The competition is organised by the Ministry of Education, the Physical Society

or another appropriate institution of one of the participating countries on whose

territory the competition is to be conducted. The organising country is obliged

to ensure equal participation of all the delegations, and to invite teams from all

those countries that participated during the last three years. Additionally, it has

the right to invite other countries. The list of such new countries must be pre-

sented to Secretariat of the IPhOs (# 8) at least six months prior to the competition.

Within two months the Secretariat has the right to remove, after consultations

with the Advisory Committee (# 8), from the suggested list the teams that in

opinion of Secretariat or Advisory Committee do not meet the criteria of partici-

pation in the IPhOs. The new countries not accepted by the Secretariat or Advisory

Committee may, however, participate as “guest teams” but such participation

does not create any commitments with respect to inviting these countries to the

158

next competition(s).

No country may have its team excluded from participation on any political

reasons resulting from political tensions, lack of diplomatic relations, lack of rec-

ognition of some country by the government of the organising country, imposed

embargoes and similar reasons. When difficulties preclude formal invitation of

the team representing a country students from such a country should be invited

to participate as individuals.

The competition is conducted in the friendly atmosphere designed to promote

future collaborations and to encourage the formation of friendship in the scien-

tific community. Therefore all possible political tensions between the participants

should not be reflected in any activity during the competition. Any political activ-

ity directed against any individuals or countries is strictly prohibited.

§3

Each participating country shall send a delegation, normally consisting of five stu-

dents (contestants) and two accompanying persons (delegation leaders) at most.

The contestants shall be students of general or technical secondary schools

i.e. schools which cannot be considered technical colleges. Students who have

finished their school examinations in the year of the competition can be members

of the team as long as they have not commenced their university studies. The age

of the contestants should not exceed twenty years on June 30th of the year of the

competition.

The delegation leaders must be specialists in physics or physics teachers, capa-

ble of solving the problems of the competition competently. Each of them should

be able to speak English.

§4

The Organisers of the Olympiad determine in accordance to the programme the

day of arrival and the day of departure as well as the place in their country from

which the delegations are supposed to arrive and depart. The costs for each dele-

gation as a result of activities connected to the Olympiad from the day of arrival

till the day of departure are covered by the Organising Committee.

§5

The competition shall be conducted over two days, one for the theoretical exam-

ination and one for the experimental examination. There will be at least one full

159

day of rest between the examinations.

The theoretical examination shall consist of three theoretical problems and

shall be of five hours total duration.

The experimental examination shall consist of one or two problems and shall

be of five hours total duration.

Contestants may bring into the examination drawing instruments and non-pro-

grammable pocket calculators. No other aids may be brought into the examination.

The theoretical problems should involve at least four areas of physics taught at

secondary school level, (see Syllabus). Secondary school students should be able

to solve the competition problems with standard high school mathematics and

without extensive numerical calculation.

The competition tasks are chosen and prepared by the host country and have

to be accepted by the International Board (§ 7).

The host country has to prepare at least one spare problem, which will be pre-

sented to the International Board if one of the first three theoretical problems is

rejected by two thirds of members of the International Board. The rejected problem

cannot be considered again.

§6

The total number of marks awarded for the theoretical examination shall be 30 and

for the experimental examination 20. The competition organiser shall determine

how the marks are allocated within the examinations.

After preliminary grading (prior to discussion of the grading with the dele-

gation leaders) the organizers establish minima (expressed in points) for Gold

Medals, Silver Medals, Bronze Medals and Honourable Mentions according to the

following rules:

Gold Medals should be awarded to 8% of the contestants (rounded up the

nearest integer).

Gold or Silver Medals should be awarded to 25% of the contestants (rounded

up the nearest integer).

Gold, Silver or Bronze Medals should be awarded to 50% of the contestants

(rounded up the nearest integer).

An Olympic Medal or Honourable Mention should be awarded to 67% of the

contestants (rounded up the nearest integer).

The minima corresponding to the above percentages should be expressed

160

without rounding. The suggested minima shall be considered carried if one

half or more of the number of the Members of the International Board cast

their vote in the affirmative.

Results of those candidates who only receive a certificate of participation

should strictly remain to the knowledge of the Members of the International

Board and persons allowed to attend its meetings.

§7

The governing body of the IPhO is the International Board, which consists of the

delegation leaders from each country attending the IPhO.

The chairman of the International Board shall be a representative of the organ-

ising country when tasks, solutions and evaluation guidelines are discussed and

the President of the IPhO in all other topics.

A proposal placed to the International Board, except Statutes, Regulations and

Syllabus (see§ 10), shall be considered carried if more than 50% of all delegation

leaders present at the meeting vote in the affirmative. Each delegation leader

is entitled to one vote. In the case of equal number of votes for and against, the

chairman has the casting vote. The quorum for a meeting of the International

Board shall be one half of those eligible to vote.

The International Board has the following responsibilities:

to direct the competition and supervise that it is conducted according to

the regulations;

to ascertain, after the arrival of the competing teams, that all their mem-

bers meet the requirements of the competition in all aspects. The Board will

disqualify those contestants who do not meet the stipulated conditions;

to discuss the Organisers’ choice of tasks, their solutions and the suggested

evaluation guidelines before each part of the competition. The Board is author-

ised to change or reject suggested tasks but not to propose new ones. Changes

may not affect experimental equipment. There will be a final decision on the

formulation of tasks and on the evaluation guidelines. The participants in the

meeting of the International Board are bound to preserve secrecy concerning

the tasks and to be of no assistance to any of the participants;

to ensure correct and just classification of the students. All grading has to

be accepted by the International Board;

to establish the winners of the competition and make a decision concerning

161

presentation of the medals and honourable mentions. The decision of the

International Board is final;

to review the results of the competition;

to select the countries which will be assigned the organisation of future

competitions;

to elect the members of the Secretariat of the IPhO.

§8

The long-term work involved in organising the Olympiads is co-ordinated by a

Secretariat for the International Physics Olympiads. This Secretariat consists of the

President and Secretary. They are elected by the International Board for a period of

five years when the chairs become vacant.

The President and the Secretary of the IPhO should be invited to the Olympiads

as the members and heads of the International Board, their relevant expenses

should be paid by the organizers of the competition. The President and the

Secretary should not be leaders of any national team.

There shall be an Advisory Committee convented at the President of the IPhOs.

The Advisory Committee consists of:

1. The President,

2. The Secretary,

3. The host of the past Olympiad,

4. The hosts of the next two Olympiads,

5. Such other persons appointed by the President.

§9

The working language of the IPhO is English.

The competition problems should be presented to the International Board in

English, Russian, German, French and Spanish.

The solutions to the problems should be presented in English.

It is the responsibility of the delegation leaders to translate the problems into

languages required by their students.

These statutes and other IPhO-documents shall be written in English.

Meetings of the International Board shall be held in English.

§10

These statutes are supplemented by Regulations concerning the details of the

162

organisation the Syllabus mentioned in § 5.

Proposals for amendment to these Statutes and the supplementing documents

may be submitted to the president or his nominee no later than December 15th

prior to consideration.

The President shall circulate, no later than March 15th, all such proposals

together with the recommendation of the President’s Advisory Committee, to

the last recorded address of each delegation leader who attended at the last IPhO.

Such proposals shall be considered by a meeting of the International Board at the

next IPhO and shall be considered carried if

in case of Statutes and Syllabus two thirds or more and

in case of Regulations more than one half

of the number of the members of the International Board present at the

meeting cast their vote in the affirmative. Such changes shall take effect

from the end of the current IPhO and cannot affect the operation of the

competition in progress. The vote can only take place if at least 2/3 of the

all leaders are present at the meeting.

§11

Participation in an International Physics Olympiad signifies acceptance of the

present Statutes by the Ministry of Education or other institution responsible for

sending the delegation.

163

rEguLaTIOns assOcIaTEd wITh ThE sTaTuTEs OF ThE InTErnaTIOnaL PhysIcs OLymPIads

Regulations to §2

The Ministry of Education, or the institution organising the competition, allots

the task of preparation and execution of the Competition to an appropriate body.

Official invitations to the participating countries should be sent at least six

months before the Olympiad. They normally are sent to the national institution

that sent the delegation to the previous Olympiad. Copies of the invitation are also

sent to the previous years’ delegation leaders. The invitation should specify the

place and time of the Competition plus the address of the organising secretariat.

Countries wishing to attend the current IPhO must reply to the invitation before

March 15, nominating a contact person. Each participating country must in addition

supply the host country with the contestants’ personal data (surname, given name,

sex, address, date of birth and address of school) by May 15 or as soon as possible.

The host country is only obliged to invite delegations from countries that partici-

pated in one of the last three competitions. It may refuse

applications for participation from any other country

applications from participating countries not belonging to the delegation

as defined in §3 (observers, guests).

Each country should, within five years of entry, declare its intention to host for

a future Olympiad, suggesting possible years. A country that is unable to organise

the competition may be prevented from participating in IPhOs by decision of the

International Board.

Regulations to §3

The accompanying persons are considered by the organisers of the next Olympiad

and by the Secretariat of the IPhO (§ 8) as contact persons until the next Olympiad

(unless new accompanying persons or other contact persons are nominated by the

participating country).

Each participating country must ensure that the contestants are all secondary

school pupils when they announce the names of the members of their delegations.

In addition to the delegations, teams may be accompanied by observers and guests.

Observers may attend all Olympiad meetings, including the meetings of the

International Board. However they may not vote or take part in the discussions.

Guests do not attend the meetings of the International Board.

164

If possible, the host country should accept as observers any of the following

persons:

the organiser(s), or nominee(s), from the host country in the subsequent

three years

a representative of any country expressing an intention to participate in

the following IPhO.

Regulations to §4

The host country must pay for organisation of IPhO, food, lodging, transport and

excursions of the delegations plus prizes.

However it is not responsible for medical costs and sundry expenses of the

participants. Observers and guests may be asked to pay the full cost of their stay

plus an attendance fee.

The host country may ask the delegations for a voluntary contribution to the

obligatory costs.

Regulations to §5

It is recommended that the Competition should last 10 days (including arrival and

departure days).

The host country is obliged to ensure that the Competition is conducted accord-

ing to the Statutes. It should provide full information for participating countries,

prior to their arrival, concerning venue, dates, accommodation, transport from

airports, ports and railway stations. The addresses, telephone, fax, e-mail of all

IPhO officers should be provided, together with information concerning relevant

laws and customs of the host country.

A program of events during the IPhO should be prepared for the leaders and

contestants. It should be sent to the participating countries, prior to the Olympiad.

The organisers of the IPhO are responsible for devising all the problems. They

must be presented in English and the other official languages of the Olympiad as

indicated in § 9. The examination topics should require creative thinking and knowl-

edge contained within the Syllabus. Factual knowledge from outside the Syllabus

may be introduced provided it is explained using concepts within the Syllabus.

Everyone participating in the preparation of the competition problems must

not divulge their content.

The standard of problems should attempt to ensure that approximately half

165

the students obtain over half marks.

The International Board shall be given time to consider the examination papers.

It may change, or reject, problems. If a problem is rejected, the alternative prob-

lem must be accepted. The host country will be responsible for grading the exam-

ination papers. The delegation leaders shall have an opportunity to discuss with

the examiners the grading of their students’ papers. If an agreement, between

graders and leaders, to the final marks cannot be reached, the International Board

has to decide.

The organisers shall provide the delegation leaders with copies of their stu-

dents’ scripts and allow at least 12 hours for them to mark the scripts.

The host country shall provide medals and certificates in accordance with the

Statutes. They must also produce a list of all contestants receiving awards with their

marks and associated award. The awards are presented at the Closing Ceremony.

The host country is obliged to publish the Proceedings of the Competition, in

English, in the subsequent twelve months. A free copy of the Proceedings should

be sent to all delegation leaders and competitors.

Regulations to §6

Special prices may be awarded. The participant who obtains the highest score

should receive a special prize.

Regulations to §7

During the meeting of the graders where the final and most detailed version of the

grading scheme is set, 3 members of the International Board will be present. They

have the right to give advice to the group of graders in order to keep the grading

scheme within the tradition of the IPhOs.

If it is found that leaders, observers or students from a country have been in col-

lusion to cheat in one of the International Olympiad examinations, the students

concerned should be disqualified from that Olympiad. In addition, the leaders,

observers and students involved should not be allowed to return to any future

Olympiad. Appropriate decisions are taken by the International Board.

166

Regulations to §8

Election of secretary

The secretary has to have been for the five years prior to the nomination:

a member of the International Board for at least three of these years,

or an observer or member of the International Board, who has attended all

these five IPhOs

The secretary will hold office for a period of five years commencing at the con-

clusion of the final meeting of the International Board at which the secretary has

been chosen.

The secretary and the president must not be appointed at the same IPhO. If this

is the case, however, the period of the secretary will have to be shortened in such

a way that the two elections can be held at different IPhOs.

The secretary and the president must not come from the same delegation.

If the term of the secretary comes to an end, the International Board has to

be informed one year in advance that there will be the ballot of a new secretary

during the following IPhO. In addition to that, the secretariat is responsible to

send a letter to all leaders of the last three IPhOs with this information and with

the question if any leader will be ready to act as secretary for the coming period

by 31st January. This is normally done by e-mail.

If someone is willing to be a candidate for the secretary-ballot, he or she will

have to tell this to the current secretary by 31st March, normally by e-mail.

A nominee has to send his/her curriculum vitae up to 31st March. A nomination

may not be made by a person from the same country as the current president.

The secretariat is responsible to collect all these answers and has to make a list

with all the names.

If the current secretary is willing to continue his/her activity as secretary, he

or she has to enter his/her name in this list and has to follow the same rules as all

the other candidates.

The list with the candidates for the new secretary has to be published on the IPhO-

home-page and the home page of the IPhO during which the ballot will be held.

If there is just one candidate, the secretary has to inform the president about

that. In that case this candidate is accepted as secretary.

The secretariat and the organisers of the IPhO during which the election will

be held are responsible for a democratic, secret ballot of the secretary during the

last meeting of the International Board.

167

If the current secretary resigns or becomes incapable of continuing his/her

work as a secretary, the president shall appoint a replacement to act as provisional

secretary up to the next IPhO. The ballot of the new secretary has to be made as

soon as possible.

168

SyllabusAppendix to the Statutes of the International Physics Olympiads

General

a. The extensive use of the calculus (differentiation and integration) and the

use of complex numbers or solving differential equations should not be

required to solve the theoretical and practical problems.

b. Questions may contain concepts and phenomena not contained in the Syllabus

but sufficient information must be given in the questions so that candidates

without previous knowledge of these topics would not be at a disadvantage.

c. Sophisticated practical equipment likely to be unfamiliar to the candidates

should not dominate a problem. If such devices are used then careful

instructions must be given to the candidates.

d. The original texts of the problems have to be set in the SI units.

A. Theoretical Part

The first column contains the main entries while the second column contains

comments and remarks if necessary.

1. Mechanics

a) Foundation of kinematics of a

point mass

Vector description of the position of

the point mass, velocity and acceler-

ation as vectors

b) Newton’s laws, inertial systems Problems may be set on changing mass

c) Closed and open systems, momen-

tum and energy, work, power

d) Conservation of energy, conserva-

tion of linear momentum, impulse

e) Elastic forces, frictional forces, the

law of gravitation, potential energy

and work in a gravitational field

Hooke’s law, coefficient of friction

(F/R = const), frictional forces, static

and kinetic, choice of zero of poten-

tial energy

f) Centripetal acceleration, Kepler’s laws

169

2. Mechanics of Rigid Bodies

a) Statics, center of mass, torque Couples, conditions of equilibrium

of bodies

b) Motion of rigid bodies, transla-

tion, rotation, angular velocity,

angular acceleration, conservation

of angular momentum

Conservation of angular momen-

tum about fixed axis only

c) External and internal forces, equa-

tion of motion of a rigid body around

the fixed axis, moment of inertia,

kinetic energy of a rotating body

Parallel axes theorem (Steiner’s

theorem), additivity of the moment

of inertia

d) Accelerated reference systems,

inertial forces

Knowledge of the Coriolis force

formula is not required

3. Hydromechanics

No specific questions will be set on this but students would be expected to know

the elementary concepts of pressure, buoyancy and the continuity law.

4. Thermodynamics and Molecular Physics

a) Internal energy, work and

heat, first and second laws of

thermodynamics

Thermal equilibrium, quantities

depending on state and quantities

depending on process

b) Model of a perfect gas, pressure

and molecular kinetic energy,

Avogadro’s number, equation of

state of a perfect gas, absolute

temperature

Also molecular approach to such

simple phenomena in liquids and

solids as boiling, melting etc.

c) Work done by an expanding gas

limited to isothermal and adiabatic

processes

Proof of the equation of the adiaba-

tic process is not required

d) The Carnot cycle, thermodynamic

efficiency, reversible and irrevers-

ible processes, entropy (statistical

approach), Boltzmann factor

Entropy as a path independent

function, entropy changes and

reversibility, quasistatic processes

170

5. Oscillations and waves

a) Harmonic oscillations, equation

of harmonic oscillation

Solution of the equation for har-

monic motion, attenuation and

resonance -qualitatively

b) Harmonic waves, propagation of

waves, transverse and longitudinal

waves, linear polarization, the clas-

sical Doppler effect, sound waves

Displacement in a progressive wave

and understanding of graphical

representation of the wave, measure-

ments of velocity of sound and light,

Doppler effect in one dimension only,

propagation of waves in homogene-

ous and isotropic media, reflection

and refraction, Fermat’s principle

c) Superposition of harmonic

waves, coherent waves, interfer-

ence, beats, standing waves

Realization that intensity of wave

is proportional to the square of its

amplitude. Fourier analysis is not

required but candidates should have

some understanding that complex

waves can be made from addition of

simple sinusoidal waves of different

frequencies. Interference due to

thin films and other simple systems

(final formulae are not required),

superposition of waves from second-

ary sources (diffraction)

6. Electric Charge and Electric Field

a) Conservation of charge,

Coulomb’s law

b) Electric field, potential, Gauss’

law

Gauss’ law confined to simple sym-

metric systems like sphere, cylinder,

plate etc., electric dipole moment

c) Capacitors, capacitance, dielec-

tric constant, energy density of

electric field

7. Current and Magnetic Field

171

a) Current, resistance, internal

resistance of source, Ohm’s law,

Kirchhoff’s laws, work and power

of direct and alternating currents,

Joule’s law

Simple cases of circuits containing

non-ohmic devices with known V-I

characteristics

b) Magnetic field (B) of a current,

current in a magnetic field, Lorentz

force

Particles in a magnetic field, simple

applications like cyclotron, mag-

netic dipole moment

c) Ampere’s law Magnetic field of simple symmetric

systems like straight wire, circular

loop and long solenoid

d) Law of electromagnetic induc-

tion, magnetic flux, Lenz’s law,

self-induction, inductance, perme-

ability, energy density of magnetic

field

e) Alternating current, resistors,

inductors and capacitors in

AC-circuits, voltage and current

(parallel and series) resonances

Simple AC-circuits, time constants,

final formulae for parameters of

concrete resonance circuits are not

required

8. Electromagnetic waves

a) Oscillatory circuit, frequency of

oscillations, generation by feedback

and resonance

b) Wave optics, diffraction from one

and two slits, diffraction grating,re-

solving power of a grating, Bragg

reflection,

c) Dispersion and diffraction spec-

tra, line spectra of gases

d) Electromagnetic waves as trans-

verse waves, polarization by reflec-

tion, polarizers

Superposition of polarized waves

172

e) Resolving power of imaging

systems

f) Black body, Stefan-Boltzmanns law Planck’s formula is not required

9. Quantum Physics

a) Photoelectric effect, energy and

impulse of the photon

Einstein’s formula is required

b) De Broglie wavelength,

Heisenberg’s uncertainty principle

10. Relativity

a) Principle of relativity, addition of

velocities, relativistic Doppler effect

b) Relativistic equation of motion,

momentum, energy, relation

between energy and mass, conser-

vation of energy and momentum

11. Matter

a) Simple applications of the Bragg

equation

b) Energy levels of atoms and

molecules (qualitatively), emission,

absorption, spectrum of hydrogen

like atoms

c) Energy levels of nuclei (quali-

tatively), alpha-, beta- and gam-

ma-decays, absorption of radiation,

halflife and exponential decay,

components of nuclei, mass defect,

nuclear reactions

173

B. Practical Part

The Theoretical Part of the Syllabus provides the basis for all the experimental

problems. The experimental problems given in the experimental contest should

contain measurements.

Additional requirements:

Candidates must be aware that instruments affect measurements.

Knowledge of the most common experimental techniques for measuring

physical quantities mentioned in Part A.

Knowledge of commonly used simple laboratory instruments and devices

such as calipers, thermometers, simple volt-, ohm- and ammeters, poten-

tiometers, diodes, transistors, simple optical devices and so on.

Ability to use, with the help of proper instruction, some sophisticated

instruments and devices such as double-beam oscilloscope, counter, ratem-

eter, signal and function generators, analog-to-digital converter connected

to a computer, amplifier, integrator, differentiator, power supply, univer-

sal (analog and digital) volt-, ohm- and ammeters.

Proper identification of error sources and estimation of their influence on

the final result(s).

Absolute and relative errors, accuracy of measuring instruments, error of a

single measurement, error of a series of measurements, error of a quantity

given as a function of measured quantities.

Transformation of a dependence to the linear form by appropriate choice

of variables and fitting a straight line to experimental points.

Proper use of the graph paper with different scales (for example polar and

logarithmic papers).

Correct rounding off and expressing the final result(s) and error(s) with

correct number of significant digits.

Standard knowledge of safety in laboratory work. (Nevertheless, if the

experimental set-up contains any safety hazards the appropriate warnings

should be included into the text of the problem.)

174

175

AppendicesTartu – The World Capital of PhysicsPrOgram

20th of July10:00 – 16:30 Career day of the universities

12:00 – 12:30 Declaration of Tartu as the World Capital of Physics

at the Town Hall Square

12:30 – 16:30 Workshops

DETAILED TIMESCHEDULE

10:00 – 16:30 Career day presentations by universities in the conference hall

of the University of Tartu history museum Address: Lossi 25

10:15 - 11:00 University of Tartu

11:10 - 11:40 University of Oxford

13:00 - 14:00 National University of Singapore

14:10 - 15:10 Massachusetts Institute of Technology

15:20 - 16:20 Tallinn University of Technology

12:00 – 12:30 Declaration of Tartu as the World Capital of Physics

at the Town Hall Square

WORKSHOPS

12:30 – 16:30 Workshops at the Cathedral ruins and the history museum

of the University of Tartu Address: Lossi 25

CONTINUOUS WORKSHOPS

200-year-old physics office

Physics equipment belonging to the physics office of the University of Tartu from

the 19th century will be introduced.

176

Making of magic disks

Make a popular optical toy from the 19th century.

Colours to the T-shirt

Design a unique T-shirt for yourself by using a chromatographic method.

Micro and macro worlds in nature

Study various animals and stones with a microscope.

Finding the thief

A fun game introducing genetics.

How special are you?

A game designed for Tallinn TV Tower based on the database of the Estonian

Genome Centre where people can test how similar or different they are compared

to other Estonians.

Circles of the woods

One tree can tell endless stories if you get to know it up close.

Making of tar

How to make tar from a block of wood?

Timber preservation workshop

People have used timber for a very long time and old buildings have preserved well.

But nowadays logs or floors tend to break up in a few years. We will find out why.

Various ways of inoculating and grafting fruit cultures

Organic farming: growth, cultures, breeds, foods

Measuring a person

Find out your height, weight, and body mass index.

Let’s go inside a head.

We will go inside a head and study how the brain works: eyesight, reading, deci-

sion-making. Will be possible to make a back-up brain and solve a brain puzzle.

Feel the dimensions

Ethnographic experiments: units of length, weight and volume in modern times.

Inventions from the University of Tartu

Inventions by university scientists that have been or will be implemented in

everyday life: glass that changes transparency, a new photosensitive material,

instrument for diagnosing glaucoma, ME-3 bacterium, myometer, robots.

Interesting facts from the botanic kingdom

Finding close or distant relatives in the botanic kingdom: which trunks and cones

match.

177

Workshops with definite start and end times

12:30 - 13:15 Glance at the past of the University of Tartu

Guided tour in the history museum13:30 - 14:15

14:30 - 15:15

15:30 - 16:15

Workshops in the Tartu observatory and in its proximity Address: Lossi 40

CONTINUOUS WORKSHOPS

Science bus

Various experiments and workshops related to physics, chemistry and biology.

Workshops with definite start and end times

12:30 - 13:00 Planetarium show

During the planetarium show “Night sky”, the planets,

their movement during the year will be shown together

with more interesting constellations from both the

northern and southern hemisphere.

13:15 - 13:45

14:00 - 14:30

14:45 - 15:15 Planetarium StarLab

15:30 - 16:00 StarLab planetarium with an inflatable tent as a dome

will be used for the planetarium show.

12:30-13:15 Guided tours in the planetarium

Observation of the sun and a workshop for making a

sun clock

Workshop for making a quadrant

Weather measurement

13:15-14:00

14:00-14:45

14:45-15:30

15:30-16:15

178

yEar OF scIEncE In EsTOnIa and TarTu — ThE wOrLd caPITaL OF PhysIcsViire Sepp

IPhO 2012 Steering Committee Secretary

Estonia hosting the IPhO 2012 became the incentive for proclaiming the period

between September 2011 and September 2012 a Year of Science in our country.

The objective was to raise youth interest in exact, natural and technical sciences

and towards career choices related to these subject areas. Under the auspices of

the Year of Science, hundreds of events have taken place in schools, towns and

counties organized by the partners of the Year of Science – schools, universities,

associations, societies, etc.

One of the main events took place on the 20th of July when Tartu was proclaimed

the World Capital of Physics within the IPhO 2012. The objectives of proclaiming

the host city as the World Capital of Physics were to highlight the importance of

natural and exact sciences and valuation of education, creativity and purposeful

self-development. Numerous public events took place during that day in Tartu

– the first “World Capital of Physics”. At the same time, the annual Hanseatic

days also commenced in Tartu. At 12 o’clock, the ceremony declaring Tartu the

World Capital of Physics took place in the Town Hall Square. The IPhO delega-

tion handed over the Declaration announcing Tartu the Capital of Physics to the

Mayor of Tartu. Subsequently, everyone was welcomed to the historic Toome Hill

where the Science Town was set up during the Hanseatic days. The Science Town

included tens of hands-on workshops, shows by the science bus of the Estonian

Physical Society, and the 2007 winner of the EU prize for Science Communication.

In addition, there was a possibility to visit the 200-year-old observatory which was

affirmed in the world science history by its director F.G.F. Struve who was the first

to measure the distance between a star and Earth in 1835. Tartu Observatory has

belonged to the UNESCO World Heritage since 2005. In the history museum of the

University of Tartu it was possible to get acquainted with the university’s history

and contemporary situation. Also, the University Fair (so called “a career day”)

featured presentations on enrolment and study opportunities by Massachusetts

Institute of Technology, University of Oxford, National University of Singapore,

Tallinn University of Technology and University of Tartu. The honorary guest of

the IPhO, 1996 Nobel Prize winner Sir Harold Kroto, gave an academic lecture at

179

the Vanemuise Concert Hall in the evening and the day was concluded with a

reception by the Mayor of Tartu.

Since the IPhO is a relatively closed event, it is essential to acknowledge the

bridge between the IPhO and the host city, make the IPhO more visible in the

urban space, strengthen international reputation, and advance potential inter-

national relations of the hosting city. The idea that the host city could bear that

title was beforehand introduced to the IPhO president Dr Hans Jordens who found

the initiative worthwhile. The Statute and the Declaration of the World Capital

of Physics were prepared by organizers of IPhO2012 and were presented to team

leaders of participating countries in order to make this event traditional in future

IPhOs. These documents are printed herewith.

sTaTuTE OF ThE IPhO wOrLd caPITaL OF PhysIcs In order to highlight cities that host the International Physics Olympiad (IPhO)

and to intensify the impact of the IPhO in extending international contacts of

these cities, IPhO2012 Estonia launches the tradition of nominating the hosting

city of the IPhO, for one day of the Olympiad, as the World Capital of Physics.

The first IPhO World Capital of Physics is Tartu (Estonia), which was the host-

ing city of the 43rd International Physics Olympiad in 2012.

The statute establishes the aims of nomination and the principles of co-op-

erative activities of the IPhO secretariat (President), Organizing Committee and

hosting city and regalia of the World Capital of Physics.

The aims of nomination are to:

Attract public and primarily youth attention to the significance of the

natural and exact sciences, promote education, creativity and purposeful

self-actualization

Acknowledge the venue of the IPhO

Extend the international reputation of the hosting city whilst promoting

its potential international relations

The nomination of the IPhO hosting city as the World Capital of Physics takes

place in a public urban space.

Regalia of the World Capital of Physics:

Parchment (Declaration of nomination)

Appendix (good wishes from the participants of the IPhO to the hosting city)

180

Symbol of the World Capital of Physics (the Chronicle Book) launched by

Tartu (Estonia, IPhO2012)

Local Organizing Committee of the Olympiad:

Arranges and conducts the World Capital of Physics nomination ceremony

in co-operation with the hosting city and IPhO secretariat (President)

Organizes Parchment/Declaration compilation and its delivery to the

Mayor of the hosting city

Arranges perpetuation of the hosting city in the Chronicle Book of the

World Capital of Physics

Hosting city:

Guarantees the presence of the Mayor or his/her representative in the nom-

ination ceremony of the World Capital of Physics

Accepts the Parchment (Declaration)

Delivers the Chronicle Book of the World Capital of Physics to the IPhO

President (or his/her representative)

IPhO secretariat/President:

Follows the continuity of the tradition of the World Capital of Physics

Delivers the Parchment to the hosting city

Accepts the Symbol (the Chronicle Book) and delivers it to the organiz-

ers of the next IPhO (at the IB meeting or at the closing ceremony of the

Olympiad)

dEcLaraTIOn Date and place

We, the participants of the …. (number) International Physics Olympiad:

Appreciating the beauty and power of science, that can take us far beyond the

realm of everyday life - from microworlds to the distant cosmos;

Understanding the importance of physics as one of the cornerstones of modern

technological society that has given birth to numerous inventions, which are of

benefit to mankind;

Acknowledging science as an essential collective effort, uniting peoples and

181Photos by Henry Teigar, Karl Veskus and Siim Pille

Tartu town hall square

geniuses in red

President of IPhO hans Jordens declares Tartu as the world capital of Physics

Estonian student kaur aare saar reads the declaration of Tartu as the world capital of Physics choir named E sTuudio presenting Estonian folk music

182

countries in solving the crucial problems that society faces, in the name of peace

and common progress;

Recognizing the everlasting curiosity of people of all ages, never stopping

asking questions and searching for answers, as the main driving force of science;

Committing to the hard work one has to do in mastering science in order to

exploit its potentialities to the fullest extent;

Being indebted to our teachers for the knowledge and excitement they provided

us with, together with the responsibility to carry this on to the next generation;

We declare the City of …….(name of city ) on the occasion of …..(number)

International Olympiad of Physics to be the World Capital of Physics for

…….(date).

We challenge everyone to discover the power and beauty of physics in order to

contribute to the progress and welfare of mankind and to share our love of science!

Signature Signature

IPhO President IPhO Secretary

183

CircularsFIrsT cIrcuLar

Dear Colleagues,The Republic of Estonia has the honour of being the host of the 43rd International

Physics Olympiad (IPhO 2012), which will be held from the 15th to the 24th of July 2012.

The Olympiad will be held in two locations, in the capital Tallinn (team lead-

ers), and in the oldest university town in Estonia, Tartu (students).

The IPhO2012 is organized under the auspices of the Estonian Ministry of

Education and Research by the Estonian Information Foundation (main organizing

institution) jointly with the University of Tartu, Tallinn University of Technology,

the National Institute of Chemical Physics and Biophysics, the Estonian Academy

of Sciences, Archimedes Foundation, and the Estonian Society of Physics. The

Steering Committee, comprizing the representatives of these institutions, governs

the activities of the Academic and Organizing Committees of the IPhO 2012.

InvitationOn behalf of the organizers of the 43rd International Physics Olympiad, it is our

pleasure and honor to invite and welcome your country to send a delegation to take

part in the IPhO 2012 in Estonia. The International Physics Olympiad is the most

prestigious international physics competition for individual secondary school stu-

dents aimed to stimulate young people’s interest in physics, to propagate natural

and exact sciences amongst school students, and to promote science education

throughout the world by means of international contacts.

Requirements for the participating teamsAccording to the IPhO Statutes, each team will comprise up to five students and two

leaders. The contestants shall be students of general or technical secondary schools,

i.e. schools which cannot be considered technical colleges. Students who have fin-

ished their school graduating examinations in the year of the competition can be

members of the team as long as they have not commenced their university studies.

The age of the students should not exceed twenty years, as of the 30th of June 2012.

The team leaders should be physics teachers or specialists who can speak English.

184

Fees and ExpensesThe IPhO2012 Organizing Committee will provide meals, accommodation and

transportation free of charge to all the participants during the duration of the

Olympiad period. However, travel expenses of each participating country will be

the liability of each participating country. Arrangements for extra accommodation

before and/or after the official period will be provided, though the related expenses

are to be covered by the participating country.

As an established custom in the International Physics Olympiad, each par-

ticipating country is asked to contribute a voluntary fee of € 3,500.Observers and

visitors are welcome, but they are expected to cover their own expenses. Observers

may attend all Olympiad meetings, including the meetings of the International

Board. However, they may not vote or take part in the discussions. Visitors do not

attend the meetings of the International Board. The fee of each observer is € 1,200

and for visitor € 1,100.

Transportation Transportation from Tallinn International Airport and Port of Tallinn to hotels

and vice versa will be arranged by the local Organizing Committee at no addi-

tional cost.

Information and contactsFor the detailed information of IPhO 2012 and the most recent update, please visit

our official homepage www.ipho2012.ee

You may contact us via e-mail: [email protected]

Phone number: +372 628 5800

Address: 43rd IPhO Organizing Committee

Estonian Information Technology Foundation

Raja 4C, 12616 Tallinn

Estonia

185

Preliminary confirmation of participationAs the first step, please complete the attached form: PRELIMINARY CONFIRMATION

FORM OF IPhO 2012 PARTICIPATION.

This information is needed for sending official letter of invitation to participate

in IPhO2012 to your Minister of Education or the equivalent official in your country.

In addition, this information is necessary for communication with the contact

person in your country.

The form should be filled in and sent as an e-mail attachment to the 43rd IPhO

Organizing Committee ([email protected]) no later than: 20th of December 2011.

Estonian people and the organizers of the Olympiad are looking forward to

meeting young physicists and their supervisors from all over the world, and to

introducing them to the innovative country, which values education, has a rich

cultural heritage and beautiful land.

Welcome to IPhO2012 in Estonia!

Sincerely,

Prof Jaak Kikas Dr Viire Sepp

Head of the Academic Committee Academic secretary of the

Steering Committee

186

Second CircularDear Colleagues,It is our pleasure to inform you about the progress regarding the 43rd International

Physics Olympiad in Tallinn/Tartu, Estonia during 15th – 24th of July 2012.

Confirmation of participation to this dateUntil the present date, the following 86 teams have confirmed their participation

in the 43rd IPhO:

Albania

Armenia

Australia

Austria

Azerbaijan

Bangladesh

Belarus

Belgium

Bolivia

Bosnia & Herzegovina

Brazil

Bulgaria

Cambodia

Canada

People`s Republic of China

Colombia

Croatia

Cuba

Cyprus

Czech Republic

Denmark

Ecuador

El Salvador

Estonia

Finland

France

Georgia

Germany

Ghana

Greece

Hong-Kong

Hungary

Iceland

India

Indonesia

Islamic Republic of Iran

Ireland

Israel

Italy

Japan

Jordan

Kazakhstan

Republic of Korea

Kuwait

Kyrgyzstan

Lao PDR

Latvia

Liechtenstein

Lithuania

Macao, China

187

Macedonia

Malaysia

Mexico

Moldova

Montenegro

Netherlands

Nigeria

Norway

Pakistan

Philippines

Poland

Portugal

Puerto Rico

Qatar

Romania

Russia

Saudi Arabia

Serbia

Republic of Singapore

Slovakia

Slovenia

South Africa

Spain

Sri Lanka

Suriname

Sweden

Switzerland

Syria

Taiwan

Tajikistan

Thailand

Ukraine

United Kingdom

United States of America

Uzbekistan

Vietnam

Official invitationDuring on beginning of February official invitation letters were sent to the

Ministers of Education or equivalent officials of the teams that had already returned

the Confirmation Form to the IPhO2012 Organizing Committee. Pdf copies of the

original invitation letters were also sent to the ministries and equivalent officials.

If you have not received these letters, please contact us ([email protected]).

Online registrationThe online registration will be available at the IPhO 2012 website: http://www.

ipho2012.ee from February 20, 2012. The contact person of each country will receive an

auto-generated e-mail message with USERNAME, PASSWORD and instructions.

The contact person is requested to register all participants in his/her respective

team (students, leaders, visitors and observers).

188

VisaComplete information on visa application and issuances is available at the website

of Estonian Ministry of Foreign Affairs: http://www.vm.ee/?q=en/node/53

Who does not need a visa to visit Estonia?

Nationals of the member states of the European Union (EU) and the European

Economic Area (EEA) and any third-country national who is a holder of a residence

permit of a Schengen state do not need a visa to enter Estonia.

EU and EEA member states are Austria, Belgium, Bulgaria, Cyprus, Czech

Republic, Denmark, Estonia, Finland, France, Germany, Greece, Hungary,

Iceland, Ireland, Italy, Latvia, Liechtenstein, Lithuania, Luxembourg, Malta,

Norway, Poland, Portugal, Romania, Slovakia, Slovenia, Spain, Sweden, The

Netherlands, United Kingdom and Switzerland.

Schengen States are Austria, Belgium, Czech Republic, Denmark, Estonia,

Finland, France, Germany, Greece, Hungary, Iceland, Italy, Latvia, Liechtenstein,

Lithuania, Luxembourg, Malta, The Netherlands, Norway, Poland, Portugal,

Slovakia, Slovenia, Spain, Sweden, Switzerland.

! Although documents are not checked when one crosses an internal Schengen

border, it is still necessary to carry a passport or ID card. Authorities (police,

immigration officials) in Schengen states do have the right to check identifying

documents, if necessary.

The holders of passports of the following countries do not need a visa to enter a

Schengen region (incl Estonia) for stays of no more than 90 days in a 6 month period:

Albania*, Andorra, Antigua and Barbuda, Argentina, Australia, Bahamas,

Barbados, Bosnia and Herzegovina*, Brazil, Brunei, Canada, Chile, Costa

Rica, Croatia, El Salvador, Guatemala, Holy See, Honduras, Hong Kong Special

Administrative Region, Israel, Japan, Macao Special Administrative Region,

Macedonia*, Malaysia, Mauritius, Mexico, Monaco, Montenegro*, New Zealand,

Nicaragua, Panama, Paraguay, San Marino, Serbia*, Seychelles, Singapore, South

Korea, St Kitts-Nevis, Taiwan**, United States of America, Uruguay, Vatican City,

Venezuela.

* Only for the biometrical passports holders

** Passports issued by Taiwan which include an identity card number

Countries requiring visa for entry to Estonia, please start the application pro-

cess early. It will take approximately 2–3 weeks for the applications to be processed.

189

Please take into consideration that it is impossible to buy the Estonian visa

at the order.

If you have questions concerning visas or need additional documents, please

contact the Organizing Committee ([email protected]) .

Fees and paymentAll participating teams (5 students + 2 leaders) are encouraged to contribute a

voluntary fee of EUR 3,500. The fee will cover: meals, accommodation and official

social events during the official period of the Olympiad. Observers and visitors are

welcome, and they are expected to cover their expenses, visitors EUR 1,100 and

observers EUR 1,200. Observers may attend all Olympiad meetings, including the

meetings of the International Board. However, they may not vote or take part in

discussions. Visitors do not attend the meetings of the International Board.

If the participating team from your country is smaller than 5 people, then the

participation fee is 500 euros per person and the total amount of the invoice will

depend on the number of people.

We kindly ask you to pay the invoice before arrival to Estonia, within 10 days

upon receiving it by e-mail. In case you wish to pay the fee on site, please inform

the Organizing Committee well in advance ([email protected])

TransportationIPhO 2012 will start and end in Tallinn. Please make all your travel arrangements

with arrival to and departure from Tallinn Airport (http://www.tallinn-airport.ee/eng)

For teams travelling via Helsinki (Finland), sea transportation from

Helsinki to Tallinn could be the alternative (http://www.portofhelsinki.fi/passengers/

departure_times_and:terminals).

Transportation from the airport and the seaport to the hotels and vice versa

will be provided by the Organizing Committee at no additional cost.

AccommodationIn Tallinn students will be staying at the Sokos Hotel Viru. During the student’s

programme in Tartu from the 16th to the 21st of July contestants will be staying at

the following hotels: Hotel Dorpat, Tartu Hotel, Hotel Pallas.Leaders, visitors

and observers will be accommodated in Radisson Blu Hotel Olümpia in Tallinn.

190

All hotels are situated in the very heart of Tallinn and Tartu. Two persons will

share a room.

If leaders, visitors and observers would like to stay in a single room, please let

us know when you register for 43rd IPhO; there is an additional charge of EUR 400

per room for 9 nights during the official period of the Olympiad.

If you plan to arrive in Estonia earlier than July 15th 2012 or leave Tallinn later

than July 24th 2012, please let us know well in advance (by noting it in the online

registration and by sending us an e-mail, indicating the number of persons and

type of the room) so that we can arrange your accommodation well in advance,

because July is the tourism high season in Estonia and the hotels need to be booked

as early as possible. The extra charge per night for Tallinn hotel double rooms are

EUR 70 for the students’ hotel and EUR 75 for the leaders’ hotel.

43rd IPhO Website and ContactsThe circulars and all information regarding the 43rd IPhO can be found at our web-

site at http://www.ipho2012.ee which will be updated regularly.

Organizers of the 43rd IPhO are looking forward to meeting you in Estonia!

Prof Jaak KIkas Dr Viire Sepp

Head of the Academic Committee Academic Secretary of the

Steering Committee

191

Third CircularDear Colleagues,We are only 20 days before the beginning of the 43rd International Physics Olympiad

in Estonia. It is our pleasure to inform you about the progress regarding the prepa-

ration and necessary information for your teams.

RegistrationThe online registration is about to complete, the following 86 teams have con-

firmed their participation in the 43rd IPhO:

1. Albania

2. Armenia

3. Azerbaijan

4. Australia

5. Austria

6. Bangladesh

7. Belarus

8. Belgium

9. Bolivia

10. Bosnia & Herzegovina

11. Brazil

12. Bulgaria

13. Canada

14. Colombia

15. Croatia

16. Czech Republic

17. Cuba

18. Cyprus

19. Denmark

20. Ecuador

21. El Salvador

22. Estonia

23. Finland

24. France

25. Georgia

26. Ghana

27. Germany

28. Greece

29. Hong Kong

30. Hungary

31. Iceland

32. India

33. Indonesia

34. Ireland

35. Islamic Republic of Iran

36. Israel

37. Italy

38. Japan

39. Kazakhstan

40. Kuwait

41. Kyrgyzstan

42. Latvia

43. Liechtenstein

44. Lithuania

45. Macao, China

46. Macedonia

47. Malaysia

48. Mexico

192

49. Moldova

50. Mongolia

51. Montenegro

52. Nepal

53. Netherlands

54. Nigeria

55. Norway

56. Pakistan

57. People`s Republic of China

58. Poland

59. Portugal

60. Puerto Rico

61. Qatar

62. Republic of Korea

63. Republic of Singapore

64. Romania

65. Russia

66. Saudi Arabia

67. Serbia

68. Slovakia

69. Slovenia

70. South Africa

71. Spain

72. Sri Lanka

73. Suriname

74. Sweden

75. Switzerland

76. Syria Arab Republic

77. Taiwan

78. Tajikistan

79. Thailand

80. Turkey

81. Turkmenistan

82. Ukraine

83. United Kingdom

84. United States of America

85. Uzbekistan

86. Vietnam

Fees and paymentAll participating teams (5 students + 2 leaders) are encouraged to contribute with a

voluntary fee of EUR 3,500. The fee will cover: meals, accommodation and official

social events during the official period of the Olympiad. Observers and visitors are

welcome, and they are expected to cover their expenses, visitors EUR 1,100 and

observers EUR 1,200.

Observers may attend all Olympiad meetings, including the meetings of

the International Board. However they may not vote or take part in discussions.

Visitors do not attend the meetings of the International Board.

If the participating team from your country is smaller than 5 people, then the

participation fee is 500 euros per person and the total amount of the invoice will

depend on the number of people.

We kindly ask you to pay the invoice before arrival to Estonia, within 10 days

upon receiving it by e-mail. In case you wish to pay the fee on site, please inform

the Organizing Committee well in advance ([email protected])

193

TransportationIPhO 2012 will start and end in Tallinn. Please make all your travel arrangements

with arrival and departure at Tallinn Airport (http://www.tallinn-airport.ee/eng)

For teams travelling via Helsinki (Finland), sea transportation from

Helsinki to Tallinn could be the alternative (http://www.portofhelsinki.fi/passengers/

departure_times_and:terminals).

Transportation from the airport and the seaport to the hotels and vice versa

will be provided by the Organizing Committee at no additional cost.

Please insert your delegation travel plans into IPhO 2012 registration environ-

ment by June 29th.

AccommodationIn Tallinn student will stay at Sokos Hotel Viru. In Tartu student will stay at the

following hotels: Hotel Dorpat, Tartu Hotel, Hotel Pallas and Hotel London.

Leaders, visitors and observers will be accommodated in Radisson Blu Hotel

Olümpia. All hotels are situated in the very heart of Tallinn and Tartu. Two persons

will share a room.

If leaders, visitors and observers would like to stay in single rooms, please let

us know when you register for the 43rd IPhO; there is an additional charge of EUR

400 for 9 nights during the official period of the Olympiad.

If you plan to arrive in Estonia earlier than July 15th, 2012 or leave Tallinn later

than July 24th, 2012, please let us know well in advance (by noting it in the online

registration and by sending us an e-mail, indicating the number of persons and

type of the room) so that we can arrange your accommodation well in advance,

because July is the tourism high season in Estonia and the hotels need to be booked

as early as possible. The extra charge per night and per person in twin rooms

in Tallinn is EUR 35 for the students’ hotel and EUR 37.50 for the leaders’ hotel.

Breakfast is included.

Between the 16th and the 19th of July, there will be no phones in the student’s

hotel rooms. Because of that it is recommended that the students take along their

own alarm clocks.

Computing facilities for team leadersComputing facilities, including laptops and printers, will be available in the

International Board Meeting Room.

194

You can bring your own laptop/notebook. Standard power supply in Estonia

is 220V/50 Hz.

CalculatorsNotice regarding the calculators which can be brought into the examination room.

According to the Statutes of the IPhO, the only tools which can be brought into

the examination room by competitors are drawing instruments and a non-pro-

grammable pocket calculator. [§5] In order to give a clear guideline about which

calculators are allowed, the Academic Committee of the IPhO 2012 provides the

following definition.

A non-programmable calculator is not allowed to have:

more than 9 memory slots for recording intermediate data

the possibility of entering user-defined functions for repetitive calculations

pre-stored physical constants

the possibility of plotting graphs

The calculator needs to be a commercial product, complete with an English

manual (printed or downloadable); it is allowed to have standard mathemati-

cal functions (trigonometric, hyperbolic etc.) and standard statistical functions

(mean, standard deviation, etc.).

All the calculators of the contestants will be checked prior to the competition

and the allowed ones will be labeled. A limited amount of replacement calculators

will be available from the organizers.

Clothing and weather During the opening and closing ceremonies and the farewell party it is required

that the boys wear long trousers (please do not wear shorts unless these are part

of the national costume).

Please take comfortable clothing and shoes for the excursions (there will be

some walking and climbing).

There will be a football tournament for the students on 22nd of July. Participating

students need to have sports clothing with them.

Temperatures in the summer vary from +15 to +25 Celsius. The weather is

changeable; shiny and rainy days are intermittent so it is advisable to prepare for

both. Please bring your umbrella or raincoat. You can find more information about

Estonian weather here: http://www.weather.ee/ ; http://www.emhi.ee/index.php?nlan=eng.

195

The Chemistry Olympiad in the United States of AmericaSome IPhO students are taking part in the Chemistry Olympiad in the USA.

Countries that need separate transportation for their student(s) to the airport,

please let us know by writing to [email protected]

Information leak during the competitionThe organizers of IPhO 2012 have taken some measures to minimize the possibility

that information about the problems will leak. We hope for your understanding

and count on your co-operation to execute the following measures:

The students are not allowed to possess electronic equipment that has the

possibility to communicate with the outer world until after the last test

(July 19th). The equipment has to be delivered to the organization of IPhO

2012 in Tallinn and will be stored in special boxes.

From July 16th at 08:00 AM till July 19th at 08:00 PM the students are not

allowed to use internet facilities or phones at the hotel, nor are they allowed

to use such facilities outside the hotel.

All communication between the students and their leaders, parents etc.

during the above indicated period has to go via the guides.

Leaders and observers are not allowed to communicate with the outer world

from the start of the International Board meetings where the discussions of

the problems take place up to the beginning of the respective examination

(in the case of the experimental round, up to the beginning of the second

shift of the experimental competition). In case of emergency the commu-

nication needs to go via the IPhO 2012 office.

All violations against what is stated above will be regarded as cheating as mentioned

in the IPhO Regulations §7 and appropriate measures will be taken accordingly.

Here we quote from the Regulations Associated with the Statutes of the

International Physics Olympiads:

If it is found that leaders, observers or students from a country have been in col-

lusion to cheat in one of the International Olympiad examinations, the students

concerned should be disqualified from that Olympiad. In addition, the leaders,

observers and students involved should not be allowed to return to any future

Olympiad. Appropriate decisions are taken by the International Board.

196

43rd IPhO Website and ContactsThe circulars and all information regarding the 43rd IPhO can be found on our

website http://www.ipho2012.ee, which will be updated regularly.

The Academic and Organizing Committee of the 43rd International Physics

Olympiad looks forward to welcoming all delegates in Estonia.

Yours sincerely,

Prof Jaak Kikas Dr Viire Sepp

Head of the Academic Committee Academic Secretary of the

Steering Committee

197

NewsletterIPhO 2012 newsletter was called “Physics is in The Air” and aimed to give a daily

update about the Olympiad. The newsletter consisted of articles, interviews, photo

galleries and so on. There were 8 issues on paper handed to all participants each

morning and 10 online numbers published on website paralelly. Online numbers

always concisted of more interviews and details whereas the paper version had

its limits on characters.

Get all the articles and pictures on the newsletter website: www.ipho2012.ee/newsletter

198

199

The process of organising the IPhO 2012 started 3 years ago. The task was highly

prioritized by the Estonian Ministry of Education and Research. Eight institutions

formed the steering committee for organisational aspects – the Estonian Ministry

of Education and Research, Archimedes Foundation, the Estonian Academy of

Sciences, the Estonian Physical Society, the National Institute of Chemical and

Biophysics, the Tallinn University of Technology, the University of Tartu and the

Estonian Information Technology Foundation. The Estonian Information Technology

Foundation was responsible for the organisational side of the IPhO 2012. Eight

people comprised the Organizing Committee: Ene Koitla - head of the Organizing

Committee, Marily Hendrikson - IPhO 2012 project manager, Annika Vihul - head of

accounting, transportation and accommodation, Malle Tragon - head of events and

catering, Kerli Kusnets - head of media, Eneli Sutt - head of information technology,

Julia Šmakova and Anna Gureeva - heads of group leaders.

623 people from 80 countries participated in the IPhO 2012, 378 of these were

contestants. 177 volunteers contributed to the success of the event.

The biggest challenge facing the IPhO 2012 Organising Committee was holding

the event in two cities – Tallinn and Tartu. Issues such as accommodation, trans-

portation and having examinations posed the biggest problems. Estonia is a small

country where such a number of participants can easily create logistical liabilities.

There was also plenty of planning concerning the events and activities during free

time. The IPhO 2012 featured several innovative IT solutions assisting in organizing

the event. All teams registered by using the special registration system. Due to the

distance between the two cities, Tallinn and Tartu, there were limitations to do

some operations in a usual way. To get the translated problem sheets to the students’

desk, the leaders had to upload ready-made files to the special IT system in Tallinn

and the sheets were printed out in Tartu. The same system was used to transfer the

solutions back to the leaders and markers. Each leader and marker could decide if

he/she wanted to print them out or download the solutions. It was made possible for

Afterword

200

the leaders and markers to use the computer whenever

they wanted. All marks were inserted in the web-based

evaluation system. The students’ meetings with leaders

were held using the Skype call.

We are very honoured that we had the chance to host

the 2012 International Physics Olympiad contestants,

leaders, observers and visitors from all over the world.

We hope that the participants have enjoyed the physics

contest and we managed to demonstrate Estonian cul-

ture and variety of nature.

Organizing Committee

IPhO 2012

Photo by Merily Salura

201

202

203

204

205

Documents

12+11+05+Kogumik+FINAL