Proceedings of the 43rd International Physics Olympiad

Physics is Love

Proceedings of the 43rd International Physics Olympiad 15th – 24th of July 2012 • Tallinn and Tartu, Estonia

Estonian Ministry of Education and Research The Estonian Information Technology Foundation Designed by loremipsum.ee Edited by OĂœ 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.

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Contents Speeches

13

Opening Ceremony

13

Closing Ceremony

19

People

25

Participants

25

Organizers

54

Programs

63

Students

63

Leaders and Observers

67

Problems and solutions Theoretical Competition Experimental Competition

Results

73 73 107

131

Gold Medalists

131

Silver Medalists

133

Bronze Medalists

135

Honorable Mentions

137

Special Prizes

139

Statistics of the competition results

140

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International board

151

Minutes

151

Statutes

157

Syllabus

168

Appendices Tartu – The World Capital of Physics

175 175

Circulars

183

Newsletter

197

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Foreword Estonia 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 delegations 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 traditions originate from an era which was very different from the current one; perhaps the most important difference lies in the communication and information technologies. 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

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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 universities 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) finding 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 developed, 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:

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“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 thinking 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 problems 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 Kalda Academic Committee of the 43rd IPhO

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Sir Harold Kroto 1996 Nobel Prize winner, the honorary guest of the IPhO2012 “The job of the scientists 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

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Speeches Opening Ceremony Ene Ergma President 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

Photo by Henry Teigar

Minister of Education of the Republic of Estonia, submitted 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 successfully 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 academic 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

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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 conditions 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!

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Hans Jordens The 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

Photo by Siim Pille

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 privilege 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 difficult 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

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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 southeast. 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, organizational 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.

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Jaak Kikas Chairman 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

Photo by Siim Pille

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

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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 forthcoming competition.

Enjoy the Olympiad!

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Closing Ceremony Jaak Aaviksoo Minister of Education and Research Dear friends of physics, I would like to start with thanking 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 contribution to a better world.

I would also like to thank you for making Estonia,

Photo by Karl Veskus

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 better 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!

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Hans Jordens The 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 Photo by Karl Veskus

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 competition 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 competition. 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.

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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.

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Niels Christian Hartling President 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 Photo by Merily Salura

100th anniversary of Niels Bohr’s theory of the hydrogen 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 students 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 battle 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,

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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.

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People Participants International Physics Olympiad Committee President

Hans Jordens

Executive Secretary

Ming Juey Lin

Albania Student

Geri Emiri

Arled Papa

Leader

Antoneta Deda

Armenia Student

Vardan Avetisyan

Aram Mkrtchyan

Virab Gevorgyan

Aleksandr Petrosyan

Razmik Hovhannisyan

Leader

Gagik Grigoryan

Bilor Kurghinyan

Azerbaijan Student

Haji Piriyev

Valeh Farzaliyev

Ramazan Ramazanov

Farid Mammadov

Ahmad Mehribanli

Leader

Mirzali Murguzov

Rana Mammadova

Observer

Rashadat Gadmaliyev

25

Australia Student

Eric Huang

Jonathan Lay

Nicholas Salmon

Siobhan Tobin

Christopher Whittle

Leader

Matthew Verdon

Bonnie Zhang

Observer

Alix Verdon

Austria Student

Oliver Edtmair

Christoph Weis

Tobias Karg

Maximilian Ruep

Martin Stadler

Leader

Helmuth Mayr

Engelbert Stuetz

Bangladesh Student

Kinjol Barua

Wasif Ahmed

Shinjini Saha

Shovon Biswas

Ahmed Maksud

Leader

Fayez Ahmed Jahangir Masud

M.Arshad Momen

26

Belarus Student

Ihar Lobach

Albert Samoilenka

Aliaksey Khatskevich

Aliaksandr Yankouski

Vadzim Reut

Leader

Anatoli Slabadzianiuk

Anton Mishchuk

Observer

Leonid Markovich

Belgium Student

Romain Falla

Leandro Salemi

Basile Rosen

Basile Vermassen

Mathias Stichelbaut

Leader

Bernadette Hendrickx

Philippe Leonard

Observer

Sophie Houard

Matthieu Dontaine

Bolivia Student

Cesar Tapia

Leader

Veronica Subieta

Bosnia and Herzegovina Student

Selver Pepic

Amer Ajanovic

Nudzeim Selimovic

Sladjan Veselinovic

Stefan Gvozdenovic

Leader

Rajfa Musemic

Rodoljub Bavrlic

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Brazil Student

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

Bulgaria Student

Katerina Naydenova

Yordan Yordanov

Veselin Karadzhov

Kaloyan Darmonev

Konstantin Gundev

Leader

Victor Ivanov

Miroslav Abrashev

Canada Student

Sepehr Ebadi

Yun Jia (Melody) Guan

Tristan Downing

Henry Honglei Wu

Simon Blouin

Leader

Andrzej Kotlicki

Jean-Franรงois Caron

Visitor

Chantal Haussmann

28

Colombia Student

Daniel Eduardo Fajardo Fajardo

Andres Rios Tascon

Andres Zorrilla Vaca

Leader

Fernando Vega Salamanca

Eduardo Zalamea Godoy

Croatia Student

Samuel Bosch

Bruno Buljan

Luka Skoric

Karlo Sepetanc

Grgur Simunic

Leader

Nikolina Novosel

Ticijana Ban

Czech Republic Student

Ondřej Bartoš

Jakub Vošmera

Stanislav Fořt

Martin Raszyk

Lubomír Grund

Leader

Jan Kříž

Bohumil Vybíral

Cyprus Student

Nicolas Shiaelis

Anastasios Stylianou

Marios Ioannou

Marios Maimaris

Fidias Ieridis

Leader

Demetrios Philippou

Emmanouil Lioudakis

29

Denmark Student

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 Salvador Student

Bryan Alexander Escalante Castro

Valerie Argentina Dominguez Rivera

Julio Carlos Chorro Huezo

Leader

Jose Roberto Dimas Valle

Raúl Alvarenga

Estonia Student

Jaan Toots

Tanel Kiis

Kaur Aare Saar

Kristjan Kongas

Andres Erbsen

Leader

Mihkel Kree

Taavi Pungas

30

Finland Student

Iiro Lehto

Matias Mannerkoski

Jyri Maanpää

Arttu Yli-Sorvari

Tapio Hautamäki

Leader

Heikki Mäntysaari

Lasse Franti

France Student

Jonathan Dong

Jean Douçot

Paul Kirchner

Theodor Misiakiewicz

Simon Pirmet

Leader

Christian Brunel

Nicolas Billy

Observer

Solene Chevalier-Thery

Georgia Student

Jemal Shengelia

Giorgi Kobakhidze

Sergi Chalauri

Saba Kharabadze

Sandro Maludze

Leader

Vladimir Paverman

Kakhaber Tavzarashvili

Observer

Mariami Rusishvili

31

Germany Student

Qiao Gu

Sebastian Linร

Vu Phan Thanh

Lorenz Eberhardt

Georg Krause

Leader

Stefan Petersen

Gunnar Friege

Observer

Jochen Krรถger

Greece Student

Stavros Efthymiou

Fotios Vogias

Emmanouil Sakaridis

Emmanouil Vourliotis

Michalis Halkiopoulos

Leader

George Kalkanis

Panagiotis Tsakonas

Hong Kong Student

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

32

Hungary Student

Péter Juhász

Áron Dániel Kovács

Zoltán Laczkó

Roland Papp

Attila Szabó

Leader

Péter Vankó

Máté Vigh

Observer

Ferenc Sarlós

Iceland Student

Hólmfríður Hannesdóttir

Atli Thor Sveinbjarnarson

Freyr Sverrisson

Pétur Rafn Bryde

Stefan Alexis Sigurðsson

Leader

Ingibjörg Haraldsdóttir

Martin Swift

India Student

Bijoy Singh Kochar

Jeevana Priya Inala

Kunal Singhal

Pulkit Tandon

Rahul Trivedi

Leader

Patrick Dasgupta

Raghavendra Maigur Krishna

Observer

Shirish Pathare

33

Indonesia Student

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

Ireland Student

John Cristopher Horatio Mulholland

Dale Alexander Hughes

Liam Tomas Mulcahy

Thomas Sherlock Wyse Jackson

Leader

Eamonn Cunningham

David Rea

Islamic Republic of Iran Student

Mohamad Ansarifard

Mehrdad Malak Mohammadi

Amir Yousefi

Ramtin Yazdanian

Sajad Khodadadian

Leader

Mehdi Saadat

Ayoub Esmailpour

Observer

Seyyed Nader Rasuli

34

Israel Student

Itay Knaan Harpaz

Chen Solomon

Yigal Zegelman

Ittai Rubinstein

Eden Segal

Leader

Eli Raz Somech

Igor Lisenker

Observer

Yoav Merhav

Italy Student

Federica Maria Surace

Roberto Albesiano

Michele Fava

Martin Vlashi

Federico Re

Leader

Dennis Luigi Censi

Paolo Violino

Observer

Giorgio Busoni

Francesco Minosso

Japan Student

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

35

Kazakhstan Student

Ivan Senyushkin

Ilya Vilkoviskiy

Kaisarbek Omirzakhov

Nurzhas Aidynov

Mussa Rajamov

Leader

Askar Davletov

Guliya Nurbakova

Observer

Yernur Rysmagambetov

Kuwait Student

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

Kyrgyzstan Student

Fedor Ignatov

Salizhan Kylychbekov

Zakirbek Mamatayir uulu

Ermek Belekov

Meder Kutbidin uulu

Leader

Raia Sultanalieva

Abdymanap Tashmamatov

Observer

Ainagyl Osmonalieva

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Latvia Student

Luka Ivanovskis

Georgijs Trenins

Kristaps Znotins

Andris Gerasimovics

Maris Serzans

Leader

Vyacheslavs Kashcheyevs

Andris Muznieks

Liechtenstein Student

Benedikt Kratochwil

Lukas Lang

David Hälg

Leader

Fritz Epple

Daniel Oehry

Lithuania Student

Mantas Abazorius

Tomas Čerškus

Daumantas Kavolis

Marius Kerys

Žygimantas Stražnickas

Leader

Pavelas Bogdanovičius

Edmundas Kuokštis

Observer

Indrė Grigaitytė

Macao, China Student

Chan Lon Wu

Wai Pan Si

Ka Fai Chan

Wai Hong Lei

Wai Hei Hoi

Leader

Iat Neng Chan

37

Macedonia Student

Vesna Bacheva

Filip Simeski

Biljana Mitreska

Ljupcho Petrov

Leader

Stanisha Veljkovikj

Malaysia Student

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

Mexico Student

Eduardo Acosta-Reynoso

Javier Mendez-Ovalle

Kevin Bustillos-Barrera

Alberto Trejo-Avila

Jorge Torres-Ramos

Leader

Victor Romero-Rochin

Raul Espejel-Morales

Moldova Student

Cristian Zanoci

Ion Toloaca

Nicoleta Colibaba

Dinis Cheian

Ilie Popanu

Leader

Victor Paginu

Igor Evtodiev

38

Mongolia Student

Tsogt Baigalmaa

Battushig Myanganbayar

Munkhtsetseg Battulga

Battsooj Bayarsaikhan

Bilguun Batjargal

Leader

Batsukh Garmaa

Baatarchuluun Tsermaa

Observer

Sharavsuren Byamba

Soyolmaa Dorjyanjmaa

Montenegro Student

Petar Tadic

Marko Petric

Nikola Potpara

Vladimir Pejovic

Janko Radulovic

Leader

Jovan Mirkovic

Nevenka Antovic

Observer

Tatijana Carapic

Netherlands Student

Koen Dwarshuis

Troy Figiel

Ruben Doornenbal

Thijs van der Gugten

Martijn van Kuppeveld

Leader

Ad Mooldijk

Enno van der Laan

39

Nigeria Student

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

Norway Student

Tiantian Zhang

Håkon Tásken

Oda Lauten

Anders Strømberg

Marius Leiros

Leader

Torbjørn Mehl

Joakim Bergli

Pakistan Student

Muhammad Suhaib Qasim

Usman Ayyaz

Usman Ali Javid

Muhammad Taimoor Iftikhar

Leader

Shahid Qamar

Muhammad Aftab Rafiq

40

People’s Republic of China Student

Wenzhuo Huang

Yijun Jiang

Hengyun Zhou

Siyuan Wei

Chi Shu

Leader

Xiaolin Chen

Kun Xun

Observer

Liangzhu Mu

Chunling Zhang

Feng Song

Poland Student

Bartlomiej Zawalski

Michal Pacholski

Kacper Oreszczuk

Filip Ficek

Jan Rydzewski

Leader

Jacek Jasiak

Jan Mostowski

Portugal Student

Francisco Machado

Pedro Paredes

Manuel Cabral

Matheus Marreiros

Simão João

Leader

Fernando Nogueira

Rui Travasso

Puerto Rico Student

Logan Abel

Leader

Hector Jimenez

41

Republic of Korea Student

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 Singapore Student

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

Romania Student

Tudor Giurgică-Tiron

Dan - Cristian Andronic

Sebastian Florin Dumitru

Tudor Ciobanu

Roberta Răileanu

Leader

Delia-Constanţa Davidescu

Adrian Dafinei

Observer

Victor Păunescu

42

Russia Student

Alexandra Vasilyeva

Nikita Sopenko

Ivan Ivashkovskiy

Lev Ginzburg

David Frenklakh

Leader

Valery Slobodyanin

Dmitry Aleksandrov

Observer

Mikhail Osin

Saudi Arabia Student

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

Serbia Student

Tamara Šumarac

Milan Kornjača

Milan Krstajić

Jovan Blanuša

Ilija Burić

Leader

Aleksandar Krmpot

Mihailo Rabasović

43

Slovakia Student

Peter Kosec

Patrik Svancara

Patrik Turzak

Andrej Vlcek

Samuel Beznak

Leader

Ivo Cap

Lubomir Mucha

Observer

Lubomir Konrad

Visitor

Klara Capova

Slovenia Student

Domen Ipavec

Matevž Marinčič

Jan Šuntajs

Jurij Tratar

Miha Zgubič

Leader

Ciril Dominko

Jurij Bajc

South Africa Student

Thiolan Prevan Naidoo

Avthar Sewrathan

Xolela Jara

Lloyd Mahadeo

Shihal Menesher Sapry

Leader

Mervlyn Moodley

Bagtyyar Jorayev

44

Spain Student

Roberto Alegre

Francesc-Xavier Gispert Sánchez

David Trillo Fernández

Marc Rodà Llordés

Aitor Azemar

Leader

Juan Leon

Esperanza García-Carpintero Romero

Sri Lanka Student

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

Suriname Student

Chaandnie Bandhoe

Raynesh Kanhai

Priya Kasimbeg

Suraj Kishoen Misier

Leader

Tjien Bing Tan

Ignaas Jimidar

Visitor

Chantal Hewitt

Lachman Jurgen

45

Sweden Student

Johan Runeson

Andréas Sundström

Carl Smed

Viktor Djurberg

Simon Johansson

Leader

Max Kesselberg

Bo Söderberg

Observer

Anne-Sofie Mårtensson

Visitor

Margareta Kesselberg

Switzerland Student

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

46

Taiwan Student

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

Tajikistan Student

Abdukhomid Nurmatov

Rabboni Bafoev

Adhamzhon Shukurov

Shakhzodi Rustamdzhon

Isfandiyor Safarov

Leader

Ilkhom Khotami

Thailand Student

Pongsapuk Sawaddirak

Puthipong Worasaran

Paphop Sawasdee

Supanut Thanasilp

Nathanan Tantivasadakarn

Leader

Sirapat Pratontep

Phichet Kittara

Observer

Suwan Kusamran

Raksapol Thananuwong

47

Turkey Student

Atinc Cagan Sengul

Oguzhan Can

Abdurrahman Akkas

Mustafa Selman Akinci

Mehmet Said Onay

Leader

Ibrahim Gunal

Onur Ozcan

Turkmenistan Student

Mekan Toyjanov

Meylis Malikov

Kemal Babayev

Agajan Odayev

Övezmyrat Övezmyradow

Leader

Halit Coşkun

Gylychmammet Orazov

Ukraine Student

Volodymyr Sivak

Vsevolod Bykov

Vladysslav Diachenko

Volodymyr Rozsokhovatskyi

Yevgen Cherniavskyi

Leader

Boris Kreminskyi

Stanislav Vilchynskyi

Observer

Bushtruk Artem

48

United Kingdom Student

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 America Student

Allan Sadun

Eric Schneider

Jeffrey Cai

Jeffrey Yan

Kevin Zhou

Leader

Paul Stanley

Andrew Lin

Vietnam Student

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

49

50

Photos by Andres Mihkeson, Siim Pille and Merily Salura

51

52

Photos by Andres Mihkeson, Siim Pille and Merily Salura

53

Organizers Steering Committee Chairman Janar Holm

Estonian Ministry of Education and Research

Secretary Viire Sepp

Gifted and Talented Development Centre

of University Tartu

Members Ergo 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

54

Academic Committee Jaak 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 Committee Ene 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

55

Markers Helle 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

56

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

57

Volunteers Agnes Vask

Eno Paenurk

Airike Jõesaar

Erik Ilbis

Allan-Cristjan Puks

Eva Mõtshärg

Anastassia Samovitš

Evelin Pihlap

Andreas Ragen Ayal

Eveli Soo

Andres Ainelo

Gerli Krjukov

Andres Allik

Gerli Vaik

Andres Mihkelson

Gertrud Metsa

Anete Merilin Leetberg

Grete Helena Roose

Anete Sammler

Grethe Aikevitšius

Anete Viise

Hanna Britt Soots

Anna Dunajeva

Hanna Kadri Metsvaht

Anna Jazõkova

Hanna Kivila

Anna Krajuškina

Hanna Moor

Annemari Sepp

Hanna-Loore Hansen

Anni Müüripeal

Hedvig Tamman

Anni Sandra Varblane

Helbe-Laura Nikitkina

Annika Lukner

Helena Ainsoo

Annika Pille

Helena Talimaa

Ants Johanson

Heli Aomets

Anu Viks

Heli Pärn

Artur Panov

Henry Teigar

Arvo Ehrstein

Ida Rahu

Ats Kurvet

Inger Kangur

Auli Relve

Ivo Kruusamägi

Ave-Stina Udam

Jaanika Jensen

Bety Mehide

Jane Lihtmaa

Brenda Rauniste

Janne Disko

Diana Oidingu

Jasper Kursk

Dmitri Lanevski

Jelizaveta Dõljova

Donatas Braziulis

Jelizaveta Žatkina

Egert Vinogradov

Johanna-Maria Muuga

Elina Libek

Joonas Jäme

Enna Elismäe

Jorma Veiderpass

58

Juhani Almers

Liisa Hunt

Julia Gavrilova

Liisa Veerus

Kadi Ainsaar

Liisi Liivalaid

Kadi Külasalu

Liisi Mõtshärg

Kadri Alumets

Liisi Sünd

Kadri Ann Rebane

Liisu Miller

Kadri Eek

Lisett Kiudorv

Kadri Tinn

Lona-Liisa Sutt

Kaisa Jõgi

Ly Pärnaste

Karl Kütt

Maksim Ivanov

Karl Veskus

Maksim Mišin

Karl-Mattias Tepp

Marek Järvik

Karolin Rõõm

Maria Krajuškina

Kaspar Märtens

Mari-Liis Jaansalu

Kati Randmäe

Mariliis Maamägi

Katrin Tuude

Mari-Liis Tamm

Keidi Suursaar

Maris Ertmann

Ken Riisalu

Maris Palo

Kerli Kalk

Marit Puusepp

Kerstin Kivila

Mark Gimbutas

Krista Kallavus

Marren Tiivits

Kristi Kartus

Mart Ernits

Kristiina Štõkova

Marta Tanaga

Kristin Ehala

Marta Vihtre

Kristin Liiksaar

Mary-Ann Kubre

Kristine Diane Liive

Merilin Kalavus

Kristine Leetberg

Merilin Vesingi

Kristjan Kalve

Merily Salura

Ksenia Kukuskina

Merle Lust

Laura Liisa Lankei

Merlin Russak

Laura Soon

Mette-Triin Purde

Liina Nõmm

Mihkel Lepik

Liis Kass

Mihkel Tali

Liis Nurmis

Minna-Triin Kohv

Liis Talimaa

Mirjam Laurimäe

59

Mirjam Mikk

Sandhra-Mirella Valdma

Morten Piibeleht

Sergei Jakovlev

Natalia Nekrassova

Siim Kaspar Uustalu

Nele Kriisa

Siim Pille

Olga Bulgakova

Siiri Mägi

Oliver Grauberg

Sille Hausenberg

Ott Kekišev

Simona Kalatšov

Paap Koemets

Sirje Kollom

Paul Liias

Sten Aus

Pille-Riin Peet

Stina Avvo

Raimo Armus

Teisi Timma

Rando Porosk

Terje Kapp

Rasmus Kuusmann

Tiina Pärtel

Reile Juhanson

Tiina Turban

Rene Rünt

Triin Rebane

Riinu Ansper

Triin Ärm

Rudolf Bichele

Triinu Hordo

Saile Mägi

Uku-Kaspar Uustalu

Sander Benga

Ulla Meeri Liivamägi

Sander Kütisaar

Urmet Paloveer

Sander Soo

Üllar Kivila

Sander Udam

60

Photos by Siim Pille, Andres Mihkeson and Henry Teigar

61

62

Programs Students Sunday, 15th of July Arrival 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 July 07:00 – 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

63

17:30 – 19:00

Free time and preparation for theory

19:00 – 21:00

Dinner – Restaurant Dorpat

Tuesday, 17th of July 06:00 – 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 July 07:00 – 09:00

Breakfast

09:00 – 20:00

Excursion: Rakvere Castle

20:00 – 23:00

Free time and preparation for experiment

Thursday, 19th of July 06:00 – 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

64

Students in Tartu Adventure Park Photo by Siim Pille

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 July Tartu – 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 July 08:00 – 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

65

Sunday, 22nd of July 07:00 – 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 July 07.00 – 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

The Ceremony will be broadcasted over the Internet

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 July Departure

66

Football award. Football tournament Photos by Henry Teigar

Leaders and Observers Sunday, 15th of July Arrival 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 July 07:00 – 09:00

Breakfast

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 – 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

67

Tuesday, 17th of July 06:00 – 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 July 07:00 – 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

19:00 – 21:00

Dinner – Restaurant Senso

21:00 – …

International Board Meeting:

Translation of experimental problems

Thursday, 19th of July 07:00 – 09:00

Breakfast

09:00 – 12:00

Free time

12:00 – 13:00

Collection of marks from Leaders (theory) – Online

13:00 – 15:00

Lunch - Restaurant Senso

68

Estonian Open Air Museum. Skype meeting with students Photos by Siim Pille, Henry Teigar

15:00 – 19:00

Free time

19:00 – 21:00

Leaders group A: Distribution of practical papers

from IPhO office

19:00 – 21:00

Dinner – Restaurant Senso

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 July Tartu – 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

69

Saturday, 21st of July 07:00 – 09:00

Breakfast

10:00 – 12:00

International Board Meeting

11:00

Distribution of marks (theory) - Online

12:00 – 14:00

Lunch - Restaurant Senso

14:00 – 21:00

International Board Meeting:

Moderation of theoretical papers

19:00

Distribution of marks (experiment) - Online

19:00 – 21:00

Dinner – Restaurant Senso

Sunday, 22nd of July 07:00 – 09:00

Breakfast

09:00 – 17:00

International Board Meeting:

Moderation of experimental papers

12:00 – 14:00

Lunch – Restaurant Senso

17:00 – 19:00

International Board Meeting:

Deciding final marks and medals

19:00 – …

Free time and dinner at Tallinn restaurants

70

Farwell Party Photo by Merily Salura

Monday, 23rd of July 07:00 – 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

The Ceremony will be broadcasted over the Internet

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 July Departure

Closing Ceremony Photo by Merily Salura

71

72

Problems and solutions The 43rd International Physics Olympiad — Theoretical Competition rd

43 —International Olympiad Tartu, The Estonia Thursday, JulyPhysics 17th 2012

— Theoretical Competit

Tartu, Estonia — Tuesday, July 17th 2012 • The examination lasts for 5 hours. There are 3 problems

each Problem, the Solution Sheets ar

worth in total 30 points. Please note that the point values of the three theoretical problems are not

the sheets according to the enumeration which Problem Part and Question

equal.

ing with. Copy the final answers into boxes of the Answer Sheets. There a

• You must not open the envelope with the problems 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

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International Physics Olympiad Competition • For— eachTheoretical problem, there are dedicated Solution Sheets header the number and pictogramme). Write Tartu, Estonia — Tuesday,(see July 17th for 2012

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Problem T1.

Focus on sketches Problem T1. Focus on sketches (13 points) (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.

(0.8 pts) By adjusting the launching angle for a ball thrown — page 1 of 5i.—

with a fixed initial speed v0 from the origin, targets can be hit within the region given by

z ≤ z0 − kx2 ;

B. Air flow For this useful. For

around a wing (4 points) Problem Part, the following a flow of liquid or gas in a tu

1 line p + ρgh + 2 ρv 2 = const., assuming is much smaller than the sound speed. H h — height, g — free fall acceleration, a pressure. Streamlines are defined as the

particles (assuming that the flow pattern that the term 12 ρv 2 is called the dynamic

In the fig. below, a cross-section of an picted together with streamlines of the wing, as seen in the wing’s reference fr (a) the air flow is purely two-dimensiona

city vectors of air lie in the figure plane) pattern is independent of the aircraft sp

you can use this fact without proving it. Find the constants z0 wind; (d) the dynamic pressure is much and k. mospheric pressure p0 = 1.0 × 105 Pa . Yo take measurements from the fig. on the a 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.

i. (0.8 pts) If the aircraft’s ground spee iii. (2.5 pts) What is the minimal launching speed vmin needed what is the speed of the air vP at the poin to hit the topmost point of a spherical building of radius R ? with respect to the ground?

ii. (1.2 pts) In the case75 of high relative hum speed of the aircraft increases over a critica

the launching angle can be adjusted as needed; the aim is to hit the topmost point of a spherical the aim is to hit the topmost point of a spherical building of radius R (see fig.) with as small as building of radius R (see fig.) with as small as possible initial speed v0 (prior hitting the target, bouncing off hit the topmost point of a spherical possible initial speed v0 (prior hitting the target, bouncing off the roof is not allowed). Sketch qualitatively the shape of the dius R (see fig.) with as small as the roof is not allowed). Sketch qualitatively the shape of the optimal trajectory of the ball (use the designated box on the off al speed v0 (prior hitting the target, bouncing optimal trajectory of the ball (use the designated box on the answer sheet).

Note: the points are given only for the sketch. t allowed). Sketch qualitatively the shape of the answer sheet). Note: the points are given only for the sketch. i. (0.8 pts) If ctory of the ball (use the designated box on the iii. (2.5 pts) What is the minimal launching speed vmin needed i. (0.8 pts) If what is the spee . 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 ? what is the spee with respect to t i. (0.8 pts)point If the ground speed is v0 = R 100 ? m/s, to hit the topmost of aircraft’s a spherical building of radius What is the minimal launching speed vmin needed with respect to t (1.2 pts) In t what is the speed of the air vP at the point P (marked inii.fig.) most point of a spherical building of radius R ? ii. (1.2 pts) In t speed of the aircr with respect to the ground? speed of the aircr of water droplet ii. (1.2 pts) In the case of high relative humidity, as the ground of water droplet emerge at a cert speed of the aircraft increases over a critical value vcrit , a stream emerge at a cert answer sheet. Ex of water droplets is created behind the wing. The droplets answer sheet. Ex emerge at a certain point Q. Mark the point Q in fig. ontext theas possible) text as possible) (2.0 pts) Es answer sheet. Explain qualitatively (using formulae and asiii. few iii. (2.0 pts) Es ing data: relative text as possible) how you determined its position. ing data: relative of air at constan iii. (2.0 pts) Estimate the critical speed vcrit using the followof air at constan heat ing data: relative humidity of the air is r = 90% , specificof saturated wat of saturated wat 3 of air at constant pressure cp = 1.00 × 10 J/kg · K , pressure ure of the unper ure of the unper at Tb = 294 K . of saturated water vapour: psa = 2.31 kPa at the temperatat Tb = 294 K . kPa also need ure of the unperturbed air Ta = 293 K and psb = 2.46may may also need at Tb = 294 K . Depending on your approximations cyou V = 0.717 × 10 cV = 0.717 × 10 defined as the rat may also need the specific heat of air at constant volume defined as the rat 3 pressure at the g cV = 0.717 × 10 J/kg · K . Note that the relative humidity is pressure at the g defined as the va defined as the ratio of the vapor pressure to the saturated vapor defined as the va withis the liquid. pressure given temperature.

Saturated vaporPart pressure La Géode, Parc deat la the Villette, Paris.

Photo: katchooo/flickr.com with the liquid.

La Géode, Parc de la Villette, Paris. Photo: katchooo/flickr.com Part defined as the vapor pressure by which vapor is in equilibrium

d on the ground level z = 0, and angle can be adjusted as needed;

with Part B. deFocus la Villette, Photo: katchooo/flickr.com Air the flowliquid. around a wing (4 points) 1.cus onParis. sketches (13 points) B. on sketches (13 points) Air flow around a wing (4 points)

— page 2 of 5 — — page 2 of 5 —

— page 2 of For 5 —this Problem Part, the following information may be

For this Problem Part, the following information may be useful. For a flow of liquid or gas in a tube, along a streamuseful. For a flow of liquid or gas in a tube, along a streamwith an initial speed v0 moves in a homogeneous line p + ρgh + 12 ρv 2 = const., assuming that the velocity v const., assuming that the velocity v n initial speed v0 moves in a homogeneous line p + ρgh + 21 ρv 2 = eld in x − z plane, where the x-axis is horizontal, is much smaller than the sound speed. Here ρ is the density, x − z plane, where the x-axis is horizontal, is much smaller than the sound speed. Here ρ is the density, cal, antiparallel to the free fall acceleration g ; h — height, g — free fall acceleration, and p — hydrostatic ntiparallel to the free fall acceleration g ; h — height, g — free fall acceleration, and p — hydrostatic drag. pressure. Streamlines are defined as the trajectories of fluid pressure. Streamlines are defined as the trajectories of fluid particles (assuming that the flow pattern is stationary). Note particles (assuming that the flow pattern is stationary). Note y adjusting the launching angle for a ball thrown 1 sting the launching angle for a ball thrown ρv 2 is called the dynamic pressure. that the 1term 2 be the term 2 ρv is2 called the dynamic pressure. nitial speed v0 from the origin, targets canthat peed v0 from the origin, targets can be In the fig. below, a cross-section of an aircraft wing is deIn the fig. below, a cross-section of an aircraft wing is deregion given by n given by picted together with streamlines of the air flow around the picted together with streamlines of the air flow around the wing, as seen in the wing’s reference frame. Assume that wing, as seen in the wing’s reference frame. Assume that (a) the air flow is purely two-dimensional (i.e. that the veloz ≤ z02 − kx2 ; (a) the air flow is purely two-dimensional (i.e. that the veloz ≤ z0 − kx ; city vectors of air lie in the figure plane); (b) the streamline city vectors of air lie in the figure plane); (b) the streamline pattern is independent of the aircraft speed; (c) there is no pattern is independent of the aircraft speed; (c) there is no is fact without proving it. Find the constants z0 wind; (d) the dynamic pressure is much smaller than the atwithout proving it. Find the constants z0 wind; (d) the dynamic pressure is much smaller than the atmospheric pressure p = 1.0 × 105 Pa . You can use a ruler to mospheric pressure p0 = 1.00 × 105 Pa . You can use a ruler to take measurements from the fig. on the answer sheet . take measurements from the fig. on the answer sheet .

Now, the launching point can be he launching point can be on the ground level z = 0, and e ground level z = 0, and angle can be adjusted as needed; can be adjusted as needed; t the topmost point of a spherical opmost point of a76spherical dius R (see fig.) with as small as (see fig.) with as small as speed v (prior hitting the target, bouncing off

tics (4.5 points) 4.5 points)

(a) the air flow is purely two-dimensional (i.e. that the velocity vectors of air lie in the figure plane); (b) the streamline pattern is independent of the aircraft speed; (c) there is no the constants z0 wind; (d) the dynamic pressure is much smaller than the atmospheric pressure p0 = 1.0 × 105 Pa . You can use a ruler to take measurements from the fig. on the answer sheet .

e

d d;

al s

get, bouncing off the shape of the

ated box on the y for the sketch.

i. (0.8 pts) If the aircraft’s ground speed is v0 = 100 m/s, peed vmin needed what is the speed of the air vP at the point P (marked in fig.) ng of radius R ? 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 following 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 temperature 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

oo/flickr.com Part

with the liquid.

— page 2 of 5 —

77

Part C. Magnetic straws (4.5 points)

ignated box of t

Consider a cylindrical tube made of a superconducting material. The length of the tube

red dots marked ii. (1.2 pts) F

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

The centre of the tube coincides with the

middle of the tu tube, z > 0 and

origin, and its axis coincides with the zaxis. There is a magnetic flux Φ through

iii. (2.5 pts) N identical and pa

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

second tube has

x2 + y 2 < r2 . magnetic field, Superconductor is a material which expels any magnetic y = l , x = z =

traws (4.5 points)

opposite sides of field (field is zero inside it). i. (0.8 pts) Sketch five such magnetic field lines onto the des- magnetic interac ignated box of the answer sheet which pass through the five

al tube made of a superThe length of the tube

red dots marked on the axial cross-section of the tube. ii. (1.2 pts) Find the z-directional tension force T in the

dius is r ; always l ≫ r. ube coincides with 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).

coincides with the zgnetic flux Φ through

iii. (2.5 pts) Now there is another tube, identical and parallel to the first one. The

tion of the tube, z = 0,

second tube has opposite direction of the

magnetic field, and its centre is placed at is a material which expels any magnetic y = l , x = z = 0 (so that the tubes form

opposite sides of a square). Determine the ide it). five such magnetic field lines onto the des- magnetic interaction force F between the two tubes.

Problem T2.

Kelvin water dropper (8 points) Problem T2. Kelvin water dropper (8 points)

Part B. Two pip

The following facts about the surface tension may turn out An apparatus ca to be useful for this problem. For the molecules of a liquid, the pipes (identical

positions at the liquid-air interface are less favourable as com- via a T-junction pared with the positions in the bulk of the liquid. Therefore, centres of two c

this interface is ascribed the so-called surface energy U = σS, meter D, L ≫ D where S is the surface area of the interface and σ — the surface droplets per uni

78

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

ductive bowls un

liquid surface pull each other with a force F = σl, where l is the length of a straight line separating the fragments.

electrodes as sho capacitance C .

A long metallic pipe with internal diameter d is pointing directly downwards; water is slowly

and electrodes. N

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

Water can be considered to be electrically con-

will cause an im charge separatio

ducting; its surface tension is σ and density —

The first dro

Problem T2. Kelvin water dropper (8 points)

to be useful for this problem. For the molecules of a liquid, the The following facts about the surface tension may turn out positions at the liquid-air interface are less favourable as comto be useful for this problem. For the molecules of a liquid, the pared with the positions in the bulk of the liquid. Therefore, positions at the liquid-air interface are less favourable as comthis interface is ascribed the so-called surface energy U = σS, pared with the positions in the bulk of the liquid. Therefore, where S is the surface area of the interface and σ — the surface this interface is ascribed the so-called surface energy U = σS, tension coefficient of the liquid. Further, two fragments of the where S is the surface area of the interface and σ — the surface liquid surface pull each other with a force F = σl, where l is tension coefficient of the liquid. Further, two fragments of the the length of a straight line separating the fragments. liquid surface pull each other with a force F = σl, where l is A long metallic pipe with internal diameter d the length of a straight line separating the fragments. is pointing directly downwards; water is slowly A long metallic pipe with internal diameter d dripping from a nozzle at its lower end, see fig. is pointing directly downwards; water is slowly Water can be considered to be electrically condripping from a nozzle at its lower end, see fig. ducting; its surface tension is σ and density — Water can be considered to be electrically conρ . Always assume that d ≪ r. Here, r is the ducting; its surface tension is σ and density — radius of the droplet hanging below the nozzle, ρ . Always assume that d ≪ r. Here, r is the which grows slowly in time until the droplet sepradius of the droplet hanging below the nozzle, arates from the nozzle due to the free fall acceleration g . which grows slowly in time until the droplet sepPart A. Single pipe (4 points) arates from the nozzle due to the free fall acceleration g . i. (1.2 pts) Find the radius rmax of a drop just before it sepPart A. Single pipe (4 points) arates from the nozzle. i. (1.2 pts) Find the radius rmax of a drop just before it sepii. (1.2 pts) Relative to the far-away surroundings, the pipe’s arates from the nozzle. electrostatic potential is ϕ . Find the charge Q of a drop when ii. (1.2 pts) Relative to the far-away surroundings, the pipe’s its radius is r . electrostatic potential is ϕ . Find the charge Q of a drop when iii. (1.6 pts) For this question, assume that r is kept conits radius is r . stant and ϕ is slowly increased. The droplet becomes unstable iii. (1.6 pts) For this question, assume that r is kept conand breaks into two pieces if the hydrostatic pressure inside stant and ϕ is slowly increased. The droplet becomes unstable the droplet becomes smaller than the atmospheric one. Find and breaks into two pieces if the hydrostatic pressure inside the critical potential ϕmax at which this will happen. the droplet becomes smaller than the atmospheric one. Find

the critical potential ϕmax at which this will happen.

Part B. Two pipes (4 points) pipes (identical to the one described in An apparatus called “Kelvin water drop via a T-junction, see fig. The ends of bo pipes (identical to the one described in centres of two cylindrical electrodes (wit via a T-junction, see fig. The ends of bo meter D, L ≫ D ≫ r); for both tubes, the centres of two cylindrical electrodes (wit droplets per unit time. Droplets fall from meter D, L ≫ D ≫ r); for both tubes, the th ductive bowls underneath the nozzles, cro droplets per unit time. Droplets fall from electrodes as shown in Fig; the electrodes ductive bowls underneath the nozzles, cro capacitance C . There is no net charge on electrodes as shown in Fig; the electrodes and electrodes. Note that the water conta capacitance C . There is no net charge on first droplet fall the willwater have micro andThe electrodes.

Notetothat conta

will cause an imbalance between the tw The first droplet to fall will have micro charge separation across the capacitor. will cause an imbalance between the tw i. (1.2separation pts) Express charge acrossthe themodulus capacitor. of the charge Q0 of the drops separi. (1.2 pts) Express the modulus ating at the instant when the capaof the charge Q0 of the drops separcitor’s charge is q in terms of rmax ating at the instant when the capa(from Part A-i). Neglect the effect citor’s charge is q in terms of rmax described in Part A-iii. (from Part A-i). Neglect the effect ii. (1.5 pts) Find the dependence described in Part A-iii.

of q on time t by approximating it ii. (1.5 pts) Find the dependence with a continuous function q(t) and of q on time t by approximating it assuming that q(0) = q0 . with a continuous function q(t) and iii. (1.3 pts) The = dropper’s functioning q0 . assuming that q(0) the effect shown in Part A-iii. Additiona iii. (1.3 pts) The dropper’s functioning the achievable voltage between the elect the effect shown in Part A-iii. Additiona electrostatic push between a droplet and the achievable voltage between the elect find Umax . electrostatic push between a droplet and

find Umax .

8 points)

Part B. Two pipes (4 points) on may turn out An apparatus called “Kelvin water dropper” consists of two es of a liquid, the pipes (identical to the one described in Part A), connected vourable as com- via a T-junction, see fig. The ends of both pipes are at the quid. Therefore, centres of two cylindrical electrodes (with height L and dia-

energy U = σS, σ — the surface fragments of the

meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n droplets per unit time. Droplets fall from height H into conductive bowls underneath the nozzles, cross-connected to the

= σl, where l is gments.

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.

d wly fig. on-

— he zle, ep-

eleration g .

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 separating at the instant when the capacitor’s charge is q in terms of rmax (from Part A-i). Neglect the effect

— page 4 of 5 — — page 4 of 5 — 79

bed the so-called surface energy U = σS, meter D, L ≫ D ≫ r); for both tubes, the dripping rate is n e area of the interface and σ — the surface droplets per unit time. Droplets fall from height H into conthe liquid. Further, two fragments of the

ductive bowls underneath the nozzles, cross-connected to the

ach other with a force F = σl, where l is ght line separating the fragments.

electrodes as shown in Fig; the electrodes are connected via a capacitance C . There is no net charge on the system of bowls

with internal diameter d downwards; water is slowly

and electrodes. Note that the water container is grounded.

zle at its lower end, see fig. dered to be electrically con-

will cause an imbalance between the two sides and a small charge separation across the capacitor.

The first droplet to fall will have microscopic charge, which

ension is σ and density — that d ≪ r. Here, r is the

i. (1.2 pts) Express the modulus of the charge Q0 of the drops separating at the instant when the capa-

t hanging below the nozzle, n time until the droplet sep-

le due to the free fall acceleration g .

(4 points)

citor’s charge is q in terms of rmax (from Part A-i). Neglect the effect

described in Part A-iii. e radius rmax of a drop just before it sepii. (1.5 pts) Find the dependence le. e to the far-away surroundings, the pipe’s of q on time t by approximating it al is ϕ . Find the charge Q of a drop when with a continuous function q(t) and assuming that q(0) = q0 .

his question, assume that r is kept con- iii. (1.3 pts) The dropper’s functioning can be hindered by increased. The droplet becomes unstable the effect shown in Part A-iii. Additionally, a limit Umax to pieces if the hydrostatic pressure inside smaller than the atmospheric one. Find

l ϕmax at which this will happen.

the achievable voltage between the electrodes is set by the electrostatic push between a droplet and the bowl beneath it; find Umax .

Problem T3. protostar formation (9 points) Problem T3. Protostar formation (9 points)

iii. (2.5 pts) As

Let us model the formation of a star as follows. A spherical the time tr→0 ne cloud of sparse interstellar gas, initially at rest, starts to col- a much smaller r

lapse due to its own gravity. The initial radius of the ball is iv. (1.7 pts) At r0 and the mass — m . The temperature of the surroundings enough to be o (much sparser than the gas) and the initial temperature of the amount of heat gas is uniformly T0 . The gas may be assumed to be ideal. radius r0 down t The average molar mass of the gas is µ and its adiabatic inv. (1 pt) For ra dex is γ > 43 . Assume that G mµ r0 ≫ RT0 , where R is the gas diation. Determ constant and G — the gravity constant. on its radius r < i. (0.8 pts) During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, vi. (2 pts) Even — page 4 i.e. of 5 the — ball stays in a thermodynamic equilibrium with its sur- sure

on the dy roundings. How many times (n) does the pressure increase r = r4 (with r4

while the radius is halved ( r1 = 0.5r0 )? Assume that the gas neglected and th density stays uniform. nuclear fusion. T

ii. (1 pt) Estimate the time t2 needed for the radius to shrink anymore, but ro from r0 to r2 = 0.95r0 . Neglect the change of the gravity field factors can still 80

at the position of a falling gas particle.

respective tempe

3

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-

0

enough to be opaque to the heat radia amount of heat Q radiated away during t radius r0 down to r3 .

v. (1 pt) For radii smaller than r3 you dex is γ > 43 . Assume that G mµ r0 ≫ RT0 , where R is the gas diation. Determine how the temperature T constant and G — the gravity constant. on its radius r < r3 . i. (0.8 pts) During much of the collapse, the gas is so transparent that any heat generated is immediately radiated away, vi. (2 pts) Eventually we cannot neglect

i.e. the ball stays in a thermodynamic equilibrium with its sur- sure on the dynamics of the gas and t roundings. How many times (n) does the pressure increase r = r4 (with r4 ≪ r3 ). However, the r

while the radius is halved ( r1 = 0.5r0 )? Assume that the gas neglected and the temperature is not yet h density stays uniform. nuclear fusion. The pressure of such a pro

ii. (1 pt) Estimate the time t2 needed for the radius to shrink anymore, but rough estimates with inacc from r0 to r2 = 0.95r0 . Neglect the change of the gravity field factors can still be done. Estimate the fin respective temperature T4 .

at the position of a falling gas particle.

points)

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

ws. A spherical the time tr→0 needed for the ball to collapse from r0 down to st, starts to col- a much smaller radius using Kepler’s Laws for elliptical orbits.

us of the ball is iv. (1.7 pts) At some radius r ≪ r , the gas becomes dense 3 0 the surroundings enough to be opaque to the heat radiation. Calculate the mperature of the amount of heat Q radiated away during the collapse from the med to be ideal. radius r0 down to r3 . its adiabatic inv. (1 pt) For radii smaller than r3 you may neglect heat rare R is the gas diation. Determine how the temperature T of the ball depends on its radius r < r3 . e gas is so transy radiated away, vi. (2 pts) Eventually we cannot neglect the effect of the pres-

ium with its sur- sure on the dynamics of the gas and the collapse stops at ressure increase r = r4 (with r4 ≪ r3 ). However, the radiation can still be

ume that the gas neglected and the temperature is not yet high enough to ignite nuclear fusion. The pressure of such a protostar is not uniform radius to shrink anymore, but rough estimates with inaccurate numerical prethe gravity field factors can still be done. Estimate the final radius r4 and the respective temperature T4 .

— page 5 of 5 —

81

Solutions Problem T1.

Focus on sketches (13 points) 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 = v02 /2g (as it follows from the energy conservation law). Comparing this with the inequality z ≤ z0 − kx2 we conclude that

z0 = v02 /2g.

82

[0.3 pts]

Note that the region where the the top with ini the building, we ing angle. On th region needs to gap, it would be where the optim ory through that hence we arrive So, with v0 co getable region to there are two to be four, and for origin at the to defined by the fo

Let us consider the asymptotics z → −∞; the trajectory of the stone is a parabola, and at this limit, the horizontal displacement (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 = v0 t and z = −gt2 /2, i.e. its trajectory is given by z = −gx2 /2v02 . Now, let us recall that z ≤ z0 − kx2 , i.e. −gx2 /2v02 ≤ z0 − kx2 ⇒ k ≤ g/2v02 .

Note that k < g/2v02 would imply that there is a gap between x2 the parabolic region z ≤ z0 − kx2 and the given trajectory z = −gx2 /2v02 . This trajectory is supposed to be optimal for Upon eliminatin hitting targets far below (z → −∞), so there should be no such

g a gap, and hence, we can exclude the option k < g/2v02 . This x4 2v02 leaves us with k = g/2v02 .

[0.5 pts] Hence the speed can be found from ii. (1.2 pts) Let us note that the

1 stone trajectory is reversible and due 2 to the energy conservation law, one can equivalently ask, what is the minBearing in mind imal initial speed needed for a stone the ground level to be thrown from the topmost point we finally obtain of the spherical building down to the ground without hitting the roof, and what is the respective trajectory. It is easy to understand that the trajectory either needs to touch the roof, or start horizontally from the topmost point Alternative with the curvature radius equal to R. Indeed, if neither were is thrown from the case, it would be possible to keep the same throwing angle different heigh and just reduce the speed a little bit — the stone would still ing speed, the reach the ground without hitting the roof. Further, if it were This fact may b tangent at the topmost point, the trajectory wouldn’t touch can be also easi nor intersect the roof anywhere else, because the curvature of ing with velocity the parabola has maximum at its topmost point. Then, it the point B, and would be possible to keep the initial speed constant, and in- possible. Now, crease slightly the throwing angle (from horizontal to slightly to it a perpendi

points)

ally upwards, it follows from the h the inequality

[0.3 pts]

to be thrown from the topmost point we finally obtain of the spherical building down to the

ground without hitting the roof, and what is the respective tra2 vmin = v0 + 4gR = 3 jectory. It is easy to understand that the trajectory either needs to touch the roof, or start horizontally from the topmost point Alternative solution using the fa with the curvature radius equal to R. Indeed, if neither were is thrown from a point A to a poi the case, it would be possible to keep the same throwing angle different heights) with the minimal and just reduce the speed a little bit — the stone would still ing speed, the initial and terminal ve reach the ground without hitting the roof. Further, if it were This fact may be known to some of the tangent at the topmost point, the trajectory wouldn’t touch can be also easily derived. Indeed, supp nor intersect the roof anywhere else, because the curvature of ing with velocity �v0 at point A, the ball the parabola has maximum at its topmost point. Then, it the point B, and |�v0 | is the minimal speed would be possible to keep the initial speed constant, and in- possible. Now, let us rotate the vector �v crease slightly the throwing angle (from horizontal to slightly to it a perpendicular small vector �u ⊥ �v0 upwards): the new trajectory wouldn’t be neither tangent at the launching velocity �v1 = �v0 + �u, the t the top nor touch the roof at any other point; now we can re- will still go almost through point B: ne duce the initial speed as we argued previously. So we conclude displacement of the trajectory cannot cha that the optimal trajectory needs to touch the roof somewhere, Indeed, a linear in |�u| displacement would as shown in Fig. sentially the same speed |�v1 | ≈ |�v0 |, it w iii. (2.5 pts) The brute force approach would be writing down throw over point B, which is in a contra the condition that the optimal trajectory intersects with the timality of the original trajectory. Henc building at two points and touches at one. This would be de- vector �r0 (τ ) − �r1 (τ ) needs to be parallel t scribed by a fourth order algebraic equation and therefore, it is to the velocity �vB of the ball at point B not realistic to accomplish such a solution within a reasonable is the radius vector of the ball as a funct was launched with velocity �v0 [�v1 ]. In a f time frame.

Note that the interior of the building needs to lie inside the — page 1 of 5 — 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 throwing 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 trajectory through that target wouldn’t touch the building anywhere, hence we arrive at a contradiction.

So, with v0 corresponding to the optimal trajectory, the targetable 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:

the trajectory of e horizontal disth respect to the abola, the larger test shape when z = −gt2 /2, i.e. ow, let us recall 2 ⇒ k ≤ g/2v02 . gx2 v2 s a gap between x2 + z 2 + 2zR = 0, z = 0 − 2 .

2g 2v0 given trajectory o be optimal for Upon eliminating z, this becomes a biquadratic equation for x: hould be no such 2

2

2 v0 g 1 gR v0 2 k < g/2v02 . This − + R = 0. + x4 + x 2v02 2 v02 4g g [0.5 pts]

Hence the speed by which the real-valued solutions disappear can be found from the condition that the discriminant vanishes: 2

1 gR 1 gR gR = + 2 =⇒ 2 = 2. − 2 2 v0 4 v0 v0

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

gR he respective tra. vmin = v02 + 4gR = 3 2 tory either needs

83

low (z → −∞), so there should be no such e can exclude the option k < g/2v02 . This k = g/2v02 .

[0.5 pts]

s note that the versible and due rvation law, one what is the mineded for a stone he topmost point ding down to the ng the roof, and what is the respective traunderstand that the trajectory either needs start horizontally from the topmost point adius equal to R. Indeed, if neither were possible to keep the same throwing angle speed a little bit — the stone would still thout hitting the roof. Further, if it were most point, the trajectory wouldn’t touch of anywhere else, because the curvature of aximum at its topmost point. Then, it o keep the initial speed constant, and inhrowing angle (from horizontal to slightly trajectory wouldn’t be neither tangent at he roof at any other point; now we can red as we argued previously. So we conclude jectory needs to touch the roof somewhere,

x4

g 2v02

2

+ x2

1 gR − 2 2 v0

+

v02 +R 4g

v02 = 0. g

Hence the speed by which the real-valued solutions disappear can be found from the condition that the discriminant vanishes:

2 1 gR 1 gR gR = + 2 =⇒ 2 = 2. − 2 2 v0 4 v0 v0 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

gR . vmin = v02 + 4gR = 3 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 launching 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 starting 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 essentially the same speed |�v1 | ≈ |�v0 |, it would be possible to rute force approach would be writing down throw over point B, which is in a contradiction with the ophe optimal trajectory intersects with the timality of the original trajectory. Hence, the displacement nts and touches at one. This would be de- vector �r0 (τ ) − �r1 (τ ) needs to be parallel to the trajectory, i.e. rder algebraic equation and therefore, it is to the velocity �vB of the ball at point B. Here, �r0 (t) [�r1 (t)] mplish such a solution within a reasonable 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 r0 (t) − �r1 (t) = (�v0 − �v1 )t. So, — page 1 reference of 5 — one can easily see that � (�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 perpendicular 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 −

84

gx2 . sin2 α

2v 2

Point P with coordinates x = R sin α and z = R(cos α − 1) be2 longs to this parabola, hence R = gR 2v 2 , 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 .

small, so we can

psa − pw ps = Ta − T T

numerically T ≈ speed to the tem ergy conservatio provided by the a good approxim needs to be mod of air and the a sider one mole o V = RT /p. Ap cels are large, s negligible. Addi we can conclude ment of a tube physical quantit end — by index the tube at one inflow carries in out 12 µv22 . The i gas equal to p1 V

and the launching velocity are parallel. Let O be the origin, provided by the Bernoulli’s law. Applyin and let us denote ∠OQP = α. Then the trajectory is given by a good approximation of the reality, but needs to be modified to take into account gx2 z = x cot α − 2 2 . of air and the associated expansion/cont 2v sin α sider one mole of air, which has the mas Point P with coordinates x = R sin α and z = R(cos α − 1) be- V = RT /p. Apparently the process is f 2 longs to this parabola, hence R = gR , from where we obtain cels are large, so that heat transfer acro 2v 2 negligible. Additionally, the process is su the previous result. we can conclude that the process is adiab Part B. Mist (4 points) ment of a tube formed by the streamlines i. (0.8 pts) In the plane’s reference frame, along the channel physical quantities at its one end by index between two streamlines the volume flux of air (volume flow end — by index 2. Then, while one mo rate) is constant due to continuity. The volume flux is the the tube at one end, as much flows out at product of speed and channel’s cross-section area, which, due inflow carries in kinetic energy 21 µv12 , and to the two-dimensional geometry, is proportional to the channel 1 2 width and can be measured from the Fig. Due to the absence of out 2 µv2 . The inflowing gas receives work wind, the unperturbed air’s speed in the plane’s frame is just v0 . gas equal to p1 V1 = RT1 , the outflowing So, upon measuring the dimensions a = 10 mm and b = 13 mm p2 V2 = RT2 . Let’s define molar heat capa (see Fig), we can write v0 a = ub and hence u = v0 ab . Since at Cp = µcp . The inflow carries in heat ener point P , the streamlines are horizontal where all the velocities outflow carries out CV RT2 . All together, 1 2 are parallel, the vector addition is reduced to the scalar addi- can be written as2 2 µv + Cp T2 = const.

1 2 a v a tion: the air’s ground speed vP = v0 − u = v0 (1 − b ) = 23 m/s. easily express ∆ 2 = C vcrit ( c2 − 1) = cp 1 2 ii. (1.2 pts) Although the dynamic pressure ρv is relatively streamline distance at the point Q, and fu 2

small, it gives rise to some adiabatic expansion and compression. 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 + 21 ρv 2 = const, the dynamic 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.

vcrit = c

where we have used c ≈ 4.5 mm and ∆T in reality, the required speed is probably s cause for a fast condensation, a considerab needed. However, within an order of magn remains valid.

Part C. Magnetic straws (4.5 points) i. (0.8 pts) Due to the superconducting walls, the magnetic field lines cannot cross the walls, so the flux is constant along the tube. For a closed contour inside the tube, there should be no circulation 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 ii. (1.2 pts) Let us consider the change of when the tube is stretched (virtually) by Note that the magnetic flux trough the tu change of flux would imply a non-zero ele and for a zero resistivity, an infinite curr Φ tion B = πr 2 . The energy density of the m

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 = psa r = 2.08 kPa. The relative change of the pressure of the saturated vapour is = (�v0 − �v1 )t. So, small, so we can linearize its temperature dependence: — page 2 of 5 —

oint O at the top mal trajctory, at e velocity is perre Q denotes the city. Hence, QP O be the origin, ctory is given by

psa − pw (1 − r)psa psb − psa =⇒ Ta − T = (Tb − Ta ) ; = Ta − T Tb − Ta 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 energy 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-

2cp ∆T ≈ 23 m a 2 − c2

85

sily see that �r0 (t) − �r1 (t) = (�v0 − �v1 )t. So, ⇒ �u � �vB ⇒ �v0 ⊥ �vB . hrowing a ball from the point O at the top notice that for the optimal trajctory, at t touches the building, the velocity is perhe radius vector QP (where Q denotes the nd to the launching velocity. Hence, QP elocity are parallel. Let O be the origin, OQP = α. Then the trajectory is given by

= x cot α −

gx2 . sin2 α

2v 2

nates x = R sin α and z = R(cos α − 1) be2 la, hence R = gR 2v 2 , from where we obtain

nts) plane’s reference frame, along the channel lines the volume flux of air (volume flow e to continuity. The volume flux is the d channel’s cross-section area, which, due al geometry, is proportional to the channel asured from the Fig. Due to the absence of ed air’s speed in the plane’s frame is just v0 . the dimensions a = 10 mm and b = 13 mm ite v0 a = ub and hence u = v0 ab . Since at ines are horizontal where all the velocities tor addition is reduced to the scalar addid speed vP = v0 − u = v0 (1 − ab ) = 23 m/s. gh the dynamic pressure 12 ρv 2 is relatively o some adiabatic expansion and compresegions the temperature will drop and hence, rated vapours will also drop. If the dew ream of droplets will appear. This process where the adiabatic expansion is maximal, tatic pressure is minimal and consequently, e Bernoulli’s law p + 21 ρv 2 = const, the dyaximal: in the place where the air speed in mal and the streamline distance minimal. arked in Fig.

small, so we can linearize its temperature dependence: psa − pw (1 − r)psa psb − psa =⇒ Ta − T = (Tb − Ta ) ; = Ta − T Tb − Ta 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 energy 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. Consider one mole of air, which has the mass µ and the volume V = RT /p. Apparently the process is fast and the air parcels 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 segment 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 21 µv12 , and the outflow carries out 12 µv22 . The inflowing gas receives work due to the pushing gas equal to p1 V1 = RT1 , the outflowing gas performs work p2 V2 = 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 21 µv 2 + Cp T = const. From this we can 2 2 2 ( ac2 − 1) = cp ∆T , where c is the easily express ∆ v2 = C1 vcrit streamline distance at the point Q, and further vcrit = c

2cp ∆T ≈ 23 m/s, a 2 − c2

where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that in reality, the required speed is probably somewhat higher, because for a fast condensation, a considerable over-saturation is needed. However, within an order of magnitude, this estimate remains valid.

Part C. Magnetic straws (4.5 points) i. (0.8 pts) Due to the superconducting walls, the magnetic field lines cannot cross the walls, so the flux is constant along the tube. For a closed contour inside the tube, there should be no circulation 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 need to calculate the dew point for the air when the tube is stretched (virtually) by a small amount ∆l.

86 nt (since the relative pressure change will Note that the magnetic flux trough the tube is conserved: any nore the dependence of the dew point on change of flux would imply a non-zero electromotive force dΦ dt ,

on and compresl drop and hence, rop. If the dew ear. This process nsion is maximal, and consequently, = const, the dye the air speed in distance minimal.

point for the air ssure change will he dew point on psa r = 2.08 kPa. urated vapour is

2cp ∆T ≈ 23 m/s, a 2 − c2

vcrit = c

where we have used c ≈ 4.5 mm and ∆T = 1.5 K. Note that in reality, the required speed is probably somewhat higher, because for a fast condensation, a considerable over-saturation is needed. However, within an order of magnitude, this estimate remains valid.

Part C. Magnetic straws (4.5 points) i. (0.8 pts) Due to the superconducting walls, the magnetic field lines cannot cross the walls, so the flux is constant along the tube. For a closed contour inside the tube, there should be no circulation 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 inducΦ B 22 tion B = πr22 . The energy density of the magnetic field is 2µ0 .

0

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

∆W =

2

B Φ ∆l. πr2 ∆l = 2µ0 2µ0 πr2

This energy increase is achieved owing to the work done by the stretching force, ∆W = T ∆l. Hence, the force

For the two straws, we have four magnet gitudinal (along a straw axis) forces cance positioned pairs of same-sign-charges pus tions). The normal force is a superpositi due to the two pairs of opposite charges

the repulsive forces of diagonal pairs, F2 attractive force will be √ 4− 2 iii. (2.5 pts) Let us analyse, what would be the change of F = 2(F1 − F2 ) = 8πµ0 the magnetic energy when one of the straws is displaced to a small distance. The magnetic field inside the tubes will remain Alternative solution based on dip constant due to the conservation of magnetic flux, but outside, lation. It is known that the axial compon the magnetic field will be changed. The magnetic field out- induction created by a solenoidal current side the straws is defined by the following condition: there is is proportional to the solid angle Ω under � (because there are no currents outside the seen from the given point: no circulation of B straws); there are no sources of the field lines, other than the B� = µ0 jΩ/4π; 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- this can be easily derived from the Biottric field of four charges ±Q. We know that if the distance distance between the tubes be a (we’ll tak between charges is much larger than the geometrical size of and let us consider a first tube’s point wh a charge, the charges can be considered as point charges (the x (with 0 ≤ x ≤ l) from where the direc electric field near the charges remains almost constant, so that point of the other tube forms an angle α the respective contribution to the change of the overall electric the tube’s axis. From that point, the open 2 field energy is negligible). Therefore we can conclude that the other tube forms a solid angle Ω = πr sin endpoints of the straws can be considered as magnetic point its contribution to the axial magnetic field T =

Φ2 .

2µ0 πr2

charges. In order to calculate the force between two magnetic Φ sin µ0 jr2 sin2 α cos α = B� = charges (magnetic monopoles), we need to establish the corres4a2 pondence between magnetic and electric quantities.

For two electric charges Q separated by a distance a, the The solenoidal current at that point form 2 1 Q force is F = 4πε 2 , and at the position of one charge, the elec0 a 2 dm = πr2 jdx = Φdx/µ tric field of the other charge has energy density w = 32π12 ε0 Q a4 ; 87 hence we can write F = 8πwa2 . This is a universal expression which has potential energy for the force (for the case when the field lines have the same

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 correspondence between magnetic and electric quantities.

For two electric charges Q separated by a distance a, the 2 1 Q force is F = 4πε 2 , and at the position of one charge, the elec0 a

B� =

The solenoidal c

2

tric field of the other charge has energy density w = 32π12 ε0 Q 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 electric 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 1 Φ magnetic flux Φ at a distance a, the induction B = 4π a2 . So, 1 Φ2 B2 the energy density w = 2µ0 = 32π2 µ0 a4 , hence F =

the magnetic energy is calculated as

=

distance between and let us consid x (with 0 ≤ x ≤ point of the oth the tube’s axis. other tube forms its contribution

2

2

B Φ πr2 ∆l = ∆l.

2µ0 2µ0 πr2

is achieved owing to the work done by the W = T ∆l. Hence, the force

which has potent

dU =

When integratin U1 =

U=

Upon taking der same result for F

1 Φ2 .

4πµ0 a2

For the two straws, we have four magnetic charges. The longitudinal (along a straw axis) forces cancel out (the diagonally positioned pairs of same-sign-charges push in opposite directions). The normal force is a superposition of the attraction 1 Φ2 due to the two pairs of opposite charges, F1 = 4πµ 2 , and 0 l √

B� =

dU

Bearing in mind tributes the sam we find

2

Φ the repulsive forces of diagonal pairs, F2 = 8πµ20 2l 2 . The net — page 3 of 5 — Φ2 attractive force will be .

T = 2µ0 πr2 √ 4 − 2 Φ2 us analyse, what would be the change of F = 2(F1 − F2 ) = .

8πµ0 l2 when one of the straws is displaced to a magnetic field inside the tubes will remain Alternative solution based on dipole energy calcuconservation of magnetic flux, but outside, lation. It is known that the axial component of the magnetic will be changed. The magnetic field out- induction created by a solenoidal current of surface density j fined by the following condition: there is is proportional to the solid angle Ω under which the current is because there are no currents outside the seen from the given point: o sources of the field lines, other than the B� = µ0 jΩ/4π; aws; each of the endpoints of the straws is nes with a fixed magnetic flux ±Φ. These e condition as those which define the elec- this can be easily derived from the Biot-Savart law. Let the arges ±Q. We know that if the distance distance between the tubes be a (we’ll take derivative over a), much larger than the geometrical size of and let us consider a first tube’s point which has a coordinate s can be considered as point charges (the x (with 0 ≤ x ≤ l) from where the direction to the one ende charges remains almost constant, so that point of the other tube forms an angle α = arctan a/x with bution to the change of the overall electric the tube’s axis. From that point, the open circular face of the 2 2 2 ible). Therefore we can conclude that the other tube forms a solid angle Ω = πr sin α cos α/a , so that its contribution to the axial magnetic field at the point x aws can be considered as magnetic point

calculate the force between two magnetic 88 onopoles), we need to establish the corresmagnetic and electric quantities.

Φ sin2 α cos α µ0 jr2 sin2 α cos α = .

4a2 4πa2

agnetic field outndition: there is rents outside the , other than the s of the straws is flux ±Φ. These h define the elect if the distance ometrical size of oint charges (the constant, so that e overall electric onclude that the magnetic point en two magnetic ablish the correstities. distance a, the charge, the elec-

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� = µ0 jΩ/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 endpoint 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� =

The solenoidal current at that point forms a magnetic dipole dm = πr2 jdx = Φdx/µ0 ,

2

y w = 32π12 ε0 Q a4 ; versal expression s have the same by modulus elec, and not related to the magnetic s a derivative of displacement of e); if the energy t one point, they ull field energies. source of a fixed 1 Φ n B = 4π a2 . So, e

Φ sin2 α cos α µ0 jr2 sin2 α cos α = .

2 4a 4πa2

which has potential energy dU = B� dm = sin2 α cos αΦ2 dx/4πµ0 a2 .

When integrating over x, α varies from arctan a/l to 0, so that U1 =

dU =

sin2 α cos αΦ2 dx = 4πµ0 a2

cos αΦ2 dα .

4πµ0 a

Bearing in mind that the other end-circle of the other tube contributes the same amount to the magnetic interaction energy, we find

1 Φ2 1 . −√ U = 2U1 = 2πµ0 a a2 + l 2 Upon taking derivative over a and using a = l we obtain the same result for F as previously.

Problem T2.

Kelvin water dropper (8 points) Problem T2. Kelvin water dropper (8 points)

— page 3 Part of 5 — 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 3 ρg pressure ∆p inside the tube. So, the gravity force 43 πrmax 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,

3σd . rmax = 3 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 1 Q . So, droplet’s potential is ϕ; on the other hand, it is 4πε 0 r

where the droplet’s capacitance Cd = 4πε0 dWel = ϕmax dq = 4πε0 ϕ2max dr. Putting d equation for ϕmax , which recovers the earl Part B. Two pipes (4 points) i. (1.2 pts) This is basically the same that the surroundings’ potential is that electrode, −U/2 (where U = q/C is the and droplet has the ground potential (0). which electrode is the positive one, oppo tential may be chosen, if done consistent the cylindrical electrode is long, it shield vironment’s (ground, wall, etc) potential surroundings, the droplet’s potential is U of Part A we obtain 89

Q = 2πε0 U rmax = 2πε0 qrm

elvin water dropper (8 points)

(4 points) write the force balance for the droplet. neglect the force π4 ∆pd2 due to the excess 3 the tube. So, the gravity force 43 πrmax ρg apillary force. When the droplet separates water surface forms in the vicinity of the ch has vertical tangent. In the horizontal “neck”, the capillary force is vertical and πσd. So,

3σd . rmax = 3 4ρg

d ≪ r, we can neglect the change of the 90 e due to the tube. On the one hand, the ϕ; on the other hand, it is 1 Q . So,

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 3 pressure ∆p inside the tube. So, the gravity force 43 πrmax ρg 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,

3σd . rmax = 3 4ρg

equation for ϕma Part B. Two pip i. (1.2 pts) Th that the surrou electrode, −U/2 and droplet has which electrode tential may be c the cylindrical e vironment’s (gro surroundings, th ii. (1.2 pts) Since d ≪ r, we can neglect the change of the of Part A we obt droplet’s capacitance due to the tube. On the one hand, the 1 Q Q droplet’s potential is ϕ; on the other hand, it is 4πε . So, 0 r Q = 4πε0 ϕr.

ii. (1.5 pts) Th iii. (1.6 pts) Excess pressure inside the droplet is caused by that of the capa the capillary pressure 2σ/r (increases the inside pressure), and the farther electr by the electrostatic pressure 21 ε0 E 2 = 12 ε0 ϕ2 /r2 (decreases the it will increase th pressure). So, the sign of the excess pressure will change, if dq = 1 2 2 2 ε0 ϕmax /r = 2σ/r, hence

where dN = nd ϕmax = 2 σr/ε0 . the time dt This The expression for the electrostatic pressure used above can is solved easily t 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, ¯ is the field at the site without the field created by the q = q0 e where E surface charge element itself. Note that this force is perpendicular to the surface, so F/S can be interpreted as a pressure. iii. (1.3 pts) T The surface charge gives rise to a field drop on the surface equal anical energy m to ∆E = σ/ε0 (which follows from the Gauss law); inside the enough to overco droplet, there is no field due to the conductivity of the droplet: at the point wher ¯ − 1 ∆E = 0; outside the droplet, there is field E = E¯ + 1 ∆E, E 2 2 the potential U/2 ¯ = 1 E = 1 ∆E. Bringing everything together, we therefore E 2 2 potential −U/2. obtain the expression used above. field, so it needs Note that alternatively, this expression can be derived by tential −U/2, res considering a virtual displacement of a capacitor’s surface and equal to U Q ≤ m comparing the pressure work p∆V with the change of the electrostatic field energy 12 ε0 E 2 ∆V .

Finally, the answer to the question can be also derived from the requirement that the mechanical work dA done for an infinitesimal droplet inflation needs to be zero. From the en∴ Um ergy conservation law, dW + dWel = σ d(4πr2 ) + 12 ϕ2max dCd , where the droplet’s capacitance Cd = 4πε0 r; the electrical work dWel = ϕmax dq = 4πε0 ϕ2max dr. Putting dW = 0 we obtain an equation for ϕmax , which recovers the earlier result.

Part B. Two pipes (4 points) 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 potential may be chosen, if done consistently. Note that — since page 4 of 5 — the cylindrical electrode is long, it shields effectively the environment’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 = 2πε0 U rmax = 2πε0 qrmax /C.

8 points)

for the droplet. due to the excess 3 ρg force 43 πrmax roplet separates e vicinity of the n the horizontal e is vertical and

he change of the e one hand, the 1 Q . So, is 4πε 0 r

where the droplet’s capacitance Cd = 4πε0 r; the electrical work dWel = ϕmax dq = 4πε0 ϕ2max dr. Putting dW = 0 we obtain an equation for ϕmax , which recovers the earlier result.

Part B. Two pipes (4 points) 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 potential may be chosen, if done consistently. Note that since the cylindrical electrode is long, it shields effectively the environment’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 = 2πε0 U rmax = 2πε0 qrmax /C.

ii. (1.5 pts) The sign of the droplet’s charge is the same as plet is caused by that of the capacitor’s opposite plate (which is connected to e pressure), and the farther electrode). So, when the droplet falls into the bowl, r2 (decreases the it will increase the capacitor’s charge by Q: e will change, if q dq = 2πε0 U rmax dN = 2πε0 rmax ndt , C

e used above can ting on a surface ¯ n by F = σS · E, d created by the force is perpened as a pressure. the surface equal law); inside the ty of the droplet: d E = E¯ + 21 ∆E, ing together, we

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

πε0 n 3 6σd 2πε0 rmax n γt = . q = q0 e , γ = C C ρg

iii. (1.3 pts) The droplets can reach the bowls if their mechanical 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 pon be derived by tential −U/2, resulting in the change of the electrostatic energy tor’s surface and equal to U Q ≤ mgH, where Q = 2πε U r 0 max (see above). So, ange of the elecmgH Umax = , 2πε0 Umax rmax lso derived from A done for an in

Hσd H 3 gσ 2 ρd2 . From the en= 6 . ∴ Umax = 2 ) + 12 ϕ2max dCd , 2ε0 rmax 6ε30

91

Problem T3. protostar formation (9 points) Problem T3. Protostar formation (9 points)

be sketched as â&#x2C6;&#x161; 2E = Ď&#x2026;, one ge â&#x2C6;&#x17E; tâ&#x2C6;&#x17E; = 4Gm (Ď&#x2026; 0

1 = 3 4Ď&#x2026; 0

i. (0.8 pts) T = const =â&#x2021;&#x2019; pV = const V â&#x2C6;? r3 â&#x2C6;´ p â&#x2C6;? râ&#x2C6;&#x2019;3 =â&#x2021;&#x2019;

p(r1 ) = 23 = 8. p(r0 )

ii. (1 pt) During the period considered the pressure is negligible. Therefore the gas is in free fall. By Gaussâ&#x20AC;&#x2122; 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,

2(r0 â&#x2C6;&#x2019; r2 ) tâ&#x2030;&#x2C6; g where

Here (after shift dΞ 1 Ξ 2 = â&#x2C6;&#x2019; Ξ , final iv. (1.7 pts) By p=

mRT0 ÂľV

Work done by gr W =â&#x2C6;&#x2019;

pd

The temperature change; hence, a compression wor v. (1 pt) The c

Gm , r2 0

2r02 (r0 â&#x2C6;&#x2019; r2 ) 0.1r03 â&#x2C6;´tâ&#x2030;&#x2C6; = .

Gm Gm gâ&#x2030;&#x2C6;

iii. (2.5 pts) Gravitationally the outer layer of the ball is influenced 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 ultraelliptical 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 Keplerâ&#x20AC;&#x2122;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 Keplerâ&#x20AC;&#x2122;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:

2

r03 r0 2Ď&#x20AC; Gm =â&#x2021;&#x2019; trâ&#x2020;&#x2019;0 = Ď&#x20AC; = .

2 2trâ&#x2020;&#x2019;0 2 (r0 /2) 8Gm

pV â&#x2C6;´ â&#x2C6;´

vi. (2 pts) Duri verted into heat. ergy can be estim (exact calculatio minal heat ener cV m Âľ T4 (the app of the previous R m â&#x2C6;&#x2020;Q = Îłâ&#x2C6;&#x2019;1 Âľ T4 â&#x2030;&#x2C6;

the result of the initial full energ obtain

Gm2 m â&#x2030;&#x2C6; R r4 Âľ

92

Alternatively, one may write the energy conservation law Therefore, Gm rË&#x2122; 2 = E (that in turn is obtainable from Newtonâ&#x20AC;&#x2122;s 2 â&#x2C6;&#x2019; r Gm II law r¨ = â&#x2C6;&#x2019; Gm r 2 ) with E = â&#x2C6;&#x2019; r0 , separate the variables

dr

2Gm

dr

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 Keplerâ&#x20AC;&#x2122;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:

2

r03 r0 2Ď&#x20AC; Gm =â&#x2021;&#x2019; trâ&#x2020;&#x2019;0 = Ď&#x20AC; = .

2trâ&#x2020;&#x2019;0 2 (r0 /2)2 8Gm

ergy can be estimated as â&#x2C6;&#x2020;Î = â&#x2C6;&#x2019;Gm (r4 (exact calculation by integration adds a p minal heat energy is estimated as â&#x2C6;&#x2020;Q cV m Âľ T4 (the approximation T4 â&#x2030;Ť T0 foll of the previous question, when combined m R m â&#x2C6;&#x2020;Q = Îłâ&#x2C6;&#x2019;1 Âľ T4 â&#x2030;&#x2C6; Âľ RT4 . For the temper

the result of the previous question, T4 = initial full energy was approximately zero obtain Gm2 m â&#x2030;&#x2C6; RT0 r4 Âľ

points)

8.

r3 r4

3Îłâ&#x2C6;&#x2019;3

Alternatively, one may write the energy conservation law Therefore, 3Îłâ&#x2C6;&#x2019;3

Gm rË&#x2122; 2 = E (that in turn is obtainable from Newtonâ&#x20AC;&#x2122;s RT0 r3 4â&#x2C6;&#x2019;3Îł 2 â&#x2C6;&#x2019; r T4 â&#x2030;&#x2C6; T0 Gm II law r¨ = â&#x2C6;&#x2019; Gm ÂľmG r 2 ) with E = â&#x2C6;&#x2019; r0 , separate the variables

dr 2Gm dr . Alternatively, one can obtain the resu ( dt = â&#x2C6;&#x2019; 2E + r ) and write the integral t = â&#x2C6;&#x2019; â&#x2C6;&#x161; 2E+ 2Gm r Gm This integral is probably not calculable during the limitted equating the hydrostatic pressure Ď r4 r42 Ď time given during the Olympiad, but a possible approach can p4 = Âľ RT4 ; the result will be exactly the = Ξ and be sketched as follows. Substituting 2E + 2Gm r â&#x2C6;&#x161; 2E = Ď&#x2026;, one gets â&#x2C6;&#x17E; dΞ tâ&#x2C6;&#x17E; = 2 4Gm â&#x20AC;&#x201D; page 5 of 5 â&#x20AC;&#x201D; (Ď&#x2026; 2 â&#x2C6;&#x2019; Ξ 2 ) 0

â&#x2C6;&#x17E; 1 Ď&#x2026; Ď&#x2026; 1 1 + dΞ. + + = 3 4Ď&#x2026; 0 (Ď&#x2026; â&#x2C6;&#x2019; Ξ)2 (Ď&#x2026; + Ξ)2 Ď&#x2026;â&#x2C6;&#x2019;Ξ Ď&#x2026;+Ξ

pressure is negliHere (after shifting the variable) one can use dΞ Ξ = ln Ξ and ussâ&#x20AC;&#x2122; theorem and dΞ 1 = â&#x2C6;&#x2019; , finally getting the same answer as by Keplerâ&#x20AC;&#x2122;s laws.

2 Ξ Ξ nt in the ball is iv. (1.7 pts) By Clapeyronâ&#x20AC;&#x201C;Mendeleyev law, mass closer to the er, while the ball mRT0 p= . n its surface does ÂľV f the outermost Work done by gravity to compress the ball is

3 0

W =â&#x2C6;&#x2019;

.

of the ball is inmpressed into a otion: the fall of eriod of an ultraa line; its foci are er of the ball (by e figure (instead ipse is depicted). onger semiaxis of semiaxis is r0 /2 s, the answer is adius r0 /2:

r03 Ď&#x20AC; .

8Gm

=â&#x2021;&#x2019; r4 â&#x2030;&#x2C6; r

p dV = â&#x2C6;&#x2019;

mRT0 Âľ

3 4 3 Ď&#x20AC;r3 4 3 3 Ď&#x20AC;r0

dV 3mRT0 r0 = ln .

V Âľ 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 Îł = const =â&#x2021;&#x2019; T V Îłâ&#x2C6;&#x2019;1 = const. â&#x2C6;´ T â&#x2C6;? V 1â&#x2C6;&#x2019;Îł â&#x2C6;? r3â&#x2C6;&#x2019;3Îł r 3Îłâ&#x2C6;&#x2019;3 3 â&#x2C6;´ T = T0 . r

vi. (2 pts) During the collapse, the gravitational energy is converted into heat. Since r3 â&#x2030;Ť r4 , The released gravitational energy can be estimated as â&#x2C6;&#x2020;Î = â&#x2C6;&#x2019;Gm2 (r4â&#x2C6;&#x2019;1 â&#x2C6;&#x2019; r3â&#x2C6;&#x2019;1 ) â&#x2030;&#x2C6; â&#x2C6;&#x2019;Gm2 /r4 (exact calculation by integration adds a prefactor 35 ); the terminal heat energy is estimated as â&#x2C6;&#x2020;Q = cV m Âľ (T4 â&#x2C6;&#x2019; T0 ) â&#x2030;&#x2C6; cV m Âľ T4 (the approximation T4 â&#x2030;Ť T0 follows from the result of the previous question, when combined with r3 â&#x2030;Ť r4 ). So, m R m â&#x2C6;&#x2020;Q = Îłâ&#x2C6;&#x2019;1 Âľ T4 â&#x2030;&#x2C6; Âľ RT4 . For the temperature T4 , we can use 3Îłâ&#x2C6;&#x2019;3 the result of the previous question, T4 = T0 rr43 . Since initial full energy was approximately zero, â&#x2C6;&#x2020;Q + â&#x2C6;&#x2020;Î â&#x2030;&#x2C6; 0, we

93

0

2r02 (r0 − r2 ) 0.1r03 = .

Gm Gm

compression work W is the heat radiated. v. (1 pt) The collapse continues adiabatically.

tationally the outer layer of the ball is injust as the rest were compressed into a ore we have Keplerian motion: the fall of r layer consists in a halfperiod of an ultraellipse is degenerate into a line; its foci are e; one focus is at the center of the ball (by d the other one is at r0 , see figure (instead se, a strongly elliptical ellipse is depicted). bit is determined by the longer semiaxis of er’s 3rd law). The longer semiaxis is r0 /2 ed in half a period. Thus, the answer is od of a circular orbit of radius r0 /2:

r03 Gm =⇒ tr→0 = π = .

2 (r0 /2) 8Gm

pV γ = const =⇒ T V γ−1 = const. ∴ T ∝ V 1−γ ∝ r3−3γ r 3γ−3 3 ∴ T = T0 . r vi. (2 pts) During the collapse, the gravitational energy is converted into heat. Since r3 ≫ r4 , The released gravitational energy can be estimated as ∆Π = −Gm2 (r4−1 − r3−1 ) ≈ −Gm2 /r4 (exact calculation by integration adds a prefactor 35 ); the terminal heat energy is estimated as ∆Q = cV m µ (T4 − T0 ) ≈ cV m µ T4 (the approximation T4 ≫ T0 follows from the result of the previous question, when combined with r3 ≫ r4 ). So, m R m ∆Q = γ−1 µ T4 ≈ µ RT4 . For the temperature T4 , we can use 3γ−3 the result of the previous question, T4 = T0 rr43 . Since initial full energy was approximately zero, ∆Q + ∆Π ≈ 0, we obtain Gm2 m ≈ RT0 r4 µ

r3 r4

3γ−3

=⇒ r4 ≈ r3

RT0 r3 µmG

1 3γ−4

.

may write the energy conservation law Therefore, 3γ−3

at in turn is obtainable from Newton’s RT0 r3 4−3γ T ≈ T .

4 0 with E = − Gm µmG r0 , separate the variables

) and write the integral t = − √ dr 2Gm . Alternatively, one can obtain the result by approximately 2E+ r Gm bably not calculable during the limitted equating the hydrostatic pressure ρr4 r42 to the gas pressure ρ he Olympiad, but a possible approach can p4 = µ RT4 ; the result will be exactly the same as given above.

— page 5 of 5 —

94

Grading scheme: Theory The 43rd International Physics Olympiad â&#x20AC;&#x201D; 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 donâ&#x20AC;&#x2122;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 bearing 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 incorrectly, 0.1 is subtracted for a minor mistake (eg. missing nondimensional factor); no mark is given if the mistake is major national Physics July 2012The same rule is applied (withOlympiad non-matching â&#x20AC;&#x201D; dimensionality). to theTheory marking of mathematical calculations: each minor misGrading scheme: take leads to a subtraction of 0.1 pts (as long as the remaining s the number of score for that particular calculation remains positive), and no formula. These marks are given in the case of dimensional mistakes. if a formula is No penalty is applied in these cases when a mistake is clearly marks of all the just a rewriting typo (i.e. when there is no mistake in the draft). la is not written If formula is written without deriving: if it is simple are written bear- enough to be derived in head, full marks, otherwise zero diagram), marks marks. are allotted for If there two solutions on Solution sheets, one correct and another incorrect: only the one wich corresponds to the Ans written incor- swer Sheets is taken into account. What is crossed out is eg. missing non- never considered. mistake is major No penalty is applied for propagating errors unless the calme rule is applied culations are significantly simplified (in which case mathemateach minor mis- ical calculations are credited partially, according to the degree as the remaining of simplification, with marking granularity of 0.1 pts).

score for that particular calculation rema marks are given in the case of dimensiona No penalty is applied in these cases whe just a rewriting typo (i.e. when there is no If formula is written without derivi enough to be derived in head, full ma marks.

If there two solutions on Solution she another incorrect: only the one wich cor swer Sheets is taken into account. Wh never considered.

No penalty is applied for propagating culations are significantly simplified (in w ical calculations are credited partially, acc of simplification, with marking granularity

95

Problem T1.

Focus on sketches (13 points) Problem T1. Focus on sketches (13 points)

(full 0.6 pts if eq

Part A. Ballistics (4.5 points) i. (0.8 pts) If the parameters are derived using the particular cases of throwing up and throwing horizontally: for throwing up, zg = v02 /2 — 0.2 pts; from where z0 = v02 /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 /2v02 — 0.2 pts; Concluding k = g/2v02 — 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:

If alternative so

Shifting the Noting that the

Angle from ce

For a brute f

distance is max. when the angle is 45◦ — 0.2 Finding this maximal distance v 2 /g — 0.2

which is

Mentioning t (equivalently, an If the parameters are derived using condition that the quad- are exactly two d ratic equation for the tangent of the throwing angle has exactly one real solution: Obtaining qua Requiring x = v cos αt — 0.1 pts; Obtaining k — 0.1

Requiring z = v sin αt − gt2 /2 — 0.2 pts; Eliminating t: z = x tan α − gx2 /v 2 cos2 α — 0.1 pts;

Requiring t

Obtaining z − x tan α + gx2 /v 2 (1 + tan2 α) = 0 — 0.1 pts;

Obta

Requiring that the discriminant is 0 — 0.2 pts;

Find

Obtaining z0 = v02 /2g, k = gx2 /2v02 — 0.1 pts; (If one of the two is incorrect — 0 pts for the last line)

Part B. Air flow i. (0.8 pts)

Usi ii. (1.2 pts)

96

taking stream (if off by 1 mm o

Trajectory hits the sphere when descending — 0.7 pts measuring u (if the top of the parabola is higher than 25 R — 0.5 pts); (if off by 1 mm o Trajectory touches the sphere when ascending — 0.5 pts. at P , stream Trajectory touches the sphere at its top or is clearly non-parabolic or starts inside the sphere or intersects the sphere: total — 0 pts. (if the speed in iii. (2.5 pts) If the analysis is based on a trajectory which is are lost for 4th wrong in principle, 0 pts.

For minimal speed, z = z0 − kx2 touches the sphere — 1.5 pts ii. (1.2 pts)

Eliminating t: z = x tan α − gx /v cos α — 0.1 pts;

Requiring that x∗ is the root of the

Obtaining z − x tan α + gx2 /v 2 (1 + tan2 α) = 0 — 0.1 pts;

Obtaining the speed as a funct

Requiring that the discriminant is 0 — 0.2 pts; Obtaining z0 = v02 /2g, k = gx2 /2v02 — 0.1 pts; (If one of the two is incorrect — 0 pts for the last line)

Finding the minimum of that

Part B. Air flow around a wing (4 point i. (0.8 pts)

Using the wing’s frame of ref ii. (1.2 pts)

taking streamline distance measurmen (if off by 1 mm or more, 0.1 pts)

Trajectory hits the sphere when descending — 0.7 pts measuring unperturbed streamline di (if the top of the parabola is higher than 25 R — 0.5 pts); (if off by 1 mm or more, 0 pts) Trajectory touches the sphere when ascending — 0.5 pts. at P , streamlines are horiz. ⇒ scalar a Trajectory touches the sphere at its top or Writing continuity con is clearly non-parabolic or starts inside the sphere Finding final a or intersects the sphere: total — 0 pts. (if the speed in wing’s frame is given as iii. (2.5 pts) If the analysis is based on a trajectory which is are lost for 4th and 6th line) wrong in principle, 0 pts.

For minimal speed, z = z0 − kx2 touches the sphere — 1.5 pts ii. (1.2 pts) Shifting the origin to the sphere’s top for simplicity — 0.1 pts

Writing down continuity cond (or stating: smaller streamline distance ⇒ gx2 v02 − 2 — 0.2 pts Then z = 2g 2v0 Writing Bernoulli’ 2 2 0 pts out of 0.4 if the Bernoulli law includ and x + z + 2zR = 0 — 0.2 pts 2

2 2 (or stating that larger speed ⇒ smaller pr v v ⇒ x4 2vg 2 + x2 12 − gR + 4g0 + R g0 = 0 — 0.1 pts 2 v 0 0 Writing adiabati

points)

— page 2 of 5 —

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

Discriminant equals 0 — 0.2 pts

ng the particular y:

2 0 /2

— 0.2 pts;

/2g — 0.1 pts;

kx2 — 0.2 pts;

2v02 — 0.2 pts;

2v02 — 0.1 pts;

f a horizontally d), a trajectory

from where v02 = 0.5gR — 0.1 pts

hence vmin = 4.5gR — 0.1 pts If alternative solution is followed: Shifting the origin to the sphere’s top for simplicity — 0.1 pts 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:

— 0.2

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

g — 0.2

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

◦

Mentioning that it has exactly one pos. root — 0.2 pts k — 0.1 (equivalently, an extrememum coincides with a root or there on that the quad- are exactly two distinct real roots.) angle has exactly

2

Obtaining quadratic eq. for x-coord. of extrema — 0.2 pts αt — 0.1 pts; /2 — 0.2 pts;

2

Finding the roots of it — 0.2 pts Selecting the larger root x∗ of it — 0.2 pts

α — 0.1 pts;

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

= 0 — 0.1 pts;

Obtaining the speed as a function of α — 0.2 pts

s 0 — 0.2 pts;

Finding the minimum of that function — 0.7 pts

97

for intersection points x (or y) — 0.5 pts

max. when the angle is 45◦ — 0.2

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

2

his maximal distance v /g — 0.2

Mentioning that it has exactly one pos. root — 0.2 pts Obtaining k — 0.1 (equivalently, an extrememum coincides with a root or there are derived using condition that the quad- are exactly two distinct real roots.) tangent of the throwing angle has exactly Obtaining quadratic eq. for x-coord. of extrema — 0.2 pts Requiring x = v cos αt — 0.1 pts;

Finding the roots of it — 0.2 pts

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

Selecting the larger root x∗ of it — 0.2 pts

z = x tan α − gx2 /v 2 cos2 α — 0.1 pts;

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

n α + gx2 /v 2 (1 + tan2 α) = 0 — 0.1 pts;

Obtaining the speed as a function of α — 0.2 pts

ng that the discriminant is 0 — 0.2 pts;

Finding the minimum of that function — 0.7 pts

ng z0 = v02 /2g, k = gx2 /2v02 — 0.1 pts; ncorrect — 0 pts for the last line)

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)

s the sphere when descending — 0.7 pts measuring unperturbed streamline distance he parabola is higher than 25 R — 0.5 pts); (if off by 1 mm or more, 0 pts) es the sphere when ascending — 0.5 pts. at P , streamlines are horiz. ⇒ scalar adding uches the sphere at its top or Writing continuity condition olic or starts inside the sphere Finding final answer r intersects the sphere: total — 0 pts.

— 0.1 pts — 0.1 pts — 0.2 pts — 0.1 pts

(if the speed in wing’s frame is given as final anser, points analysis is based on a trajectory which is are lost for 4th and 6th line) pts.

z = z0 − kx2 touches the sphere — 1.5 pts ii. (1.2 pts)

o the sphere’s top for simplicity — 0.1 pts

Writing down continuity condition — 0.2 pts (or stating: smaller streamline distance ⇒ larger speed) gx2 v02 − 2 — 0.2 pts Then z = 2g 2v0 Writing Bernoulli’s law — 0.4 pts and x2 + z 2 + 2zR = 0 — 0.2 pts 0 pts out of 0.4 if the Bernoulli law includes ρgh

2 2 (or stating that larger speed ⇒ smaller pressure) v v + 4g0 + R g0 = 0 — 0.1 pts x2 12 − gR v02 Writing adiabatic law — 0.4 pts that smaller pressure ⇒ lower temperature) — page 2 (or of 5stating — Finding Q as in the figure in the solutions — 0.2 pts (if Q marked below the wing’s tip — 0.1 pts)

(Introducing a le

E

(Same marks if W

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 21 µv 2 + cp T = const: 1 mole of gas carries kin. en. 21 µv 2 — 0.2 pts 1 mole of gas carries heat en. CV T — 0.2 pts

iii. (2.5 pts)

Idea

(Sketch with a q

work done on 1 mole of gas: p1 V1 − p2 V2 — 0.4 pts en.:

1 2 2 µ(v2

− v12 ) + CV (T2 − T1 ) = p1 V1 − p2 V2 — 0.2 pts

using ideal gas law obtaining 12 v 2 + cp T = const — 0.2 pts 98

Alternative approximate approach: Bernoulli’s law 21 ρv 2 + p = const — 0.3 pts

Finding the forc and magn. quan follows:

Expres

Using the ob

2µ0

(if Q marked below the wing’s tip — 0.1 pts)

(Same marks if W expressed for entire tub

Equating ∆W = T ∆ iii. (2 pts)

Expressing T =

Φ2 2µ0 πr

Finding the dew point: idea of linearization — 0.2 pts Expression and/or numerical value for dew point — 0.2 pts iii. (2.5 pts)

Deriving 21 µv 2 + cp T = const: 1 mole of gas carries kin. en. 12 µv 2 — 0.2 pts 1 mole of gas carries heat en. CV T — 0.2 pts

Idea of using analogy with el.

(Sketch with a quadrupole conf. of el. cha

work done on 1 mole of gas: p1 V1 − p2 V2 — 0.4 pts en.:

1 2 2 µ(v2

− v12 ) + CV (T2 − T1 ) = p1 V1 − p2 V2 — 0.2 pts

using ideal gas law obtaining 12 v 2 + cp T = const — 0.2 pts

Finding the force between two magn. cha and magn. quantities is worth 1 pts in follows:

Expressing el. stat. force via en.

Alternative approximate approach: Bernoulli’s law 12 ρv 2 + p = const — 0.3 pts 0 pts out of 0.3 if the Bernoulli law includes ρgh Adiabatic law pV ⇒p

1−γ

Leading to

+

= const — 0.3 pts

γ

T = const — 0.2 pts

Approximation ∆p = 1 2 2v

γ

R cp µ cp −cV

γ p γ−1 T

∆T — 0.2 pts

T = const — 0.1 pts

Leading to 12 v 2 + cp T = const — 0.1 pts

Using the obtained Eq. to obtain F =

Any other matching scheme is graded an ing such a Q which has the same en. den expressing the force between magn. char between the matching el. charges (Q) — 1 1 ↔ 4πµ w matching pairs Q ↔ Φ and 4πε 0 0 motivation gives only 0.5 pts.

Noting that � to tubes comp. of fo

When expressing F = 2(F1 − F2 ): for fa

And further (for either approach): 2

and for fin 2

2 Bringing it to ∆ v2 = 12 vcrit ( ac2 − 1) = cp ∆T — 0.1 pts

Obtaining vcrit

(if wrong sign for F2 , subtract 0.1)

Measuring a and c — 0.2 pts

2cp ∆T =c ≈ 23 m/s — 0.1 pts a 2 − c2

Part C. Magnetic straws (4.5 points) i. (0.8 pts)

If alternative solution is followed

Lines straight and parallel inside — 0.6 pts

Plan to express inter. energy as a function

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

intending to find F as a deriva

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

Calculating B

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)

If estimated without dependance on x, 0.4

considering a tube as an array of dip

expressing dm = S

relating j

Expressing U = B(x

ii. (1.2 pts) Expressing induction as B = Φ/πr2 — 0.2 pts Stating that w =

B2 2µ0

If estimated as BSjl, 0.2

— 0.2 pts

Finding F = dU

Idea of using virtual lengthening — 0.2 pts

Taking into account the factor

— page 3 of 5 —

99

ler pressure ⇒ lower temperature)

the figure in the solutions — 0.2 pts the tip⇒—lower 0.1 pts) ler wing’s pressure temperature)

the figure in the solutions — 0.2 pts the wing’s tip — 0.1 pts)

(Introducing a lengthening ∆l is enough) 2

B πr2 ∆l — 0.2 pts Expressing ∆W = 2µ 0 (Introducing ∆lfor is entire enough) (Same marks aiflengthening W expressed tube) B2 2 πr pts Expressing ∆W = Equating ∆W ∆l — — 0.2 0.2 pts 2µ0= T ∆l 2 (Same marks if W expressed for Tentire . — 0.2 pts Expressing = Φtube) 2µ0 πr 2

ew point: idea of linearization — 0.2 pts

Equating ∆W = T ∆l — 0.2 pts

numerical value for dew point — 0.2 pts

ew point: idea of linearization = const: numerical value for dew point le of gas carries kin. en. 12 µv 2 = const: le of gas carries heat en. CV T

— 0.2 pts — 0.2 pts — 0.2 pts

1 2 le of gas carries on 1 mole of gas:kin. p1 Ven. 1 − 2pµv 2 V2

0.2 pts — 0.4

— 0.2 pts

le heat + of CVgas (T2 carries − T1 ) = p1 Ven. p2VVT2 — 0.2 pts 1 −C

2 on 1 mole of12 vgas: p2 V2 — 0.4 obtaining + cpp1TV1=−const 0.2 pts

+ CV (T2 − T1 ) = p1 V1 − p2 V2 — 0.2 pts mate approach: obtaining 12 v 2 + cp T = const — 0.2 pts rnoulli’s law 12 ρv 2 + p = const — 0.3 pts mate approach: e Bernoulli law includes ρgh

ρv 2pV + γp = const — 0.3 pts rnoulli’s law 21law Adiabatic

e Bernoulli⇒ law includes ρgh — 0.2 pts T γ = const p1−γ

γ p Adiabatic law = const — 0.3 pproximation ∆ppV = γγ−1 0.2 pts T ∆T

cp T γ = const — 0.2 pts p1−γ ng to 12 v 2 +⇒R µ cp −cV T = const — 0.1 pts γ p pproximation T ∆T — + c=p Tγ−1 = const — 0.2 0.1 pts pts Leading to 12 v 2∆p c p ng to 21 v 2 + R µ cp −cV T = const — 0.1 pts her approach): Leading to 12 v 2 + cp T = const — 0.1 pts 2 2 2 ∆ v2 = 12 vcrit ( ac2 − 1) = cp ∆T — 0.1 pts her approach): Measuring a and c — 0.2 pts a2 v2 1 2 ( c2p ∆T − 1) = cp ∆T — 0.1 pts ∆ 2 = 2 vcrit2c ng vcrit = c ≈ 23 m/s — 0.1 pts a2 − c2 a and c — 0.2 pts Measuring

traws (4.5 points) 2cp ∆T ng vcrit = c ≈ 23 m/s — 0.1 pts a 2 − c2

traws (4.5 points) straight and parallel inside — 0.6 pts

he entire length, subtract 0.2)

straight andtube’s parallel inside — 0.6 pts ed close the end)

he entire length, subtract 0.2) wards slightly after the exit — 0.2 pts

oreclose thanthe onetube’s closedend) loop on both sides is ed s out of 0.2) wards slightly after the exit — 0.2 pts

ore than one closed loop on both sides is s out of 0.2)

nduction as B = Φ/πr2 — 0.2 pts 2

B — 0.2 pts Stating that w = 2µ 0 2 nduction as B = Φ/πr sing virtual lengthening — — 0.2 0.2 pts pts 100 B2 Stating that w = 2µ0 — 0.2 pts

Expressing T =

iii. (2.5 pts)

Φ2 2µ0 πr 2 .

— 0.2 pts

Idea of using analogy with el. charges — 1 pt iii. (2.5 pts) (Sketch with a quadrupole conf. of el. charges is enough) Idea of using analogy with el. charges — 1 pt (Sketch a quadrupole of el. charges via is enough) Finding with the force between conf. two magn. matching el. and magn. quantities is worth 1 pts in total, split down as follows: Finding the force between two magn. charges via matching el. force via in en.total, density — 0.5 ptsas and magn.Expressing quantitiesel.is stat. worth 1 pts split down 1 Φ2 follows: Using the obtained Eq. to obtain F = — 0.5 pts 4πµ0 a2

stat. is force via en. density — e.g.

0.5 pts Any otherExpressing matching el. scheme graded analogously; find2 1 Φ ing Using such athe Q which hasEq. the to same en. Fdensity 0.5pts pts; obtained obtain = 4πµ0asa2Φ——0.5 expressing the force between magn. charges (Φ) as the force Any other matching scheme is graded analogously; e.g. findbetween the matching el. charges (Q) — 0.5 pts. Declaring ing such a Q which has the same en.1 density as Φ — 0.5 pts; 1 ↔ 4πµ without energy-basedmatching pairs Q ↔ Φ and 4πε 0 0 expressing the force between magn. charges (Φ) as the force motivation gives only 0.5 pts. between the matching el. charges (Q) — 0.5 pts. Declaring 1 comp.1 of force = 0 — 0.2 pts Noting that tubes matching pairs Q ↔ �Φto and 4πε0 ↔ 4πµ0 without energy-basedmotivation gives only 0.5 pts.

When expressing F = 2(F1 − F2 ): for factor “2” — 0.1 pts Noting that � to tubes comp. force =F02 — 0.2 pts and foroffinding F2 ): for factor “2” — 0.1 pts = 2(F1 −0.1) (ifWhen wrongexpressing sign for F2F, subtract and for finding F2 — 0.2 pts (if wrong sign for F2 , subtract 0.1)

If alternative solution is followed 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 Plan to express inter. energyCalculating as a function of a— 0.8 pts B� (x) to find F ason a derivative — 0.2 pts If estimatedintending without dependance x, 0.4 Calculating B� (x) — considering a tube as an array of dipoles — 0.8 0.3 pts pts If estimated without dependance x,=0.4 expressingon dm Sjdx — 0.3 pts considering a tube as an array of dipoles 0.3 pts relating j to Φ — 0.2

expressing = Sjdx — 0.4 0.3 pts Expressing U =dmB(x)dm

relating j to Φ — If estimated as BSjl, 0.2

Expressing U — Finding=F =B(x)dm dU/da — If estimatedTaking as BSjl, into0.2 account the factor “ 2”—

0.2 pts

0.4 pts pts 0.2

0.1 pts

Finding F = dU/da — 0.2 pts

If alternative solution is followed

de — 0.6 pts

Plan to express inter. energy as a function of a

0.2)

intending to find F as a derivative — 0.2 pts Calculating B� (x) — 0.8 pts

it — 0.2 pts

on both sides is

— 0.2 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

— 0.2 pts

Finding F = dU/da — 0.2 pts

— 0.2 pts

Taking into account the factor “ 2”— 0.1 pts

— page 3 of 5 —

Problem T2.

Kelvin water dropper (8 points) Problem T2. Kelvin water dropper (8 points)

(if not stated but used correctly — full m are allowed to be chosen consistently) Part A. Single pipe (4 points) i. (1.2 pts) For the terms entering the force balance of a stating that the droplet’s pote droplet immediately before separation from the nozzle, the (if not stated but used correctly — full ma points are given as follows: Applying formula mg — 0.2 pts; Using the result of Part A with m = ρV — 0.1 pts; ϕ = U/2 to obtain the fin 4 3 — 0.1 pts; V = πrmax the solutions with U instead of U/2 wil 3 for the first two lines (i.e. in total up to πσd — 0.4 pts.

(if geometrically obtained cos α is included, 0.2 pts) Force balance equation including these terms — 0.2 pts;

ii. (1.5 pts)

For expressing rmax from the equation — 0.2 pts.

Stating that a droplet will increase the ca

ii. (1.2 pts)

charge by its own charge Q = 2πε0 qrm

Stating that the droplet’s potential is ϕ — 0.2 pts (if used correctly, this does not need to be explicitly stated).

1 Q — 0.8 pts Expressing the droplet’s potential as 4πε0 r (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;

(0 pts if wrong sign) (0.2 pts if redundant factor “2”) Expressing dq =

Substituting dN =

solving the obtained differential equ

determining the integration constant from

the initial cond

substituting rmax from a

1 ε0 E 2 — 0.4 pts; 2

iii. (1.3 pts) Equation for the energy bal

1 ε0 ϕ2 /r2 — 0.2 pts.

2 Noticing that the two effects have opposite sign — 0.2 pts

expressing the droplet’s electrostatic e the f

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

expressing the droplet’s gravitational ene

bringing it to the form

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

101

ii. (1.2 Stating pts) that the droplet’s potential is ϕ — 0.2 pts (if used Stating correctly, thisthe does not need to be explicitly stated). that droplet’s potential is ϕ — 0.2 pts 1 Q (if Expressing used correctly, does not need to stated).

— 0.8 pts the this droplet’s potential as be explicitly 4πε0 r 1 Q — 0.8 Expressing the droplet’s(without potentialcorrect as 0.6 pts pts).

4πεsign r 0

Expressing Q from the obtained equation — 0.6 0.2 pts). pts. (without correct sign —

(0 pts if wrong charge by its

(0.2pts pts redun (0 if ifwrong

(0.2 pts if redun

solving th

determining solvingthe th Expressing from the obtained equation — 0.2 iii. (1.6 pts) TheQ components of the excess pressure are pts. graded determining the as follows. iii. (1.6 pts) The components of the excess pressure are graded as follows.

2σ/r — 0.5 pts; 1 iii. (1.3 pts) Eq 2 ε2σ/r 0.4 pts; 0 E — 0.5 2 iii. expressing (1.3 pts) Eq 1 1 E 22 — 0.4 pts; th — 0.2 pts. bringing it to the form ε02ϕε20/r 2 1 expressing th 2 2 ε0 ϕsign /r — —0.2 0.2pts pts. bringing it to have the form Noticing that the two effects opposite 2 expressing the (ifNoticing used correctly, nothave needopposite to be stated that thethis twodoes effects sign separately).

— 0.2 pts Equation stating pressure is 0separately).

— 0.1 pts; expressing the (if used correctly, thisthat doesthe notexcess need to be stated noticing that Expressing ϕmax that fromthe theexcess obtained equation — 0.1 0.2 pts; pts.

Equation stating pressure is 0 — noticing that expressin from the obtainedsolution, equation the — 0.2 pts.

InExpressing the case ofϕmax energy-balance-based distribution of marks is as follows.

In the case of energy-balance-based solution, the distribution of marksenergy is as follows. surface change as 4πσd(r2 ) = 8πσrdr — 0.5 pts; electrostatic energy change as 22πε ϕ2max dr ) =0 8πσrdr surface energy change as 4πσd(r 2 = 2πε 4πε00 ϕ electrostaticenergy work change as dAel as electrostatic ϕ2max dr dr max

— — — —

0.5 0.5 0.3 0.5

pts; pts; pts; pts;

equation stating thatwork en. as change work — 0.3 0.1 pts; pts; = 4πε0to ϕ2max dr — electrostatic dAel equals from obtained expressing ϕmax equation stating that en. the change equalsequation to work — — 0.2 0.1 pts. pts;

elvin water dropper (8 points)

expressin (if bowl con

(if bowl con

If instead of (which is incorre If instead of equation is given (which is incorre equation is given

from the(without obtainedvirtual equation — 0.2 pts. electric fi maxequated 0 ptsexpressing if energy ϕare displacement).

(0.1 If factor 12 missing before the electrostat. force (but otherelectric fi 0 pts if energy are equated (without virtual displacement). wise correct), -0.2 pts).

(0.1 If factor 12 missing before the electrostat. force (but otherw Part B. Two pipes (4 points) wise correct), -0.2 pts). i. (1.2 pts) w Part B. Two pipes (4 points) (if bowl conside i.Stating (1.2 pts) that the surroundings’ potential is − U/2 — 0.4 pts; (if bowl conside Stating that the surroundings’ potential is − U/2 — 0.4 pts; (if not stated but used correctly — full marks; opposite —signs page 4 of 5 — are allowed to be chosen consistently)

(4 points) — page 4 of 5 — he terms entering the force balance of a stating that the droplet’s potential is 0 — 0.4 pts y before separation from the nozzle, the (if not stated but used correctly — full marks) ollows: Applying formula U = q/C — 0.2 pts mg — 0.2 pts; Using the result of Part A with m = ρV — 0.1 pts; ϕ = U/2 to obtain the final result — 0.2 pts.

4 3 — 0.1 pts; V = πrmax the solutions with U instead of U/2 will qualify for 0.4 pts 3 for the first two lines (i.e. in total up to 0.8) πσd — 0.4 pts.

tained cos α is included, 0.2 pts)

uation including these terms — 0.2 pts;

ii. (1.5 pts)

sing rmax from the equation — 0.2 pts.

Stating that a droplet will increase the capacitor’s charge by its own charge Q = 2πε0 qrmax /C — 0.4 pts; the droplet’s potential is ϕ — 0.2 pts

is does not need to be explicitly stated).

102 1 Q — 0.8 pts oplet’s potential as 4πε0 r

(0 pts if wrong sign) (0.2 pts if redundant factor “2”) Expressing dq = Q dN — 0.2 pts;

mg — 0.2 pts;

Using the result of Part A with

V — 0.1 pts;

ax

— 0.1 pts;

σd — 0.4 pts.

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)

0.2 pts)

ms — 0.2 pts;

ii. (1.5 pts)

on — 0.2 pts.

Stating that a droplet will increase the capacitor’s charge by its own charge Q = 2πε0 qrmax /C — 0.4 pts;

ϕ — 0.2 pts

plicitly stated).

Q — 0.8 pts r n — 0.6 pts).

n — 0.2 pts.

essure are graded

(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.

σ/r — 0.5 pts;

0E

2

2

— 0.4 pts;

/r2 — 0.2 pts.

ign — 0.2 pts

ted separately). is 0 — 0.1 pts;

tion — 0.2 pts.

on, the distribu-

rdr — 0.5 pts;

x dr

— 0.5 pts;

x dr — 0.3 pts;

work — 0.1 pts;

ion — 0.2 pts.

displacement). orce (but other-

U/2 — 0.4 pts;

iii. (1.3 pts) Equation for the energy balance of a droplet: expressing the droplet’s electrostatic energy change during the fall as U Q — 0.6 pts (0.3 for U Q/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. expressing the final answer — 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 —

103

Problem T3. protostar formation (9 points) Problem T3. Protostar formation (9 points) i. (0.8 pts) In thermodynamic equilibrium T = const — 0.2 pts. (if not stated but used correctly — full marks)

[for mentioning sibly in a wrong

W =

(with either “+”

pV = const — 0.3 pts.

V ∝ r3 — 0.1 pts. p ∝ r−3 — 0.1 pts. p(r1 ) = 8 — 0.1 pts. p(r0 )

Calculating t v. (1 pt)

ii. (1 pt) t≈

2(r0 − r2 ) — 0.4 pts. g

Gm g ≈ 2 — 0.4 pts. r 0 0.1r03 t≈ — 0.2 pts.

Gm iii. (2.5 pts) First solution: Understanding that we have effectively interaction 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)

Th (If used, but no

vi. (2 pts)

In case of using fewer approxima

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

Period of the elliptical orbit is equal to the period of the circular orbit of the same longer semiaxis — 0.4 pts. (if not stated but used correctly — full marks) 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.

r03 Final answer tr→0 = π — 0.1 pts.

8Gm Alternative solution: Understanding that we have effectively interaction 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)

104

Energy conservation as a differential equation — 0.3 pts.

2 GMm (if only expressed through v (like mv = E) — 0.1 pts.; 2 − r r = − Gm if the differential equation of Newton’s 2nd law (¨ r 2 ) is given instead — 0.2 pts.)

Final an

Final ans

In case of using p mitted):

(if not stated but used correctly â&#x20AC;&#x201D; full marks)

cV â&#x2030;&#x2C6;

The longer semiaxis is r0 /2 â&#x20AC;&#x201D; 0.1 pts.

Equation(s) for the period â&#x20AC;&#x201D; 0.3 pts.

We need half a period â&#x20AC;&#x201D; 0.1 pts.

r03 Final answer trâ&#x2020;&#x2019;0 = Ď&#x20AC; â&#x20AC;&#x201D; 0.1 pts.

8Gm

Final answer r4 â&#x2030;&#x2C6; r3

RT0 r3 ÂľmG

Final answer T4 â&#x2030;&#x2C6; T0

RT0 r3 ÂľmG

1 3Îłâ&#x2C6;&#x2019;

3Îłâ&#x2C6;&#x2019; 4â&#x2C6;&#x2019;3

Alternative solution:

In case of using pressures (again, fewer app mitted): Understanding that we have effectively interaction of two point masses â&#x20AC;&#x201D; 0.5 pts.

3Îłâ&#x2C6;&#x2019; (equivalently one may mention the â&#x2C6;? râ&#x2C6;&#x2019;2 force coming from r3 T4 = T0 Gaussâ&#x20AC;&#x2122; law; if not stated, but used correctly â&#x20AC;&#x201D; full marks) r4 Energy conservation as a differential equation â&#x20AC;&#x201D; 0.3 pts.

2 GMm (if only expressed through v (like mv = E) â&#x20AC;&#x201D; 0.1 pts.; 2 â&#x2C6;&#x2019; r r = â&#x2C6;&#x2019; Gm if the differential equation of Newtonâ&#x20AC;&#x2122;s 2nd law (¨ r 2 ) is given instead â&#x20AC;&#x201D; 0.2 pts.)

dr

t= (whichever the sign is) â&#x20AC;&#x201D; 0.4 pts.

2E + 2Gm r Integration and final answer â&#x20AC;&#x201D; 1.3 pts.

iv. (1.7 pts)

points)

t â&#x20AC;&#x201D; 0.2 pts. )

t â&#x20AC;&#x201D; 0.3 pts. â&#x20AC;&#x201D; 0.1 pts. â&#x20AC;&#x201D; 0.1 pts.

8 â&#x20AC;&#x201D; 0.1 pts.

Radiated heat equals the compression work â&#x20AC;&#x201D; 1 pt. [for mentioning or using the 1st law of thermodynamics â&#x20AC;&#x201D; (pospage 5 of 5 â&#x20AC;&#x201D; sibly in a wrong way) â&#x20AC;&#x201D; 0.5 pts.]

W = â&#x2C6;&#x2019; p dV or W = â&#x2C6;&#x2019; p â&#x2C6;&#x2020;V â&#x20AC;&#x201D; 0.3 pts.

p4 = phydrostat Ď p4 = RT Âľ

phydrostatic â&#x2030;&#x2C6; Ď g Gm gâ&#x2030;&#x2C6; 2 r4 1 3Îłâ&#x2C6;&#x2019;

RT0 r3 Final answer r4 â&#x2030;&#x2C6; r3 ÂľmG 3Îłâ&#x2C6;&#x2019;

RT0 r3 4â&#x2C6;&#x2019;3 Final answer T4 â&#x2030;&#x2C6; T0 ÂľmG

(with either â&#x20AC;&#x153;+â&#x20AC;? or â&#x20AC;&#x153;â&#x2C6;&#x2019;â&#x20AC;? â&#x20AC;&#x201D; give full marks)

mRT0 â&#x20AC;&#x201D; 0.2 pts. ÂľV 3mRT0 r0 ln Calculating the integral, W = â&#x20AC;&#x201D; 0.2 pts. Âľ r3 p=

v. (1 pt) The collapse continues adiabatically.

(If used, but not written down, give full marks.) â&#x20AC;&#x201D; 0.3 pts. pV Îł = const â&#x20AC;&#x201D; 0.3 pts.

T â&#x2C6;? V 1â&#x2C6;&#x2019;Îł â&#x20AC;&#x201D; 0.2 pts. r 3Îłâ&#x2C6;&#x2019;3 3 T = T0 â&#x20AC;&#x201D; 0.2 pts. r vi. (2 pts)

ively asses â&#x20AC;&#x201D; 0.5 pts. rce coming from â&#x20AC;&#x201D; full marks)

orbit â&#x20AC;&#x201D; 1 pt.

eriod iaxis â&#x20AC;&#x201D; 0.4 pts. ) r0 /2 â&#x20AC;&#x201D; 0.1 pts.

eriod â&#x20AC;&#x201D; 0.3 pts.

eriod â&#x20AC;&#x201D; 0.1 pts.

In case of using energies (marks must not be subtracted for fewer approximations, even in the final answer): T4 = T0

r3 r4

3Îłâ&#x2C6;&#x2019;3

â&#x20AC;&#x201D; 0.1 pts.

â&#x2C6;&#x2020;Q + â&#x2C6;&#x2020;Î â&#x2030;&#x2C6; 0 â&#x20AC;&#x201D; 0.4 pts. â&#x2C6;&#x2020;Î â&#x2030;&#x2C6; â&#x2C6;&#x2019;Gm2 /r4 â&#x20AC;&#x201D; 0.6 pts. â&#x2C6;&#x2020;Q = mcV T4 â&#x20AC;&#x201D; 0.4 pts.

R â&#x20AC;&#x201D; 0.3 pts. cV â&#x2030;&#x2C6; Âľ 1

RT0 r3 3Îłâ&#x2C6;&#x2019;4 Final answer r4 â&#x2030;&#x2C6; r3 â&#x20AC;&#x201D; 0.1 pts. ÂľmG

105

The collapse continues adiabatically.

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

2(r0 − r2 ) — 0.4 pts. g

pV γ = const — 0.3 pts.

Gm g ≈ 2 — 0.4 pts. r 0 0.1r03 t≈ — 0.2 pts.

Gm

T ∝ V 1−γ — 0.2 pts. r 3γ−3 3 T = T0 — 0.2 pts. r vi. (2 pts)

anding that we have effectively nteraction of two point masses — 0.5 pts. ay mention the ∝ r−2 force coming from ated, but used correctly — full marks)

In case of using energies (marks must not be subtracted for fewer approximations, even in the final answer): T4 = T0

3γ−3

— 0.1 pts.

∆Π ≈ −Gm2 /r4 — 0.6 pts.

cal orbit is equal to the period bit of the same longer semiaxis — 0.4 pts. ed correctly — full marks)

∆Q = mcV T4 R cV ≈ µ 1 3γ−4

RT0 r3 Final answer r4 ≈ r3 µmG 3γ−3

RT0 r3 4−3γ Final answer T4 ≈ T0 µmG

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.

r03 Final answer tr→0 = π — 0.1 pts.

8Gm

— 0.4 pts.

— 0.3 pts.

— 0.1 pts.

— 0.1 pts.

In case of using pressures (again, fewer approximations are permitted):

ation as a differential equation — 0.3 pts.

2 GMm rough v (like mv = E) — 0.1 pts.; 2 − r r = − Gm uation of Newton’s 2nd law (¨ r 2 ) is pts.) dr (whichever the sign is) — 0.4 pts. + 2Gm r Integration and final answer — 1.3 pts.

quals the compression work — 1 pt.

— page 5 of 5 —

106

r3 r4

∆Q + ∆Π ≈ 0 — 0.4 pts.

otion as an ultraelliptical orbit — 1 pt.

anding that we have effectively nteraction of two point masses — 0.5 pts. ay mention the ∝ r−2 force coming from ated, but used correctly — full marks)

T4 = T0

r3 r4

3γ−3

— 0.1 pts.

p4 = phydrostatic — 0.4 pts. ρ p4 = RT4 — 0.5 pts. µ phydrostatic ≈ ρgr4 Gm g≈ 2 r4 1 3γ−4

RT0 r3 Final answer r4 ≈ r3 µmG

3γ−3 RT0 r3 4−3γ Final answer T4 ≈ T0 µmG

— 0.4 pts.

— 0.4 pts.

— 0.1 pts.

— 0.1 pts.

The 43rd International Physics Olympiad — Experimental Competition rd

43 —International Physics Olympiad Tartu,The Estonia Thursday, July 19th 2012

— Experimental Competi

Tartu, Estonia — Thursday, July 19th 2012 • The examination lasts for 5 hours. There are 2 problems 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 Problem 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).

• You are not allowed to leave your working place without permission. If you need any assistance (malfunctioning 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 organizer arrives.

to be graded. If you have written som don’t want to be graded onto the Solut as initial and incorrect solutions), cross

• If you need more paper for a certain prob the flag “help” and tell an organizer t ber; you are given two Solution sheets more than once).

• You should use as little text as p explain your solution mainly with equ tables, symbols and diagrams. When tion is unavoidable, you are encouraged lish translation alongside with the tex language (if you mistranslate, or don’t your native language text will be used d ation).

• Avoid unnecessary movements during t examination and do not shake the walls laser experiment requires stability.

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

• Do not look into the laser beam or its re permanently damage your eyes.

• 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 dealing with. Copy the final answers into the appropriate boxes of the Answer Sheets. There are also Draft papers; use these for writing things which you don’t want

• The first single sound signal tells you t min of solving time left; the second dou means that 5 min is left; the third tri marks the end of solving time. After t signal you must stop writing imme the papers into the envelope at your des allowed to take any sheet of paper o If you have finished solving before the fi please raise your flag.107

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 organizer arrives.

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

• For each problem, there are dedicated Solution Sheets Write solutions onto the appropriate Solution Sheets. For Tartu, Estonia — Thursday,your July 19th 2012 each Problem, the Solution Sheets are numbered; use asts for 5 hours. There are 2 problems to be graded.

If youtohave written something what you the sheets according the enumeration.

Always mark points. There are two tables in your don’t want to be graded ontoQuestion the Solution Sheets (such which Problem Part and you are dealInternational Olympiad Experimental Competition eighbouring cubicles),Physics the apparatus of as— initial and incorrect solutions), out. ing with.

Copy the final answerscross into these the appropriate th one table and the apparatus the Answer Tartu, Estoniaof—ProbThursday,boxes Julyof 19 2012 Sheets. There are also Draft pa• pers; If youuse need morefor paper for athings certainwhich problem, raise ther table; you can move freely between these writing you please don’t want the flag “help”Ifand anwritten organizer the problem asts foryou 5 hours.

There are 2 problems to be graded. youtell have something whatnumyou wever, are not allowed to move ber; aretogiven two Solution (youSheets can do(such this points. There arefrom two tables in your don’tyou want be graded onto thesheets Solution erimental setup one table to more thanand once). eighbouring cubicles), the apparatus of as initial incorrect solutions), cross these out.

International Physics Olympiad(see—header Experimental Competition for the number and pictogramme).

one table and the apparatus of Proberimental equipment on one table is ther table; you can move freely between he other table is boxed. You must wever, you are not allowed to move the cover nor open the box nor erimental setup from one table to ope with the problems before the f the beginning of competition erimental equipment on one table is nals). he other table is boxed. You must owed to leave your working place the cover nor open the box nor sion. If you need any assistance (malope with the problems before the uipment, broken calculator, need to f the beginning of competition tc), please raise the corresponding flag nals). et” with a long handle at your seat) owed toand leave your working ox walls keep it raised until place an orsion. If you need any assistance (maluipment, broken calculator, need to side of the sheets of paper. tc), please raise the corresponding flag therewith are dedicated Solution Sheets et” a long handle at your seat) he pictogramme).

Write ox number walls andand keep it raised until an oro the appropriate Solution Sheets. For e Solution Sheets are numbered; use side of the sheets of paper. ng to the enumeration. Always mark there dedicated Solution Sheets Part are and Question you are dealhe and pictogramme).

Write thenumber final answers into the appropriate o theSheets. appropriate Solution wer There are alsoSheets.

Draft For paSolutionthings Sheets are you numbered; use re writing which don’t want ng to the enumeration. Always mark Part and Question you are dealthe final answers into the appropriate wer Sheets. There are also Draft par writing things which you don’t want

108

• Avoid unneces examination a laser experime

• Do not look in permanently d

• The first singl min of solving means that 5 marks the end signal you m the papers int allowed to ta If you have fin please raise yo

• You use as little text as possible: to • If youshould need more paper for a certain problem, pleasetry raise explain solution mainly with equations, numbers, the flag your “help” and tell an organizer the problem numWhen (you textual tables, andtwo diagrams. ber; yousymbols are given Solution sheets canexplanado this tion isthan unavoidable, you are encouraged to provide Engmore once). lish translation alongside with the text in your native • You should usemistranslate, as little text as possible: to language (if you or don’t translate try at all, explain yourlanguage solutiontext mainly with equations, numbers, your native will be used during the Modertables, ation). symbols and diagrams. When textual explanation is unavoidable, you are encouraged to provide Eng• lish Avoidtranslation unnecessary movements theinexperimental alongside with during the text your native examination not shake the your box: language (if and you do mistranslate, or walls don’t of translate at the all, laser native experiment requires your language text stability. will be used during the Moderation).

• Do not look into the laser beam or its reflections! It may permanently damage your eyes. during the experimental • Avoid unnecessary movements — page examination and do not shake the walls of your box: the 1 of 4 — • The first single sound signal tells you that there are 30 laser experiment requires stability. min of solving time left; the second double sound signal means 5 min left; beam the third sound Itsignal • Do not that look into theislaser or itstriple reflections! may marks the enddamage of solving permanently yourtime. eyes.After the third sound signal you must stop writing immediately. Put all • The first single sound signal at tells you thatYou thereare arenot 30 the papers into the envelope your desk. min of solving time left; the of second double sound signal allowed to take any sheet paper out of the room. means that finished 5 min issolving left; the third signal If you have before thetriple final sound signal, marks the end of flag. solving time. After the third sound please raise your 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.

Problem E1.

The magnetic permeability of water (10 points) Problem E1. The magnetic permeability of water WARNINGS: (10 points) The effect of a magnetic field on most of the substances but ferromagnetics is rather weak. This is because the energy density of the magetic field in substances of relative magnetic 11 22 permeability µ is given by formula w = 2µµ 2µµ00 B , and typically, µ is very close to 1. Still, with suitable experimental techniques such effects are firmly observable. In this problem we study the effect of a magnetic field, created by a permanent neodymium magnet, on water and use the results to calculate the magnetic permeability of water. You are not asked to estimate any uncertainties throughout this problem and you do not need to take into account the effects of the surface tension.

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 cylindrical permanent 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 fastened to the stand; the caliper allows horizontal displacement of the laser. The on-off button of the laser can be kept down with the help of 13 the white conical tube. The depth of the water above the magnet should be reasonably close to 1 mm (if shallower, the water surface becomes so curved that it will be difficult 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 of graph paper (the “screen”) is to be fixed to the vertical plate with 14 small magnetic tablets. If the laser spot on the screen becomes smeared, check for a dust on the water surface (and blow 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 button for setting the origin of the caliper reading. Beneath the laser pointer, there is one more button of the caliper, which temporarily re-sets the origin (if you pushed it inadvertently, push it once again to return to the normal measuring mode).

⋄ The laser orientation is pre-adjusted, d

⋄ Do not look into the laser beam or its

⋄ Do not try to remove the strong neod

⋄ Do not put magnetic materials close t

⋄ Turn off the laser when not used, batt

Part A. Qualitative shape of the water s When a cylindrical magnet is placed belo latter becomes curved. Observe, what is th surface above the magnet; according to thi if the water is diamagnetic (µ < 1) or par

Write the letter corresponding to the cor Answer Sheet, together with an inequality this part, you don’t need to motivat Part B. Exact shape of the water surfac Curving of the water surface can be checke ity by measuring the reflection of the laser We use this effect to calculate the depend the water on the horizontal position above abov i. ( 1.6 pts) Measure the dependence of laser spot on the screen on the caliper re You should cover all the usable range of c Fill the results into the table in the Answ ii. (0.7 pts) Draw the graph of the obtai iii. ( 0.7 pts) Using the obtained graph, α00 between the beam and the horizonta horizont surface. iv. ( 1.4 pts) Please note that the slope surface can be expressed as follows: tan β ≈ β ≈

cos22 α00 y − y00 − (x − · 2 L00 + x

where y00 is the height of the laser spot the beam is reflected from the water surfa magnet, and x00 — the respective position Calculate the values of the water surfac into the Table on the Answer Sheet. Plea Ple be possible to simplify your calculations if combination of terms in the expression fro with a reading from the last graph. v. ( 1.6 pts) Calculate the height of the w to the surface far from the magnet as a fun fu into the Table on the Answer Sheet.

109 Numerical values for your calculations: horizontal distance between the magnet’s centre and the screen L00 = 490 mm. vi. (1 pt) Draw the graph of the latter d on it the region where the beam hits the w

e magnetic permeability of water

magnetic field on most of the substances s rather weak. This is because the energy ic field in substances of relative magnetic 1 B 2 , and typically, en by formula w = 2µµ 0 Still, with suitable experimental techniques y observable. In this problem we study the field, created by a permanent neodymium d use the results to calculate the magnetic r. You are not asked to estimate any ughout this problem and you do not account the effects of the surface ten-

s of 1 a stand (the highlighted numbers mbers in the fig.), 3 a digital caliper, 4 a dish with water and 7 a cylindrical t in it (the magnet is axially magnetised). he base of the stand by the magnet’s pull. the caliper, the base of which is fastened iper allows horizontal displacement of the utton of the laser can be kept down with white conical tube. The depth of the waet should be reasonably close to 1 mm (if surface becomes so curved that it will be ings from the screen). 15 A cup of water an be used for the water level adjustment 1 mm, add 13 ml of water). 2 A sheet of creen”) is to be fixed to the vertical plate etic tablets. If the laser spot on the screen heck for a dust on the water surface (and

egend for the figure is as follows: 6 the beam hits the screen; 11 the LCD screen he button which switches the caliper units and inches; 8 on-off switch; 9 button of the caliper reading. Beneath the laser 110 caliper, which tempormore button of the in (if you pushed it inadvertently, push it to the normal measuring mode).

ter setting above the be reasonably close to 1the mm (if for the magnet origin ofshould the caliper reading. Beneath laser shallower, the is water surface becomes so caliper, curved that will be there one more pointer, button of the whichittemporarily re-sets the readings origin (iffrom you the pushed it inadvertently, it 15 A cup ofpush water difficult to take screen). once to return to be theused normal measuring mode). and again 16 a syringe can for the water level adjustment (to raise the level by 1 mm, add 13 ml of water). 2 A sheet of graph paper (the “screen”) is to be fixed to the vertical plate with 14 small magnetic tablets. If the laser spot on the screen becomes smeared, check for a dust on the water surface (and blow 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 button for setting the origin of the caliper reading. Beneath the laser pointer, there is one more button of the caliper, which temporarily re-sets the origin (if you pushed it inadvertently, push it once again to return to the normal measuring mode).

surface. this part, you iv.

1.4Exact pts) sP Part( B. surface can be ex Curving of the w ity by measuring We use this effec tanonβ the ≈β the water i. ( 1.6 pts) M laser spot the where y0 isonthe You shouldis cover the beam refle Fill the results magnet, and x0in ii. (0.7 pts) Dr Calculate the iii.

0.7Table pts) oU into(the α betweentothe be0 possible sim surface. combination of t with reading iv. ( a1.4 pts) fr P surface be Ca ex v. ( 1.6can pts) to the surface far into the Table on Numerical values for your calculations: horizontal distance tan Draw β≈β between the magnet’s centre and the screen L0 = 490 mm. vi. (1 pt) Check (and adjust, if needed) the alignment of the centre of on it the region w the magnet in two perpendicular directions. The vertical axis above the magne where y is the of the magnet must intersect with the laser beam, and it must Part C. 0Magnet the beam is refle also intersect with 12 the black line on the support plate. Using theand result magnet, x0 Magnetic induction at the magnet’s axis, at the height of so-called magnet Calculate the 1 mm from the flat surface B0 = 0.50 T; Density of water netic permeabilit into the Table o ρw = 1000 kg/m3 ; free-fall acceleration g = 9.8 m/s2 ; va- the numerical re −7 be possible to sim cuum permeability µ0 = 4π × 10 H/m. of t of 4 — — page 2 combination with a reading fr WARNINGS: v. ( 1.6 pts) Ca ⋄ The laser orientation is pre-adjusted, do not move it! to the surface far into the Table on Numerical values your calculations: horizontal distance ⋄ Do not look intoforthe laser beam or its reflections! between the magnet’s centre and the screen L0 = 490 mm. vi. (1 pt) Draw ⋄Check Do not tryadjust, to remove the strong neodymium magnet! (and if needed) the alignment of the centre of on it the region w the magnet in two perpendicular directions. The vertical axis above the magne ⋄of Do put magnetic materials close to beam, the magnet! the not magnet must intersect with the laser and it must Part C. Magnet also intersect with 12 the black line on the support plate. Using the result ⋄Magnetic Turn off the laser when notmagnet’s used, batteries in 1 h!of induction at the axis, atdrain the height so-called magnet 1Part mmA.from the flat surface B = 0.50 T; Density of water netic permeabilit 0the water surface (1 points) Qualitative shape of ρw = 1000 kg/m3 ; free-fall = 9.8 m/s2 ; the va- the numerical re When a cylindrical magnet isacceleration placed below gwater surface, −7 cuum permeability µ = 4π × 10 H/m.

0 latter becomes curved. Observe, what is the shape of the—water page 2 of 4 — surface above the magnet; according to this observation, decide if the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).

Write the letter corresponding to the correct option into the Answer Sheet, together with an inequality µ > 1 or µ < 1. For this 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 sensitivity by measuring the reflection of the laser beam by the surface.

We use this effect to calculate the dependence of the depth of the water on the horizontal position above the magnet. i. ( 1.6 pts) Measure the dependence of the height y of the laser 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 angle α0 between the beam and the horizontal area of the water surface. iv. ( 1.4 pts) Please note that the slope (tan β) of the water surface can be expressed as follows:

gital caliper, 4 7 a cylindrical ally magnetised). he magnet’s pull. which is fastened placement of the kept down with depth of the walose to 1 mm (if ed that it will be A cup of water level adjustment r). 2 A sheet of he vertical plate pot on the screen ater surface (and

follows: 6 the the LCD screen the caliper units witch; 9 button Beneath the laser r, which temporvertently, push it mode).

surface above the magnet; according to this observation, decide if the water is diamagnetic (µ < 1) or paramagnetic (µ > 1).

Write the letter corresponding to the correct option into the Answer Sheet, together with an inequality µ > 1 or µ < 1. For this 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 sensitivity by measuring the reflection of the laser beam by the surface.

We use this effect to calculate the dependence of the depth of the water on the horizontal position above the magnet. i. ( 1.6 pts) Measure the dependence of the height y of the laser 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 angle α0 between the beam and the horizontal area of the water surface. iv. ( 1.4 pts) Please note that the slope (tan β) of the water surface can be expressed as follows: tan β ≈ β ≈

zontal distance L0 = 490 mm. of the centre of The vertical axis eam, and it must e support plate. at the height of nsity of water = 9.8 m/s2 ; va-

cos2 α0 y − y0 − (x − x0 ) tan α0 , · 2 L0 + x − x0

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

Calculate the values of the water surface slope and fill them into the Table on the Answer Sheet. Please note that it may be possible to simplify your calculations if you substitute some combination of terms in the expression from the previous task with a reading from the last graph. v. ( 1.6 pts) Calculate the height of the water surface relative to the surface far from the magnet as a function of x and fill it into the Table on the Answer Sheet. vi. (1 pt) Draw the graph of the latter dependence. Indicate on it the region where the beam hits the water surface directly above the magnet.

Part C. Magnetic permeability (2 points) Using the results of Part B, calculate the value of µ − 1 (the so-called magnetic susceptibility), where µ is the relative magnetic permeability of the water. Write your final formula and the numerical result into the Answer Sheet.

— page 2 of 4 —

111

Problem E2.

Nonlinear Black Box (10 points) Problem E2. Nonlinear Black Box (10 points) In simple problems, electrical circuits are assumed to consist of linear elements, for which electrical characteristics are directly 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, we examine a circuit containing nonlinear elements, enclosed into a black box, for which the assumption of proportionality no Problem E2. Nonlinear Black Box (10 points) longer holds.

The setupproblems, comprises of a circuits multimeter (labelled In simple electrical are assumed to “IPhOconsist of linear elements, for which electrical arenondirmeasure”), a current source, a blackcharacteristics box containing ectly proportional to four eachtest other.

Examples includeconnectors resistance linear elements, and leads with stackable ˙ for = wiring. careful not on the black (V RI), Be capacitance (Qto=break CV ) the andseal inductance (V =box.

LI), TheR,multimeter canconstants. measure current and voltage simultanwhere C and L are In this problem, however, we eously. You can record withnonlinear it up to 2000 data enclosed points, each examine a circuit containing elements, into of: for voltage V ,the current I, powerofPproportionality = IV , resistance aconsisting black box, which assumption no longer holds.

R = V /I, voltage time-derivative V˙ , current time-derivative I˙ The setupt. See comprises of for a multimeter “IPhOand time its manual details. If you(labelled go beyond 2000 data points,athe oldest data will abeblack overwritten. measure”), current source, box containing nonlinear elements, and four test leads with stackable connectors IN break the seal OUTon the black GND box. for wiring. Be careful not to IN can measure current OUT GND The multimeter and voltage simultaneously. You can record with it up to 2000 data points, each consisting of: voltage V , current I, power P = IV , resistance R = V /I, voltage time-derivative V˙ , current time-derivative I˙ and time t. See its manual for details.

A If you go V beyond 2000 data points, the oldest data will be overwritten.

IN

A

Multimeter OUT

GND

V

The constant current source supplies stable current as long as the voltage across its terminals stays between −0.6125 V and 0.6125 V. When switched off, the constant current source behaves as a large (essentially infinite) A resistance.

V

Multimeter Multimeter

-

The constant current source supplies stable current as long as the voltage across its terminals stays between −0.6125 V and+0.6125 V. When switched off, the constant current source behaves as a large (essentially infinite) resistance.

-

I=6mA U=-612.5mV...612.5mV

Current source

112

The black box contains an electric double layer capacitor + is a slightly nonlinear high capacitance capacitor), an (which unknown nonlinear element and an inductor L = 10 µH of negligible resistance,I=6mA switchable as indicated on the circuit diagram.

U=-612.5mV...612.5mV The nonlinear element can be considered as a resistance with a nonlinear dependence between the voltage and the current [ I is a continuousCurrent functionsource of V with I(0) = 0]. Likewise, for the capacitor, the differential capacitance C(V ) = dQ/dV is not exactly say that the voltage on the black The constant. black box We contains an electric double layer capacitor box is positive when the potential on its red terminalan is (which is a slightly nonlinear high capacitance capacitor), higher the potential blackL terminal.

Posunknownthan nonlinear element andon anthe inductor = 10 µH of negligible resistance, switchable as indicated on the circuit diagram.

The nonlinear element can be considered as a resistance with a nonlinear dependence between the voltage and the current [ I is a continuous function of V with I(0) = 0]. Likewise, for the

itive voltage w matching colo source are con voltages).

itive voltage w + matching colo source are con voltages).

-

C(

B

+ Here it is safe

shorting its inpu terminals on mu citor is enough t You are C( no throughout th

Part A. Circuit B In this Herepart, it is kee safe down), shortingsoitsthat input Please on notemu t terminals time, citor istherefore enough it t of part to avo YouA are no i.

(1 pt) Confirm throughout th is approximately Part A. Circuit it varies for vol circuit In this diagram. part, kee ii.

(1.2 so pts) down), thatSht in the black box Please note t for a single volta time, therefore it circuit diagram. of part A to avo iii.(1(2.2 pts) N i. pt) Confirm [C(V ) ≈ C0 ], de is approximately it for vol thevaries nonlinear el circuitfor diagram. curve obtain ii. (1.2 pts)sheet Sh the answer in iv.the (2.6black pts)box Us for a single volta of obtainable vo circuit diagram. obtainable posit iii.

(2.2 pts) N sheet. Write dow [C(V ) capacitanc ≈ C0 ], de ential the nonlinear el Part B. curve forCircuit obtain Enable the sheet induc the answer (push “0” down) iv.

(2.6 pts) Us ure and plot the of obtainable vo ear element. De obtainable positA curves of parts sheet. Write dow arguments. ential capacitanc Here you nee ally a capac Partalso B. Circuit to the nonlinear Enable the induc (push “0” down) ure and plot the ear element. De curves of parts A

[C(V ) ≈ C0 ], determine the current–volt the nonlinear element used in the black curve for obtainable positive voltages on the answer sheet. Document the circuit d I=6mA iv. (2.6 pts) Using measurements taken f U=-612.5mV...612.5mV of obtainable voltages, calculate and plot obtainable positive voltages on the black Current source sheet. Write down the minimal and max The black box contains an electric double layer capacitor ential capacitance Cmin , Cmax . Document (which is a slightly nonlinear high capacitance capacitor), an Part B. Circuit with inductance (3 point unknown nonlinear element and an inductor L = 10 µH of negli- Enable the inductance by opening the swi gible resistance, switchable as indicated on the circuit diagram. (push “0” down). Using the same method The nonlinear element can be considered as a resistance with a ure and plot the current–voltage charact nonlinear dependence between the voltage and the current [ I ear element. Describe any significant diff is a continuous function of V with I(0) = 0]. Likewise, for the curves of parts A and B and suggest a rea capacitor, the differential capacitance C(V ) = dQ/dV is not arguments. exactly constant. We say that the voltage on the black Here you need to know that the nonline box is positive when the potential on its red terminal is ally also a capacitance (ca 1 nF) which is higher than the potential on the black terminal. Pos- to the nonlinear resistance.

10 points) itive voltage will be acquired when the terminals of umed to consist matching colours on the black box and the current cteristics are dir- source are connected (you are allowed to use negative nclude resistance voltages). ˙ tance (V = LI), em, however, we ts, enclosed into oportionality no

+

+ labelled “IPhOcontaining nonkable connectors Nonlinear the black box. device C(V) voltage simultan— page 3 of 4 — ata points, each = IV , resistance Black box ime-derivative I˙ go beyond 2000 Here it is safe to discharge the capacitor in the black box by n. shorting its inputs either by itself or through the IN and OUT terminals on multimeter: the internal resistance of this capaGND citor is enough to keep the current from damaging anything.

You are not asked to estimate any uncertainties throughout this problem.

V

er

e current as long ween −0.6125 V nt current source nce.

e layer capacitor e capacitor), an = 10 µH of neglie circuit diagram. resistance with a d the current [ I Likewise, for the = dQ/dV is not e on the black red terminal is

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 considerable time, therefore it is recommended to read through all the tasks of part A to avoid unnecessary work. i. (1 pt) Confirm that the output current of the current source is approximately 6 mA, and determine the range within which it varies for voltages between 0 and +480 mV. Document the circuit diagram. ii. (1.2 pts) Show that the differential capacitance C(V ) used in the black box is approximately 2 F by measuring its value for a single voltage of your choice C(V0 ) = C0 . Document the circuit diagram. iii. (2.2 pts) Neglecting the nonlinearity of the capacitance [C(V ) ≈ C0 ], determine the current–voltage characteristic of the nonlinear element used in the black box. Plot the I(V ) curve for obtainable positive voltages on the black box onto the answer sheet. Document the circuit diagram. iv. (2.6 pts) Using measurements taken from the whole range of obtainable voltages, calculate and plot the C(V ) curve for obtainable positive voltages on the black box to the answer sheet. Write down the minimal and maximal values of differential 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, measure and plot the current–voltage characteristic of the nonlinear element. Describe any significant differences between the curves of parts A and B and suggest a reason using qualitative arguments.

Here you need to know that the nonlinear element has actually also a capacitance (ca 1 nF) which is connected in parallel

113

[C(V ) ≈ C0 ], determine the current–voltage characteristic of the nonlinear element used in the black box. Plot the I(V ) curve for obtainable positive voltages on the black box onto the answer sheet. Document the circuit diagram.

A iv. (2.6 pts) Using measurements taken from the whole range 12.5mV...612.5mV of obtainable voltages, calculate and plot the C(V ) curve for obtainable positive voltages on the black box to the answer ent source sheet. Write down the minimal and maximal values of differontains an electric double layer capacitor ential capacitance Cmin , Cmax . Document the circuit diagram. nonlinear high capacitance capacitor), an Part B. Circuit with inductance (3 points) lement and an inductor L = 10 µH of negli- Enable the inductance by opening the switch on the black box tchable as indicated on the circuit diagram. (push “0” down). Using the same method as in pt. A-iii, measnt can be considered as a resistance with a ure and plot the current–voltage characteristic of the nonline between the voltage and the current [ I ear element. Describe any significant differences between the tion of V with I(0) = 0]. Likewise, for the curves of parts A and B and suggest a reason using qualitative ential capacitance C(V ) = dQ/dV is not arguments.

We say that the voltage on the black Here you need to know that the nonlinear element has actuen the potential on its red terminal is ally also a capacitance (ca 1 nF) which is connected in parallel otential on the black terminal. Pos- to the nonlinear resistance.

IPhO-measure: short manual

Display

IPhO-measure is a multimeter capable of measuring voltage V and current I simultaneously. It also records their time derivat˙ their product P = V I, ratio R = V /I, and time ives V˙ and I, t of the sample. Stored measurements are organized into separate sets; every stored sample is numbered by the set number s and a counter n inside the set. All saved samples are written to an internal flash memory and can later be retrieved.

Electrical behaviour The device behaves as an ammeter and a voltmeter connected as follows.

Internal Range resistance — page 3 of 4 — Voltmeter 0 . . .

2 V 1 MΩ A V Voltmeter . . .

10 2 V 57 kΩ Multimeter Ammeter 0...1A 1Ω IN

OUT

GND

Basic usage • Push “Power” to switch the IPhO-measure on. The device is not yet measuring; to start measuring, push “Start”. Alternatively, you can now start browsing your stored data, see below.

• To browse previously saved samples (through all sets), press “Previous” or “Next”. Hold them down longer to jump directly between sets.

• While not measuring, push “Start” to start measuring a new set.

• While measuring, push “Sample” to store a data sample (with the current readings).

• While measuring, you can also browse other samples of the current set, using “Previous” and “Next”.

• Press “Stop” to end a set and stop measuring. The device is still on, you are ready to start a new measuring session, or browsing stored data.

A displayed sam 1.

2.

3.

4.

5.

6.

7.

index n of index s of time t sinc voltmeter rate of cha cannot be shown; ammeter o rate of cha cannot be shown; product P ratio R =

• Pushing “Power” turns the device off. The device will show the text “my mind is going . . . ”; don’t worry, all the 8. data measurements will be stored and you will be able to 9. browse them after you switch the device on, again. Saved samples will not be erased.

If any of the va shows “+inf” or

114

Display

suring voltage V heir time derivat= V /I, and time ganized into sepy the set number mples are written retrieved.

meter connected Internal ange resistance .2V 1 MΩ 57 kΩ 10 V .1A 1Ω

easure on. The measuring, push rt browsing your

hrough all sets), em down longer start measuring

o store a data

other samples of “Next”.

measuring. The a new measuring

The device will n’t worry, all the ou will be able to on, again. Saved

A displayed sample consists of nine variables: 1.

2.

3.

4.

5.

6.

7.

8.

9.

index n of the sample in the set; index s of the set; time t since starting the set; voltmeter output V ; rate of change of V (the time derivative V˙ ); if derivative cannot be reliably taken due to fluctuations, “+nan/s” is shown; ammeter output I; ˙ if derivative rate of change of I (the time derivative I); cannot be reliably taken due to fluctuations, “+nan/s” is shown; product P = V I; ratio R = V /I.

If any of the variables is out of its allowed range, its display shows “+inf” or “-inf”.

115

Solutions

Problem E1.

The magnetic permeability of water Problem E1. The magnetic permeability of water is essentially unp (10 points) (10 points) the graph, we ob

face — the red l Part A. Qualitative shape of the water surface (1 points) Observing reflections from the water surface (in particular, can also easily ca 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, the correct answer is of “Option (full Problem E1. Theie.magnetic permeability water D” is essentially unperturbed; connec marks(10 arepoints) given also for Option B). This profile implies the that graph, we obtain a line corre A. Qualitative the water surface (1 points) water Part is pushed away shape fromofthe magnet, which means face µ <— 1the red line. Using these Observing reflections from the water surface (in particular, can also easily calculate the angle (recall that ferromagnets with µ > 1 are pulled). those of straight lines, such as the edge of a sheet of paper), Part B. water surface (7and points) it isExact easy toshape see thatof thethe profile has one minimum has a flat bottom, is “Option (full i. (1.6relatively pts) The height ie. ofthe thecorrect spotanswer on the screenD” y is tabulated marks are given also for Option B). This profile implies that below water as a isfunction of the horizontal position x of the pushed away from the magnet, which means µ < 1 caliper. (recallthe thatvalues ferromagnets µ > 1 are pulled).

Note that of y with in millimetres can be rounded to inExact shape of the water surface (7 points) tegers Part (thisB. series of measurements aimed as high as possible i. (1.6 pts) The height of the spot on the screen y is tabulated precision). below as a function of the horizontal position x of the caliper. x (mm) 25 30 32 36 Note10 that the15 values of20 y in millimetres can be rounded to34 integers (this series as possible y (mm) 11.5 15.6 of measurements 19.8 24.3 aimed 30.2as high 33.2 37.2 40.5 precision). x (mm) 38 40 42 44 46 48 50 52 x (mm) 10 15 20 25 30 32 34 36 y (mm) 42.211.541.4 40.3 30.240.8 43.2 44.4 y (mm) 15.6 40.3 19.8 24.3 33.2 42 37.2 40.5 x (mm) 42 44 50 52 x (mm) 54 38 5640 58 60 46 62 48 64 66 68 iv. (1.4 pts) Fo y (mm) 41.4 45.4 40.3 40.3 43.2 44.4 y (mm) 45.342.245.8 44.4 40.843.642 46.2 50 53.6 (appearing in the x (mm) 54 56 58 60 62 64 66 68 (1.4 pts) For faster calculat x (mm) 70 45.3 7245.8 74 76 43.67846.2 80 85 iv.90 y (mm) 45.4 44.4 50 53.6 graph as the disc (appearing in the formula given) y (mm) 56.7 70 59.572 61.6 63.5 70.9 74.9 x (mm) 74 76 7865.380 67 85 90 graph as theisdistance re given between by equa y (mm) 56.7

ii. (0.7 pts)

59.5

61.6

63.5

65.3

67

70.9

74.9

ii. (0.7 pts)

116

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 ∆y

( is given by precalculate equation yr = y01+co precalculate 21 cos2 α0 ≈ 0.31.2 Th table lowing tablelowing (with z = tan β (wit · 105 the competition, the competition, lesser precision is sufficient).is sufficient).

x (mm) 10 z 0 x (mm) 38 z 597 x (mm) 54

15 10 40 428 56

20 27 42 239 58

25 66 44 128 60

ility34of water 36 2 37.2 40.5 ace (150points) 52 e (in43.2 particular, 44.4 sheet66of paper), 68 has a 2mum50and 53.6 “Option (full 85 D” 90 file 70.9 implies74.9 that ch means µ < 1 d). points) en y is tabulated x of the caliper. e rounded to inhigh as possible

is essentially unperturbed; connecting the respective points on the graph, we obtain a line corresponding to a flat water surface — the red line. Using these two extreme data points we ◦ can also easily calculate the angle α0 = arctan 74.9−11.5 90−10 ≈ 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 thegraph, distance and blue line; thelevels red line On as this the between data of tored two different water are 15 20 25 30 = y + (x − x ) tan α .

One can also x (mm) 10 is given by equation y r 0 0 0 depicted; blue curve corresponds to a water depth of ca 2 mm 1 2 z 0 10 27 66 204 cos table α0 ≈ above); 0.31. The the folprecalculate (data given in2 the thecalculations violet one —lead to 1tomm.

5 x (mm) 38 40 42 44 46 ; as mentioned above, during lowing table (with z = tan β · 10 iii. (0.5 pts) If the water surface were flat, the dependence of x z 597 428 239 128 53 theycompetition, lesser precision with of two numbers on would be linear, and the tangent thesignificant angle α0 would be x (mm) 54 56 58 60 62 ∆y is sufficient). , where ∆x is a horizontal displacement of given by tan α0 = ∆x z -72 -145 -278 -449 -606 the pointer, and ∆y — the respective displacement of the spot x (mm) 70 72 74 76 78 height. For the extreme positions of the pointer, the beam hits z -154 -74 -40 -20 -6 the water surface so far from the magnet that there, the surface is essentially unperturbed; connecting the respective points on 1 of 4 — ility34of water — page 36 the graph, we obtain a line corresponding to a flat water sur2 37.2 40.5 face — the red line. Using these two extreme data points we ace (150points) 52 ◦ can also easily calculate the angle α0 = arctan 74.9−11.5 e (in43.2 particular, 90−10 ≈ 38 .

44.4 sheet of paper), 66 68 mum and has a iv. (1.4 pts) For faster calculations, y − y0 − (x − x0 ) tan α0 2 50 53.6 (appearing in the formula given) can be read from the previous “Option D” (full 85 90 file implies that graph as the distance between red and blue line; the red line 70.9 74.9 ch means µ < 1 is given by equation yr = y0 + (x − x0 ) tan α0 . One can also precalculate 21 cos2 α0 ≈ 0.31. The calculations lead to the fold). lowing table (with z = tan β · 105 ; as mentioned above, during points) the competition, lesser precision with two significant numbers en y is tabulated is sufficient). x of the caliper. water levels are e rounded to in- x (mm) 10 15 20 25 30 32 34 36 epth of ca 2 mm high as possible z 0 10 27 66 204 303 473 591 — to 1 mm. x (mm) 38 40 42 44 46 48 50 52 dependence of x 34 36 z 597 428 239 128 53 26 0 -26 ngle α0 would be 2 37.2 40.5 x (mm) 54 56 58 60 62 64 66 68 l displacement of 50 52 z -72 -145 -278 -449 -606 -536 -388 -254 ment of the spot 43.2 44.4 x (mm) 70 72 74 76 78 80 85 90 er, the beam hits 66 68 z(1.4 -154 -74faster-40 -20 -6 2 -2 0α0 iv. pts) For calculations, y − y − (x − x ) tan 0 0 here, the surface 2 50 53.6 (appearing in the formula given) can be read from the previous page 1 of 4 — 85 — 90 graph the The distance red be andobtained blue line; line v. (1.6aspts) waterbetween height can as the the red integral Similarly to the previous figure, blue c

70.9 74.9 = ycalculate x0 ) tan α0 . height One can also a water depth of ca 2 mm, (data given in t is giventan byβdx. equation yr we h = Thus, water row-by0 + (x − the 1 α0 ≈height 0.31. in The lead the fol- the violet one — to 1 mm. precalculate row, by adding to2 the thecalculations previous row thetoproduct 2 cos 5 ; as mentioned lowing table (withdisplacement z = tan β · 10 −x averageduring slope of the horizontal xi+1 The position of the magnet can be fou i with the above, 1 the competition, lesser precision with two significant numbers (tan β + tan β ). caliper (find the positions when the laser b i+1 i 2 is sufficient). the magnet and determine the distance bet x (mm) 10 15 20 25 30 32 34 36 — the result is ca 24 mm), and using the −h (µm) 0 0 1 4 10 15 23 34 water levels are x (mm) x10 25 44 3046 32 is placed symmetrically with respect to t (mm) 15 38 20 40 42 48 5034 52 36 epth of ca 2 mm z 66 6620468 303 473 69 591 curve. −h0 (µm)10 46 27 56 63 69 69 — to 1 mm. x (mm) x38 44 60 4662 48 Part C. Magnetic permeability (2 points (mm) 40 54 42 56 58 64 6650 68 52 dependence of x z 597 428 239 128 53 26 0 17 -26 Water surface takes an equipotential shape −h (µm) 68 66 61 54 44 32 23 ngle α0 would be x (mm) x54 60 76 6278 64 water, the potential energy associated wit (mm) 56 70 58 72 74 80 8566 90 68 l displacement of B2 −1 2 1−µ z -72 -145 -278 -449 -606 -536 −h (µm) 12 10 9 8 8 8 8-388 8 -254 action is 2µ0 (µ −1) ≈ B 2µ0 ; the potent ment of the spot x (mm) 70 72 74 76 78 80 85 90 with the Earth’s gravity is ρgh. At the wa er, the beam hits Note that the water level height at the end of the table should z -154 -74 -40 -20 -6 2 -2 0 of those two needs to be constant; for a p here, the surface be also 0 (this corresponds also to an unperturbed region); the surface, this expression equals to zero, so — page 1 non-zero of 4 — result is explained by the measurement uncertainties. and hence, µ − 1 = 2µ0 ρgh/B 2 . Here, h One can improve the result by subtracting from h a linear trend the depth of the water 117 surface at the axis mm 8 µm · x−10 80 mm . that we have compensated the cumulative If the water level above the magnet is 1 mm, the water level the end of the previous task and obtained

x (mm) −h (µm) x (mm) −h (µm) x (mm) −h (µm)

38 46 54 68 70 12

40 56 56 66 72 10

42 63 58 61 74 9

44 66 60 54 76 8

46 68 62 44 78 8

48 69 64 32 80 8

50 69 66 23 85 8

52 69 68 17 90 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 mm 8 µm · x−10 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)

is placed symme curve.

Part C. Magnet Water surface ta water, the poten B2 (µ−1 action is 2µ 0 with the Earth’s of those two nee surface, this exp and hence, µ − 1 the depth of the that we have com the end of the p between the dep depth at the rig the numbers, we

ater height can be obtained as the integral Similarly to the previous figure, blue curve corresponds to us, we calculate the water height row-by- a water depth of ca 2 mm, (data given in the table above), and he height in the previous row the product the violet one — to 1 mm. placement xi+1 − xi with the average slope 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 10 15 20 25 30 32 34 36 — the result is ca 24 mm), and using the symmetry: magnet 0 0 1 4 10 15 23 34 — page 2 of 4 — is placed symmetrically with respect to the surface elevation 38 40 42 44 46 48 50 52 curve.

46 56 63 66 68 69 69 69 Part C. Magnetic permeability (2 points) 54 56 58 60 62 64 66 68 Water surface takes an equipotential shape; for a unit volume of 68 66 61 54 44 32 23 17 water, the potential energy associated with the magnetic inter70 72 74 76 78 80 85 90 B2 (µ−1 −1) ≈ B 2 1−µ action is 2µ 12 10 9 8 8 8 8 8 2µ0 ; the potential energy associated 0 with the Earth’s gravity is ρgh. At the water surface, the sum level height at the end of the table should of those two needs to be constant; for a point at unperturbed ponds also to an 118 unperturbed region); the surface, this expression equals to zero, so B 2 µ−1 2µ0 + ρgh = 0 plained by the measurement uncertainties. and hence, µ − 1 = 2µ0 ρgh/B 2 . Here, h = 120 µm stands for

row the product the violet one — to 1 mm. he average slope 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 2 34 36 — the result is ca 24 mm), and using the symmetry: magnet 5 23 34 is placed symmetrically with respect to the surface elevation 8 50 52 curve.

9 69 69 Part C. Magnetic permeability (2 points) 4 66 68 Water surface takes an equipotential shape; for a unit volume of 2 23 17 water, the potential energy associated with the magnetic inter0 85 90 B2 −1 −1) ≈ B 2 1−µ action is 8 8 8 2µ0 (µ 2µ0 ; the potential energy associated with the Earth’s gravity is ρgh. At the water surface, the sum the table should of those two needs to be constant; for a point at unperturbed rbed region); the surface, this expression equals to zero, so B 2 µ−1 2µ0 + ρgh = 0 ent uncertainties. and hence, µ − 1 = 2µ0 ρgh/B 2 . Here, h = 120 µm stands for m h a linear trend the depth of the water surface at the axis of the magnet; note that we have compensated the cumulative error as described at m, the water level the end of the previous task and obtained h as the difference xis of the magnet between the depth at the magnet’s axis (121 µm) and the halfdepth at the right-hand-side of the graph (1 µm). Putting in the numbers, we obtain µ − 1 = −1.2 × 10−5 .

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 ± 40 mVs varies from one experimental setup to another. When the current source is switched off or disconnected, the capacitor will discharge through the nonlinear element.

Multimeter IN OUT

GND

• When the voltage on the black box through the nonlinear element is I0 rent source off, we will have the c with the same current.

C0 = −I0 /V˙ ↓ (V = V

• We can also measure the capacitance voltage as in A-iv.

iii. (2.2 pts) If we neglect the nonlinear there are (at least) two ways to obtain the c acteristic of the nonlinear element in the b

• Applying Kirchhoff’s I law to the ch

Current source Switch − +

O I

I(V ) = Ic − C0 V˙ ↑ (V

Black box Switch

O I

An I(V ) characteristic obtained by ch is shown on the following figure.

• Applying Kirchhoff I law to the disc

i. (1 pt) During charging of the capacitor from V = 0 to — page 2 Vof = 4— Vmax we note that the output of the current source is constant (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.

I(V ) = −C0 V˙ ↓ (V

6

119 Part A Part B

• Applying K

Switch

An I(V ) ch is shown o

O I

O I

• Applying K i. (1 pt) During charging of the capacitor from V = 0 to V = Vmax we note that the output of the current source is constant (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.

6

5

There are several ways to determine the capacitance used in the black box based on chosen voltage.

4 I (mA)

dQ dV dQ = = C(V )V˙ Ic = dt dV dt

• 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.

Par Par

3

2

1

C0 = I0 /V˙ ↑ (V = 0) 0

0 0 This can be measured more precisely after first reversing the polarity of the current source and charging the capacitor backwards, because the multimeter does not display iv. (2.6 pts) In derivatives when they change sharply (as in few moments we solve a system after switching the current source on).

Example measurements taken this way follow.

I0 = V˙↑

V↑ (0) (mV/s) C0 (F)

3.51 1.71

3.32 1.81

out inductance (7 points) ke all the measurements needed for this e circuit as shown in the figure. While the tched on, we are charging the capacitor in the current I(Vmax ) through the nonlinear e output current I0 of the current source. varies from one experimental setup to anrent source is switched off or disconnected, scharge through the nonlinear element.

timeter OUT

GND

Black box

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 characteristic of the nonlinear element in the black box.

I(V ) = Ic − C0 V˙ ↑ (V ).

Switch

O

Therefore we nee and discharging voltages. A grap

• When the voltage on the black box is Vmax , the current — page 3 of 4 — through the nonlinear element is I0 . Switching the current source off, we will have the capacitor discharging with the same current.

• Applying Kirchhoff’s I law to the charging capacitor, 120

I(V ) =

3.55 1.69

C0 = 1.74 F

nlinear Black Box (10 points)

tch

Black box

Current source Switch − +

An I(V ) characteristic obtained by charging the capacitor

figure. While the the capacitor in gh the nonlinear e current source. ntal setup to anor disconnected, near element.

D

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 characteristic of the nonlinear element in the black box.

• Applying Kirchhoff’s I law to the charging capacitor, I(V ) = Ic − C0 V˙ ↑ (V ).

Black box

An I(V ) characteristic obtained by charging the capacitor is shown on the following figure.

• Applying Kirchhoff I law to the discharging capacitor,

from V = 0 to ent source is conthe multimeter. tial capacitance, citor in the black n the black box.

I(V ) = −C0 V˙ ↓ (V ).

6

Part A Part B

5

acitance used in I (mA)

lose to zero, the is also close to hing the current 0 will at first go

4

3

2

1

0

0 0.1 0.2 0.3 0.4 0.5 er first reversing V (V) harging the capadoes not display iv. (2.6 pts) In order to obtain the differential capacitance, in few moments we solve a system of linear equations by eliminating I(V ):

ollow.

3.55 1.69

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 —

121

V (V)

the current source and charging the capas, because the multimeter does not display iv. (2.6 pts) In order to obtain the differential capacitance, en they change sharply (as in few moments we solve a system of linear equations by eliminating I(V ): the current source on).

urements taken this way follow.

I0 = V˙↑ C(V ) + I(V ) I0 . =⇒ C(V ) = V˙↑ − V˙↓ I(V ) = −V˙↓ C(V ); 0) (mV/s) 3.51 3.32 3.55

C0 (F)

1.71

1.81

1.69 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.

C0 = 1.74 F

— page 3 of 4 —

Part B. Circuit Measuring and the nonlinear el obtain a graph t istance (I ′ (V ) < This is the regio lations, the non Ohmic resistanc LC circuit whos dampened) by th resonant frequen

2.1 2.05 2

C (F)

1.95 1.9 1.85 1.8 1.75 1.7

0

0.1

0.2

0.3 V (V)

0.4

0.5

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 resistance (I ′ (V ) < 0) region, in our case 70 mV < V < 330 mV. This is the region where, when we look at small-signal oscillations, 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 1 ∼ 30 MHz (with Cp being the resonant frequency ω = LC p 0.2

0.3 V (V)

0.4

122

0.5

capacitance of the nonlinear element) is high, we actually measure the average current through the nonlinear element, while the real current oscillates all over the region of negative differential resistance.

capacitance of th ure the average the real current ential resistance

Grading scheme: Experiment The 43rd International Physics Olympiad â&#x20AC;&#x201D; July 2012 Grading scheme: Experiment

General rules This grading scheme describes the number of points allotted for the design of the experiments, measurements, plotting, and formulae used for calculations. In the case of a formula, points are allotted for each term entering it. If a certain term of a useful formula is written incorrectly, 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). Points for the data measurements and calcuational Physics Olympiad July 2012 lations are not givenâ&#x20AC;&#x201D; automatically: the data which are clearly wrong are not credited.

Grading scheme: Experiment If the numerical data miss units, but the units can be s the number of guessed, 25% of points will be subtracted for the corresponding s, measurements, line in this grading scheme (rounded to one decimal place). The the case of a for- same rule applies if there is a typo with units with a missing g it. If a certain or redundant prefactor (millli-, micro-, etc); no mark is given y, 0.1 is subtrac- if the mistake is dimensional (e.g Ampere instead of Volt). mensional factor); No penalty is applied in these cases when a mistake is clearly th non-matching just a rewriting typo (i.e. when there is no mistake in the draft). ments and calcuNo penalty is applied for propagating errors unless the calwhich are clearly culations are significantly simplified (in which case mathematical calculations are credited partially, according to the degree he units can be of simplification, with marking granularity of 0.1 pts).

guessed, 25% of points will be subtracted f line in this grading scheme (rounded to one same rule applies if there is a typo with or redundant prefactor (millli-, micro-, et if the mistake is dimensional (e.g Ampere No penalty is applied in these cases whe just a rewriting typo (i.e. when there is no No penalty is applied for propagating culations are significantly simplified (in w ical calculations are credited partially, acc of simplification, with marking granularity

123

Problem E1.

The magnetic permeability of water Problem E1. The magnetic permeability of water (10 points) points) Problem E1. The magnetic permeability of water (10 (10 points) Part A. Qualitative of the water surface (1of points) Problem E1. Theshape magnetic permeability water Part points) A. Qualitative shape of the water surface (1 points) (10

Correct shape choice of (Bthe or D) — surface 0.5 pts; (1 points) Part A. Qualitative water Correct choice (B or D) — 0.5 pts; Correct sign (µ < 1) — 0.5 pts; Correct sign(B(µor<D) 1) — 0.5 pts; Correct choice Part B. Exact shape of the water surface points) Correct sign (µ < 1) — 0.5(7pts; Part B.pts) Exact shape of the water surface (7 points) i.

(1.6 Data i.

(1.6 Data Part B.pts) Exact shape of the water surface (7 points) Sufficient number i. (1.6 pts) Data of reasonably accurate ±2 mm Sufficient number of reasonably accurate ±2 mm data points: (n − 9)/10 pts but ≥ 0 and ≤ 0.9 — ≤ 0.9 pts; data points: (n −of9)/10 pts butaccurate ≥ 0 and ±2 ≤ 0.9 — ≤ 0.9 pts; Sufficient number reasonably (-0.1 if there is a sign error throughout themm whole measure(-0.1 if there is a sign error throughout the whole data points: (n − 9)/10 pts but ≥ 0 and ≤ 0.9 — ≤measure0.9 pts; ments) ments) (-0.1 if there is a sign error throughout the whole measureSufficient range of horizontal displacements: (x − 45 mm)/50 Sufficient ments) range of horizontal displacements: (x − 45 mm)/50 rounded to one decimal but ≥ 0 and ≤ 0.7 — ≤ 0.7 pts. rounded to of one decimal but ≥ 0 and ≤ 0.7 —45≤mm)/50 0.7 pts.

Sufficient range horizontal displacements: (x −

rounded to one decimal but ≥ 0 and ≤ 0.7 — ≤ 0.7 pts. ii. (0.7 pts) Graph ii. (0.7 pts) Graph Axes supplied with units — 0.1 pts; ii. (0.7 pts) Graph Axes supplied with units — 0.1 pts; Data points correctly plotted — 0.4 pts; DataAxes points correctly plotted supplied with units — 0.4 0.1 pts; (each clearly wrong point on graph: −0.1 pt down to total 0) (each clearly wrong point on graph: −0.1— pt 0.4 down Data points correctly plotted pts;to total 0) More densely spaced data points More clearly denselywrong spacedpoint data on points (each graph: −0.1 pt down to total 0) in the regions of fast change — 0.2 pts. in the regions of fast change — 0.2 pts.

More densely spaced data points in the regions of fast change — 0.2 pts. iii. (0.7 pts) Angle (≈ 38 ◦ ) iii. (0.7 pts) Angle (≈ 38 ◦ ) Idea:(0.7 using theAngle flat regions iii. pts) (≈ 38 ◦ ) Idea: using the flat regions far from the magnet — 0.5 pts from the magnet — 0.5 pts Idea: using the flat far regions if the central part of the magnet is used: 0.1 if the central part offar thefrom magnet is used: — 0.10.5 pts the magnet Correct value: ±5 ◦ / ±10 ◦ / > 10 ◦ — 0.2/0.1/0 pts; ◦ ◦ ◦ Correct value: ±5 / ±10 / > 10 — if the central part of the magnet is used: 0.10.2/0.1/0 pts; Correct value: ±5 ◦ / ±10 ◦ / > 10 ◦ — 0.2/0.1/0 pts; iv. (1.4 pts) Calculated table iv. (1.4 pts) Calculated table According to the number of correctly calculated data points: iv. (1.4 pts) the Calculated table According 0.1n up to to n = 10;number of correctly calculated data points: 0.1n up to to n =the 10;number of correctly calculated data points: According 0.1n up to n = 10;

124

1 + 0.05(n − 10) 1 + 0.05(n (but ≤ 1.4);− 10) (but ≤ 1.4);− 10) 1 + 0.05(n (but ≤ 1.4); v. (1.6 pts) He v. (1.6 pts) He calculat v.Idea (1.6ofpts) He Idea of calculat (if not stated bu (ifIdea not of stated bu calculat ∆x multiplie ∆x stated multiplie (if not For nbuc

For n c (but∆x nomultiplie more tha (but no more For ntha c

(but no more tha vi. (1 pt) Grap vi. (1 pt) Grap

vi. (1 pt) Grap

(each clearly wro (each clearly wro Correct qualitat Correct qualitat (each clearly in wings less wro th

in wings lessregi th Correct qualitat flat central flat central regi in wings less th region spanning region spanning flat central regi

region spanning Magnet indicat Magnet indicat

Magnet indicat

Part C. Magnet Part C. Magnet

Part C. Magnet Form Form Formula Formula Form height at the mi

height at the mi Formula Value calc heightValue at thecalc mi

Valueif calc (no marks not (no marks if not

(no marks if not

bility of water 1 + 0.05(n − 10) rounded down one decimal place for n > 10 (but ≤ 1.4);

ace (1 points)

pts;

pts;

7 points)

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

mm

.9 — ≤ 0.9 pts;

For n correctly calculated data points: — n/30 pts (but no more than 0.6; rounded down to one decimal place).

whole measurevi. (1 pt) Graph

x − 45 mm)/50

.7 — ≤ 0.7 pts.

1 pts;

4 pts;

down to total 0)

2 pts.

5 pts

2/0.1/0 pts;

Units on axes — 0.1 pts Data points correctly plotted — 0.4 pts (each clearly wrong point on graph: −0.1 pt down to total 0) 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

Part C. Magnetic permeability (2 points) 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

ated data points: (no marks if not obtained from the experimental data) correct sign — 0.1 pts

125

Problem E2.

Nonlinear Black Box (10 points)

Problem E2. Nonlinear Black Box (10 points)

(-0.1 for measuri

Part A. Circuit without inductance (7 points)

Using the fac 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 ) linear extrapo 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 iii. (2.2 pts) Fo circuit diagram is not indicated, take -0.1 pts from the marks If charging the c for the circuit.

(-0.1 for measuri

i. (1 pt)

(-0.1 for wrong p Correct circuit — 0.3 pts

Realisi

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)

If discharging th

Correct

(-0.1 for wrong p

Re

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 documented, 0.7 pts in total; if only Imin , Imax , 0 pts in total; if a For measuremen scheme without the black box and a single I measurement, Total # of data 0.2 pts in total.

10 ≤ n < 1

ii. (1.2 pts) Measuring C0 at V = 0 Correct circuit — 0.2 pts (-0.1 for measuring voltage on ammeter + black box) (-0.1 for wrong polarity of the black box) Realising that Ic = C0 V˙ — 0.2 pts

Additionaly for c

In range 0mV

In range 80mV

In range 200m

In range 400mV

Using the fact that Ic = I0 when V = 0 — 0.2 pts 126

Correct result — 0.2 pts -0.1 for error between ±30% and ±50%

For plotting:

0

Circuit diagram — 0.1 pts Measurement with the correct result — 0.1 pts

If only Imin , Imax together with a correct scheme documented, 0.7 pts in total; if only Imin , Imax , 0 pts in total; if a For measurements: scheme without the black box and a single I measurement, Total # of data correct data points 0.2 pts in total.

10 ≤ n < 15; 15 ≤ n < 20; n ≥ 20 —

ii. (1.2 pts) Measuring C0 at V = 0

Additionaly for correct data points:

Correct circuit — 0.2 pts (-0.1 for measuring voltage on ammeter + black box)

In range 0mV - 80mV at least 4/5 dat

In range 80mV - 200mV at least 4/5 dat

In range 200mV - 400mV at least 3 dat

(-0.1 for wrong polarity of the black box) Realising that Ic = C0 V˙ — 0.2 pts

In range 400mV - 550mV at least 4/5 dat

Using the fact that Ic = I0 when V = 0 — 0.2 pts Correct result — 0.2 pts

For plotting:

-0.1 for error between ±30% and ±50%

Units on axes —

linear extrapolation to obtain V˙ at V = 0 — 0.4 pts Alternatively instead of the last line

Data points correctly plotted —

(each clearly wrong point on graph: −0.1

Correct qualitative shape —

Three or more measurements — 0.2 pts Precharging the capacitor to a negative voltage — 0.2 pts

10 points)

)

(single maximum, single minimum with fl by fast rise)

Alternate solution C0 at Vmax (-0.1 for measuring voltage on ammeter + black box) — 0.2 0.2 pts pts Realising Correct that Ic =circuit C0 V˙ —

10 points)

(-0.1 for measuring voltage + black box) pts −Iammeter Using the fact that Ic = on 0 when V = Vmax — 0.2 — page 3 of 4 — und in solutions ) Realising that I C0 V˙ — 0.2 pts c =result Correct — 0.1 pts e I(V ) and C(V ) −I0 when V = =V Vmax — — 0.2 0.2 pts pts Using extrapolation the fact that Ito c = linear obtain V˙ at V max varies bit from und inasolutions Correct result — ticular can) Three or more measurements —0.1 0.2pts pts e I(V ) setup and C(V ˙ ry. linear extrapolation to obtain V at V = V — 0.2 pts max varies a bit from switches on can the iii. (2.2 pts) For method: ticular setup Three or more measurements — 0.2 pts from the marks ry.

If charging the capacitor: switches on the iii. (2.2 pts) For method: Correct circuit — 0.2 pts from the marks If charging the capacitor: (-0.1 for measuring voltage on ammeter + black box) Correct (-0.1 for wrong polarity of the blackcircuit box) — 0.2 pts — 0.3 pts — 0.3 pts — 0.2 0.3 pts pts — — 0.2 0.3 pts pts — — 0.2 pts — 0.2 pts

black box:

— 0.1 pts black box: — 0.1 pts 0.1 pts docut —scheme

(-0.1 for measuring voltage ˙ ↑ black — 0.2box) pts Realising that I(Von ) =ammeter I0 − C0 V+ (-0.1 for wrong polarity of the black box) Realising that I(V ) = I0 − C0 V˙ ↑ — 0.2 pts If discharging the capacitor: Correct circuit (there are several) — If discharging capacitor: (-0.1 for wrongthe polarity of the black box) Correct circuit (there — Realising that I(Vare ) =several) −C0 V˙ ↓ — (-0.1 for wrong polarity of the black box) Realising that I(V ) = −C0 V˙ ↓ —

— 0.1 pts if a For measurements: pts in total; I measurement, t scheme docu- Total # of data correct data points

0.2 pts 0.2 pts pts 0.2 0.2 pts 127

Additionaly for correct data points: Correct circuit — 0.2 pts

In range 0mV - 80mV at least 4/5 data pts — 0.1/0.2 pts In range 80mV - 200mV at least 4/5 data pts — 0.1/0.2 pts

oltage on ammeter + black box)

In range 200mV - 400mV at least 3 data pts — 0.1 pts

ity of the black box) Realising that Ic = C0 V˙ — 0.2 pts

In range 400mV - 550mV at least 4/5 data pts — 0.1/0.2 pts

fact that Ic = I0 when V = 0 — 0.2 pts Correct result — 0.2 pts

For plotting:

n ±30% and ±50%

Units on axes — 0.1 pts

olation to obtain V˙ at V = 0 — 0.4 pts

d of the last line

Data points correctly plotted — 0.4 pts (each clearly wrong point on graph: −0.1 pt down to total 0) Correct qualitative shape — 0.3 pts

Three or more measurements — 0.2 pts

apacitor to a negative voltage — 0.2 pts

(single maximum, single minimum with flat bottom, followed by fast rise)

n C0 at Vmax Correct circuit — 0.2 pts iv. (2.6 pts) — page 3 of 4 —

Correct circuit — 0.2 pts

(-0.1 for measuring voltage on ammeter + black box) (-0.1 for wrong polarity of the black box) 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

Part B. Circuit

For correct data In range

In range 80

In range 200

In range 400 For plotting:

(0.3 pts if intermediate values read from graph) For plotting: Units on axes — 0.1 pts

(each clearly wro

Data points correctly plotted — 0.4 pts (each clearly wrong point on graph: −0.1 pt down to total 0)

(two sharp falls,

For correct data points:

For detecting di

In range 0mV - 80mV at least 4 data pts — 0.1 pts In range 80mV - 200mV at least 4 data pts — 0.1 pts In range 200mV - 400mV at least 3 data pts — 0.1 pts

Noti

In range 400mV - 550mV at least 4 data pts — 0.1 pts

Notin

For finding Cmax and Cmin :

128

(or something eq

Finding reasonable Cmax — 0.1 pts

Men

Finding reasonable Cmin — 0.1 pts

Noting

it — 0.2 pts

ck box) iii— 0.4 pts

V )— 0.4 pts

V )— 0.1 pts

les— 0.4 pts

h)

Part B. Circuit with inductance (3 points) For correct data points: In range 0mV - 80mV at least 4 data pts — 0.1 pts In range 80mV - 200mV at least 4 data pts — 0.1 pts In range 200mV - 400mV at least 3 data pts — 0.1 pts In range 400mV - 550mV at least 4 data pts — 0.1 pts For plotting: Units on axes — 0.1 pts Data points correctly plotted — 0.4 pts

xes — 0.1 pts

ed — 0.4 pts

down to total 0)

(each clearly wrong point on graph: −0.1 pt down to total 0) Correct qualitative shape — 0.3 pts (two sharp falls, plateau in between) For detecting differences:

ts — 0.1 pts

Correct range for V :— 0.2 pts

ts — 0.1 pts

Correct cond. for I(V ) — 0.5 pts

ts — 0.1 pts

Noting that we have now LC-circuit— 0.2 pts

ts — 0.1 pts

Noting that neg. resist. → instability— 0.4 pts (or something equivalent)

ax

— 0.1 pts

Mentioning emergence of oscillations— 0.2 pts

in

— 0.1 pts

Noting that I(V ) is the average current— 0.3 pts

129

130

Results Gold Medalists

Siyuan Wei, People’s Republic of China

Hengyun Zhou, People’s Republic of China

Chi Shu, People’s Republic of China

Yijun Jiang, People’s Republic of China

Wonseok Lee, Republic of Korea

Chien-An Wang, Taiwan

Wenzhuo Huang, People’s Republic of China

Kai-Chi Huang, Taiwan

Yu-Ting Liu, Taiwan

Paphop Sawasdee, Thailand

Wei-Jen Ko, Taiwan

Eric Schneider, United States of America

Rahul Trivedi, India

Jun-Ting Hsieh, Taiwan

Lev Ginzburg, Russia

Photo by Henry Teigar

Attila Szabó, Hungary

131

Tudor GiurgicÄ&#x192;-Tiron, Romania

Jaan Toots, Estonia

Allan Sadun, United States of America

Nathanan Tantivasadakarn, Thailand

Kazumi Kasaura, Japan

Phi Long Ngo, Vietnam

Nikita Sopenko, Russia

Yuichi Enoki, Japan

Ding Yue, Republic of Singapore

Ihar Lobach, Belarus

Lam Ho Tat, Hong Kong

Veselin Karadzhov, Bulgaria

Sooshin Kim, Republic of Korea

Ivan Tadeu Ferreira Antunes Filho, Brazil

Jaemo Lim, Republic of Korea

Ilya Vilkoviskiy, Kazakhstan

Lorenz Eberhardt, Germany

Kevin Zhou, United States of America

Ivan Ivashkovskiy, Russia

Ang Yu Jian, Republic of Singapore

Michal Pacholski, Poland

Fung Tsz Chai, Hong Kong

Qiao Gu, Germany

Kuan Jun Jie, Joseph, Republic of Singapore

Adrian Nugraha Utama, Indonesia

Ngoc Hai Dinh, Vietnam

Albert Samoilenka, Belarus

Huan Yan Qi, Republic of Singapore

Puthipong Worasaran, Thailand

132

Gold Medalists Photo by Henry Teigar

Alexandra Vasilyeva, Russia

Jakub Vošmera, Czech Republic

Bijoy Singh Kochar, India

Stanislav Fořt, Czech Republic

Jeevana Priya Inala, India

Pongsapuk Sawaddirak, Thailand

Zoltán Laczkó, Hungary

Suyeon Choi, Republic of Korea

Dan-Cristian Andronic, Romania

Yordan Yordanov, Bulgaria

Ramtin Yazdanian, Islamic Republic of Iran

Hiromasa Nakatsuka, Japan

Vladysslav Diachenko, Ukraine

Aliaksey Khatskevich, Belarus

Sebastian Linß, Germany

Peter Kosec, Slovakia

Kunal Singhal, India

Chan Cheuk Lun, Hong Kong

Kohei Kawabata, Japan

Itay Knaan Harpaz, Israel

Volodymyr Sivak, Ukraine

Tasuku Omori, Japan

David Frenklakh, Russia

Vadzim Reut, Belarus

Vsevolod Bykov, Ukraine

Filip Ficek, Poland

Theodor Misiakiewicz, France

Daumantas Kavolis, Lithuania

Woojin Kweon, Republic of Korea

Tanel Kiis, Estonia

Kaisarbek Omirzakhov, Kazakhstan

Katerina Naydenova, Bulgaria

Huy Quang Le, Vietnam

Jeffrey Yan, United States of America

Ondřej Bartoš, Czech Republic

Kristjan Kongas, Estonia

Photo by Henry Teigar

Silver Medalists

133

Silver Medalists Photo by Karl Veskus

Georg Krause, Germany

Matias Mannerkoski, Finland

Yevgen Cherniavskyi, Ukraine

Soo Wah Ming, Wayne, Republic of Singapore

Jonathan Dong, France

Mussa Rajamov, Kazakhstan

Kaur Aare Saar, Estonia

Lai Kwun Hang, Hong Kong

Abdurrahman Akkas, Turkey

Tudor Ciobanu, Romania

Jeffrey Cai, United States of America

Roland Papp, Hungary

Georgijs Trenins, Latvia

Eric Wieser, United Kingdom

Henry Honglei Wu, Canada

Nurzhas Aidynov, Kazakhstan

Patrik Svancara, Slovakia

Adam Brown, United Kingdom

Oliver Edtmair, Austria

Žygimantas Stražnickas, Lithuania

Amir Yousefi, Islamic Republic of Iran

Paul Kirchner, France

Lubomír Grund, Czech Republic

Atli Thor Sveinbjarnarson, Iceland

Patrik Turzak, Slovakia

Sebastian Florin Dumitru, Romania

Milan Krstajić, Serbia

Péter Juhász, Hungary

Atinc Cagan Sengul, Turkey

Christoph Schildknecht, Switzerland

Ittai Rubinstein, Israel

Wai Hong Lei, Macao, China

Mehmet Said Onay, Turkey

Andres Erbsen, Estonia

Jean Douçot, France

134

Bronze Medalists Milan Kornjača, Serbia

Simon Pirmet, France

Petar Tadic, Montenegro

Simon Blouin, Canada

Cristian Zanoci, Moldova

Roberto Albesiano, Italy

Nicholas Salmon, Australia

Viet Thang Dinh, Vietnam

Oguzhan Can, Turkey

Pulkit Tandon, India

Mustafa Selman Akinci, Turkey

Sajad Khodadadian, Islamic Republic of Iran

Volodymyr Rozsokhovatskyi, Ukraine

Christoph Weis, Austria

Richard Thorburn, United Kingdom

Kacper Oreszczuk, Poland

Andrej Vlcek, Slovakia

Yigal Zegelman, Israel

Xuan Hien Bui, Vietnam

Frank Bloomfield, United Kingdom

Ivan Senyushkin, Kazakhstan

Tamara Šumarac, Serbia

Chen Solomon, Israel

Vu Phan Thanh, Germany

Martijn van Kuppeveld, Netherlands

Federica Maria Surace, Italy

Jonathan Lay, Australia

Christopher Whittle, Australia

Supanut Thanasilp, Thailand

Konstantin Gundev, Bulgaria

Jan Rydzewski, Poland

Martin Stadler, Austria

Johan Runeson, Sweden

Martin Raszyk, Czech Republic

Iiro Lehto, Finland

Thanh Phong Lê, Switzerland

Mehrdad Malak Mohammadi,

Aliaksandr Yankouski, Belarus

Islamic Republic of Iran

Sepehr Ebadi, Canada

Hólmfríður Hannesdóttir, Iceland

Domen Ipavec, Slovenia

Photo by Henry Teigar

Mohamad Ansarifard, Islamic Republic of Iran Francisco Machado, Portugal

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Ruben Doornenbal, Netherlands

Giorgi Kobakhidze, Georgia

Yun Jia (Melody) Guan, Canada

Troy Figiel, Netherlands

Ion Toloaca, Moldova

Áron Dániel Kovács, Hungary

Wai Pan Si, Macao, China

Ilija Burić, Serbia

Eden Segal, Israel

Nicoleta Colibaba, Moldova

Jose Luciano De Morais Neto, Brazil

Abdullah Alsalloum, Saudi Arabia

Tobias Karg, Austria

Haji Piriyev, Azerbaijan

Sergi Chalalauri, Georgia

Javier Mendez-Ovalle, Mexico

Meylis Malikov, Turkmenistan

Ka Fai Chan, Macao, China

Virab Gevorgyan, Armenia

Guilherme Renato Martins Unzer, Brazil

Tristan Downing, Canada

Muhammad Taimoor Iftikhar, Pakistan

Luqman Fathurrohim, Indonesia

Emmanouil Vourliotis, Greece

Aitor Azemar, Spain

Jemal Shengelia, Georgia

Dinis Cheian, Moldova

Lara Timbo Araujo, Brazil

Lo Hei Chun, Hong Kong

Ilie Popanu, Moldova

Eric Huang, Australia

Francesc-Xavier Gispert Sánchez, Spain

Jorge Torres-Ramos, Mexico

Battsooj Bayarsaikhan, Mongolia

Roberta Răileanu, Romania

Samuel Bosch, Croatia

Jovan Blanuša, Serbia

Koen Dwarshuis, Netherlands

Michele Fava, Italy

Shinjini Saha, Bangladesh

Selver Pepic, Bosnia and Herzegovina

Leandro Salemi, Belgium

Vardan Avetisyan, Armenia

Chan Lon Wu, Macao, China

Maximilian Ruep, Austria

Adhamzhon Shukurov, Tajikistan

Saba Kharabadze, Georgia

Peter Budden, United Kingdom

Sebastian Käser, Switzerland

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Bronze Medalists Photo by Henry Teigar

Honorable Mentions Jyri Maanpää, Finland

Nudzeim Selimovic, Bosnia and Herzegovina

Sandro Maludze, Georgia

Ahmed Maksud, Bangladesh

Siobhan Tobin, Australia

Arttu Yli-Sorvari, Finland

Simão João, Portugal

Farid Mammadov, Azerbaijan

Tomas Čerškus, Lithuania

Karlo Sepetanc, Croatia

Osama Yaghi, Syrian Arab Republic

Romain Falla, Belgium

Koay Hui Wen, Malaysia

Fotios Vogias, Greece

Kasper Tolborg, Denmark

Miha Zgubič, Slovenia

Ramadhiansyah Ramadhiansyah, Indonesia Tsogt Baigalmaa, Mongolia Mohamed Alrazzouk, Syrian Arab Republic

Dominic Schwarz, Switzerland

Battushig Myanganbayar, Mongolia

Bruno Buljan, Croatia

Aram Mkrtchyan, Armenia

Bartlomiej Zawalski, Poland

Munkhtsetseg Battulga, Mongolia

Shovon Biswas, Bangladesh

Michalis Halkiopoulos, Greece

Homoud Alharbi, Saudi Arabia

Nikolaj Theodor Thams, Denmark

Bilguun Batjargal, Mongolia

Federico Re, Italy

Mantas Abazorius, Lithuania

Stefan Alexis Sigurðsson, Iceland

Photo by Henry Teigar

Ooi Chun Yeang, Malaysia

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Martin Vlashi, Italy

Dale Alexander Hughes, Ireland

Mathias Stichelbaut, Belgium

Luka Ivanovskis, Latvia

Tapio Hautamäki, Finland

I Made G.N. Kumara, Indonesia

Jurij Tratar, Slovenia

Mohammad Alhejji, Saudi Arabia

Wai Hei Hoi, Macao, China

David Trillo Fernández, Spain

Ali Alhulaymi, Saudi Arabia

Aleksandr Petrosyan, Armenia

Kaloyan Darmonev, Bulgaria

Vladimir Pejovic, Montenegro

Pedro Paredes, Portugal

Andres Rios Tascon, Colombia

Laura Gremion, Switzerland

Eduardo Acosta-Reynoso, Mexico

Yeoh Chin Vern, Malaysia

Werdi Wedana Gunawan, Indonesia

Grgur Simunic, Croatia

Ghadeer Shaaban, Syrian Arab Republic

Matheus Marreiros, Portugal

Maris Serzans, Latvia

Thijs van der Gugten, Netherlands

Pétur Rafn Bryde, Iceland

Salizhan Kylychbekov, Kyrgyzstan

Kemal Babayev, Turkmenistan

Molte Emil Strange Andersen, Denmark

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Honorable Mentioned students Photo by Henry Teigar

Special Prizes IPhO 2012 Absolute winner Attila Szabó, Hungary

Special Prizes for the Best Solution of the Experimental Christoph Schildknecht, Switzerland Ivan Ivashkovskiy, Russia Chi Shu, People’s Republic of China

Special Prizes for the Best Solution of the Theoretical Eric Schneider, United States of America

Best in Experiment

Absolute winner Attila Szabó with Jaan Kalda Photo by Henry Teigar

Kai-Chi Huang, Taiwan

Best in Theory Attila Szabó, Hungary

Prizes of Association of Asia Pacific Physical Societies Jeevana Priya Inala, India Hengyun Zhou, People’s Republic of China

Prize of European Physical Society Attila Szabó, Hungary

Special Price for the Best Girl IPhO 2012 Alexandra Vassiljeva, Russia

Special Price for the best Estonian student Jaan Toots, Estonia

Special Price for the best innovative solution Lev Ginzburg, Russia 139

Statistics of the competition results Jaan Kalda Academic Committee of IPhO 2012 Let us analyze the overall difficulty of the problem set using the score-rank relationship 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 lastplace-owner with 0 points. Bearing in mind that the performance of the contestants 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 questions 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.

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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 moderation. 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 sometimes 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 experimental marks separately.

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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 rightmost 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.

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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 totalling 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 Szab贸 (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);

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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 corner. 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.

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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 contestants with top scores are as follows (all gold medalists, unless otherwise noted).

8.8: Chi Shu (CHN); 8.5: Kai-Chi Huang (TWN);

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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 contestants, 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

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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 education 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 contestants 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,

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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.

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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 phenomenon (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.)

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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 comparable. 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 â&#x20AC;&#x201C; 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.

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International board Minutes Minutes 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

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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 Mr. Attila Szabó, Hungary 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 Mr. Attila Szabó, Hungary 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 :

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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 committee 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

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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 preparatory 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)

(signature)

(signature)

......................

......................

......................

Prof. Hans Jordens

Prof. Ming-Juey Lin

Prof. Jaak Kikas

President of the IPhO

Secretary of the IPhO

Head of the Academic Committee of IPhO2012 Tallinn, Estonia

July 22nd, 2012

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Photos by Merily Salura, Henry Teigar and Karl Veskus

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Statutes Statutes 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 education 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 presented 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 participation 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

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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 recognition 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 scientific community. Therefore all possible political tensions between the participants should not be reflected in any activity during the competition. Any political activity directed against any individuals or countries is strictly prohibited. §3 Each participating country shall send a delegation, normally consisting of five students (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, capable 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 delegation 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 examination and one for the experimental examination. There will be at least one full

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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-programmable 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 presented 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 delegation 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

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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 organising 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 members 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 authorised 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

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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

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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.

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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 participated 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.

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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 according 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 knowledge 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

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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 problem must be accepted. The host country will be responsible for grading the examination 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 students’ 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 collusion 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.

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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 conclusion 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 IPhOhome-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.

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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.

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Syllabus

Appendix 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

Vector description of the position of

point mass

the point mass, velocity and acceleration 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

Hooke’s law, coefficient of friction

law of gravitation, potential energy

(F/R = const), frictional forces, static

and work in a gravitational field

and kinetic, choice of zero of potential energy

f) Centripetal acceleration, Kepler’s laws

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2. Mechanics of Rigid Bodies a) Statics, center of mass, torque

Couples, conditions of equilibrium of bodies

b) Motion of rigid bodies, transla-

Conservation of angular momen-

tion, rotation, angular velocity,

tum about fixed axis only

angular acceleration, conservation of angular momentum c) External and internal forces, equa-

Parallel axes theorem (Steiner’s

tion of motion of a rigid body around

theorem), additivity of the moment

the fixed axis, moment of inertia,

of inertia

kinetic energy of a rotating body d) Accelerated reference systems,

Knowledge of the Coriolis force

inertial forces

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

Thermal equilibrium, quantities

heat, first and second laws of

depending on state and quantities

thermodynamics

depending on process

b) Model of a perfect gas, pressure

Also molecular approach to such

and molecular kinetic energy,

simple phenomena in liquids and

Avogadro’s number, equation of

solids as boiling, melting etc.

state of a perfect gas, absolute temperature c) Work done by an expanding gas

Proof of the equation of the adiaba-

limited to isothermal and adiabatic

tic process is not required

processes d) The Carnot cycle, thermodynamic

Entropy as a path independent

efficiency, reversible and irrevers-

function, entropy changes and

ible processes, entropy (statistical

reversibility, quasistatic processes

approach), Boltzmann factor

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5. Oscillations and waves a) Harmonic oscillations, equation

Solution of the equation for har-

of harmonic oscillation

monic motion, attenuation and

b) Harmonic waves, propagation of

Displacement in a progressive wave

waves, transverse and longitudinal

and understanding of graphical

waves, linear polarization, the clas-

representation of the wave, measure-

sical Doppler effect, sound waves

ments of velocity of sound and light,

resonance -qualitatively

Doppler effect in one dimension only, propagation of waves in homogeneous and isotropic media, reflection and refraction, Fermat’s principle c) Superposition of harmonic

Realization that intensity of wave

waves, coherent waves, interfer-

is proportional to the square of its

ence, beats, standing waves

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 secondary sources (diffraction)

6. Electric Charge and Electric Field a) Conservation of charge,

Coulomb’s law b) Electric field, potential, Gauss’

Gauss’ law confined to simple sym-

law

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

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a) Current, resistance, internal

Simple cases of circuits containing

resistance of source, Ohm’s law,

non-ohmic devices with known V-I

Kirchhoff’s laws, work and power

characteristics

of direct and alternating currents, Joule’s law b) Magnetic field (B) of a current,

Particles in a magnetic field, simple

current in a magnetic field, Lorentz

applications like cyclotron, mag-

force

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, permeability, energy density of magnetic field e) Alternating current, resistors,

Simple AC-circuits, time constants,

inductors and capacitors in

final formulae for parameters of

AC-circuits, voltage and current

concrete resonance circuits are not

(parallel and series) resonances

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,resolving power of a grating, Bragg reflection, c) Dispersion and diffraction spec-

tra, line spectra of gases d) Electromagnetic waves as trans-

Superposition of polarized waves

verse waves, polarization by reflection, polarizers

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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

Einstein’s formula is required

impulse of the photon 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, conservation 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 (qualitatively), alpha-, beta- and gamma-decays, absorption of radiation, halflife and exponential decay, components of nuclei, mass defect, nuclear reactions

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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, potentiometers, 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, ratemeter, signal and function generators, analog-to-digital converter connected to a computer, amplifier, integrator, differentiator, power supply, universal (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.)

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Appendices Tartu – The World Capital of Physics PROGRAM 20th of July 10:00 – 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.

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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.

Letâ&#x20AC;&#x2122;s go inside a head.

We will go inside a head and study how the brain works: eyesight, reading, decision-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.

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Workshops with definite start and end times 12:30 - 13:15

Glance at the past of the University of Tartu

13:30 - 14:15

Guided tour in the history museum

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

13:15 - 13:45

During the planetarium show â&#x20AC;&#x153;Night skyâ&#x20AC;?, the planets,

14:00 - 14:30

their movement during the year will be shown together with more interesting constellations from both the northern and southern hemisphere.

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

13:15-14:00

Observation of the sun and a workshop for making a

14:00-14:45

sun clock

14:45-15:30

Workshop for making a quadrant

15:30-16:15

Weather measurement

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Year of Science in Estonia and Tartu — the World Capital of Physics Viire 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 delegation 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

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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 international 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 hosting city of the 43rd International Physics Olympiad in 2012.

The statute establishes the aims of nomination and the principles of co-operative 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)

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●● 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 nomination 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 organizers 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

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Tartu town hall square

President of IPhO Hans Jordens declares Tartu as the World Capital of Physics

Geniuses in red

Estonian student Kaur Aare Saar reads the Declaration of Tartu as the World Capital of Physics

Choir named E STuudio presenting Estonian folk music

Photos by Henry Teigar, Karl Veskus and Siim Pille

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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

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Circulars First 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 leaders), 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.

Invitation On 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 students aimed to stimulate young peopleâ&#x20AC;&#x2122;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 teams According 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 finished 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.

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Fees and Expenses The 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 participating 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 additional cost.

Information and contacts For 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: ipho2012@eitsa.ee Phone number: +372 628 5800 Address: 43rd IPhO Organizing Committee Estonian Information Technology Foundation Raja 4C, 12616 Tallinn Estonia

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Preliminary confirmation of participation As 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 (ipho2012@eitsa.ee) 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

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Second Circular Dear 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 date Until the present date, the following 86 teams have confirmed their participation in the 43rd IPhO: ●● Albania

●● France

●● Armenia

●● Georgia

●● Australia

●● Germany

●● Austria

●● Ghana

●● Azerbaijan

●● Greece

●● Bangladesh

●● Hong-Kong

●● Belarus

●● Hungary

●● Belgium

●● Iceland

●● Bolivia

●● India

●● Bosnia & Herzegovina

●● Indonesia

●● Brazil

●● Islamic Republic of Iran

●● Bulgaria

●● Ireland

●● Cambodia

●● Israel

●● Canada

●● Italy

●● People`s Republic of China

●● Japan

●● Colombia

●● Jordan

●● Croatia

●● Kazakhstan

●● Cuba

●● Republic of Korea

●● Cyprus

●● Kuwait

●● Czech Republic

●● Kyrgyzstan

●● Denmark

●● Lao PDR

●● Ecuador

●● Latvia

●● El Salvador

●● Liechtenstein

●● Estonia

●● Lithuania

●● Finland

●● Macao, China

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●● Macedonia

●● Republic of Singapore

●● Malaysia

●● Slovakia

●● Mexico

●● Slovenia

●● Moldova

●● South Africa

●● Montenegro

●● Spain

●● Netherlands

●● Sri Lanka

●● Nigeria

●● Suriname

●● Norway

●● Sweden

●● Pakistan

●● Switzerland

●● Philippines

●● Syria

●● Poland

●● Taiwan

●● Portugal

●● Tajikistan

●● Puerto Rico

●● Thailand

●● Qatar

●● Ukraine

●● Romania

●● United Kingdom

●● Russia

●● United States of America

●● Saudi Arabia

●● Uzbekistan

●● Serbia

●● Vietnam

Official invitation During 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 (ipho2012@eitsa.ee).

Online registration The 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).

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Visa Complete 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 process early. It will take approximately 2â&#x20AC;&#x201C;3 weeks for the applications to be processed.

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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 (ipho2012@eitsa.ee) .

Fees and payment All 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 (ipho2012@eitsa.ee)

Transportation IPhO 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.

Accommodation In Tallinn students will be staying at the Sokos Hotel Viru. During the studentâ&#x20AC;&#x2122;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 OlĂźmpia in Tallinn.

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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 studentsâ&#x20AC;&#x2122; hotel and EUR 75 for the leadersâ&#x20AC;&#x2122; hotel.

43rd IPhO Website and Contacts The circulars and all information regarding the 43rd IPhO can be found at our website 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

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Third Circular Dear 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 preparation and necessary information for your teams.

Registration The online registration is about to complete, the following 86 teams have confirmed their participation in the 43rd IPhO: 1. Albania

25. Georgia

2. Armenia

26. Ghana

3. Azerbaijan

27. Germany

4. Australia

28. Greece

5. Austria

29. Hong Kong

6. Bangladesh

30. Hungary

7. Belarus

31. Iceland

8. Belgium

32. India

9. Bolivia

33. Indonesia

10. Bosnia & Herzegovina

34. Ireland

11. Brazil

35. Islamic Republic of Iran

12. Bulgaria

36. Israel

13. Canada

37. Italy

14. Colombia

38. Japan

15. Croatia

39. Kazakhstan

16. Czech Republic

40. Kuwait

17. Cuba

41. Kyrgyzstan

18. Cyprus

42. Latvia

19. Denmark

43. Liechtenstein

20. Ecuador

44. Lithuania

21. El Salvador

45. Macao, China

22. Estonia

46. Macedonia

23. Finland

47. Malaysia

24. France

48. Mexico

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49. Moldova

68. Slovakia

50. Mongolia

69. Slovenia

51. Montenegro

70. South Africa

52. Nepal

71. Spain

53. Netherlands

72. Sri Lanka

54. Nigeria

73. Suriname

55. Norway

74. Sweden

56. Pakistan

75. Switzerland

57. People`s Republic of China

76. Syria Arab Republic

58. Poland

77. Taiwan

59. Portugal

78. Tajikistan

60. Puerto Rico

79. Thailand

61. Qatar

80. Turkey

62. Republic of Korea

81. Turkmenistan

63. Republic of Singapore

82. Ukraine

64. Romania

83. United Kingdom

65. Russia

84. United States of America

66. Saudi Arabia

85. Uzbekistan

67. Serbia

86. Vietnam

Fees and payment All 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 (ipho2012@eitsa.ee)

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Transportation IPhO 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 environment by June 29th.

Accommodation In 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 leaders Computing facilities, including laptops and printers, will be available in the International Board Meeting Room.

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You can bring your own laptop/notebook. Standard power supply in Estonia is 220V/50 Hz.

Calculators Notice 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-programmable 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 mathematical 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.

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The Chemistry Olympiad in the United States of America Some 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 ipho2012@eitsa.ee

Information leak during the competition The 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 communication 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 collusion 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.

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43rd IPhO Website and Contacts The 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

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Newsletter

IPhO 2012 newsletter was called â&#x20AC;&#x153;Physics is in The Airâ&#x20AC;? 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

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Afterword 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, transportation 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

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Photo by Merily Salura

the leaders and markers to use the computer whenever they wanted. All marks were inserted in the web-based evaluation system. The studentsâ&#x20AC;&#x2122; 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 culture and variety of nature.

Organizing Committee IPhO 2012

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