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Students who are winners of the respective national competitions are invited to participate in the IESO, and all interested countries are encouraged to contribute to the IESO. It is one of the major activities of the International Geoscience Education Organization — IGEO that aims at raising student interest in and public awareness of Earth sciences, as well as to enhance Earth science learning of students. The international competition was adopted as one of major activities of International Geoscience Education Organization later that year.

In November in Seoul, rep. The school offers both 4-year and 8-year study programs, and students can specialize in mathematics and physics or in foreign languages. Look up iao in Wiktionary, the free dictionary. The Czech Astronomical Olympiad is organised for pupils of elementary and secondary schools in the Czech Republic. It is very popular among younger pupils because of the opportunity it provides to use all informational resources with the exception of the help of other persons. Practical observational tasks are required in the correspondence round. After the final round of categories "EF" and "CD", a two-day workshop is held to select candidates for the International Astronomy Olympiad.

Over the years, the institute has evolved research programmes in various areas, with an emphasis on technology and applied research. History IIIT Hyderabad was set up in under the public private partnership model,[3] with the state government supplying a grant of land and buildings. Rajeev Sangal designed the syllabus and served as the first director of the institute until 10 April Olympiad is being conducted in one theoretical stage by correspondence.

The languages of Olympiad are English and Russian. A deadline of paper submission is February, 11th Year Number of participants Number of countries 26 2 46 7 44 12 54 10 See also International Astronomy Olympiad External links The official website. There are about students with most going on to university in the second year. All students live in the dormitory, two students to a room. IAPT[1] conducts the National Standard Examination in Physics[2] in November of each year throughout the country, with approximately 31, participants.

Students from class 8 to class 10 can participate in the competition. Competition Pattern There are 4 levels in NSO, an online examination conducted on the website, a written examination and two level of Quiz competition related to astrophysics, space science etc. The online examination and written examination consists of multiple choice questions. Students must pass the online examination to attend the event. Students of any recognized school. Korean Astronomical Society is a non-profit corporation that aims a contribution toward an astronomy scholarship, technological development, education, education, and the spread of astronomical knowledge.

It operates Journal of the Korean Astronomical Society. It was founded on 21 March , and has over members. Retrieved 18 February Nakamura, Tsuko; Orchiston, Wayne, eds. It is ranked number one at International Science Olympiads by the number of medals won by its students more than There are approx.

More than half of the professors are former students of the School. Retrieved Science Olympiad logo Science Olympiad is an American team competition in which students compete in 23 events pertaining to various scientific disciplines, including earth science, biology, chemistry, physics, and engineering. Over 7, middle school and high school teams from 50 U. Invitational tournaments, run by high schools and universities, are unofficial tournaments and serve as practice for regional and state competitions. Winners later receive several kinds of awards, including medals, trophies and plaques, as well as scholarships.

The school serves as a national educational institution to nurture talented Vietnamese students who excelled at natural sciences. The largest percentage of its graduates attend the most prestigious universities in Vietnam. The department of Mathematics was established first in , followed by the department of Physics; the department of Chemistry and Biology was established in The Philippine Science High School System is a research-oriented and specialized public high school system in the Philippines that operates as an attached agency of the Philippine Department of Science and Technology.

Admission to the PSHS is by competitive examination, and only Filipino citizens are eligible to attend. Graduates of the PSHS are bound by law to major in the pure and applied sciences, mathematics, or engineering on entering college. The system is known to have a very challenging curriculum which produces the best professionals in the country.

It was founded in The broad goals of the institute are to promote equity and excellence in science and mathematics education from primary school to undergraduate college level, and encourage the growth of scientific literacy in the country. It is India's nodal centre for Olympiad programmes in mathematics, physics, chemistry, biology[3] and astronomy. HBCSE runs a graduate sch. It is well renowned for producing the highest number of fresh graduates to Indonesia's top notch universities, most notably the Bandung Institute of Technology more than any other high school in the country on an annual basis.

Wolff Schoemaker. Construction began in Fifteenth Gymnasium Croatian: XV. It specializes in mathematics and computer science and it is considered to be the best high school in the capital along with Fifth Gymnasium V. The Meissel—Mertens constant is analogous to the Euler—Mascheroni constant, but the harmonic series sum in its definition is only over the primes rather than over all integers and the logarithm is taken twice, not just once.

Mertens's theorems are three results related to the density of prime numbers. The competition pits students from the thirty-four provinces of Indonesia, and winners of the competition are further selected to represent Indonesia in their respective subjects' International Science Olympiad. The first ever national-stage competition was held in Yogyakarta, and in it was held in Balikpapan with improved rules and procedures.

The streams offered are Science and Commerce. Love house- green 2. Peace hous. In a ranking of Chinese high schools that send students to study in American universities, it ranked number 21 in mainland China in terms of the number of students entering top American universities. It also has three other campuses, including two in Guangzhou, one in Foshan and one in Shanwei.

The main campus is a provincial public school, while the other campuses are private schools. History HSFZ is the result of a series of ren. The streams offered are Science, Commerce and Arts. The chain of schools was established by Dr. Jagdish Gandhi and his wife Mrs. Bharti Gandhi. The school houses many contemporary facilities, including a fully functional physics, chemistry, biology, computer and biotechnology laboratory.

For extra-curricular activities, City Montessori School provides facilities like swimming pool, basketball court, auditorium and tennis court for students. The Pre-Primary section admits boys and girls between the ages of 2 and 5 as follows: Montessori: 2 to 3 years of age Play Group Nursery: 4 years of age Kinder. The present Principal is Mrs Swapna Mansharamani. The Vice Principal of the school is Mrs. Indu Mohan, who also heads the Senior wing of the branch. The Junior in charge is Mrs.

Poonam Arora. Victoria is the Primary in charge. The Pre-Primary section admits boys and girls between the ages of 2 and 5 as follows: Montessori: 2 to 3 yea. Artur Avila Cordeiro de Melo born 29 June is a Brazilian and French mathematician working primarily on dynamical systems and spectral theory. He is one of the winners of the Fields Medal,[2] being the first Latin American to win such an award.

Since September he is a professor at the University of Zurich. S in mathematics. That same year he moved abroad to France to. Mathematical Gymnasium Belgrade is a special school for gifted and talented students of mathematics, physics and informatics, ranked number one at International Science Olympiads by the number of medals won by its students. His previou doctoral wor. It was named by King Vajiravudh and established on July 3, The School's area holds 5 rai, 2 ngan, square metres approx.

In , Wat Suthiwararam School the system was changed from the old school move to the Western. This school was for girls and name is Satri Si Suriyothai School. The school grounds are about 3 a. It is located south of Interstate 80, and east of Watt Avenue. It is a part of the San Juan Unified School District with a student body of approximately students. Mira Loma has encouraged implementation of the International Baccalaureate program in other school districts with noted success. Because of the Doppler effect, in the laboratory system this frequency is observed as c denotes the velocity of light : a f c v f.

The arithmetic mean value of these frequencies is equal to f. Otherwise it would be necessary to use relativistic formulae for the Doppler effect. Also we neglected the recoil of atom s in the emission process.

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One should notice that for the visible radiation or radiation not too far from the visible range the recoil cannot change significantly the numerical results for the critical velocity 0 v and the ratio f f. The recoil is important for high-energy quanta, but it is not this case. The solutions were marked according to the following scheme draft : 1. Energy of excitation up to 3 points 2.

Correct description of the physical processes up to 4 points 3. Doppler effect up to 3 points Problem 2 Consider a parallel, transparent plate of thickness d Fig. Its refraction index varies as R x n n. Find the refraction index B n at the point B. Find B x i. Find the thickness d of the plate. Data: 2. Solution Fig. So, we may make use of it also in case of continuous dependence of the refractive index in one direction in our case in the x direction.

Consider the situation shown in Fig. The refractive index at this point is 0 n. Numerically 3. B The value of B x can be found from the dependence x n given in the text of the problem. The answer to the third question requires determination of the trajectory of the light ray. According to considerations described at the beginning of the solution we may write see Fig. It means that at the point C the ray must be tangent to the circle. Moreover, the ray that is tangent to the circle at some point must be tangent also at farther points.

Therefore, the ray cannot leave the circle as long as it is inside the plate! Thus, the ray must propagate along the circle shown in Fig. The shape of the trajectory x y can be determined also by using more sophisticated calculations. But tg x is the derivative of x y. It means that the ray moves in the plate along to the circle as found previously.

R,0 y x Fig. We draw a number of straight lines inside the plate close to each other and passing trough the point R,0 - Fig. From the formula given in the text of the problem it follows that the refraction index on each of these lines is inversely proportional to the distance to the point R,0. Now we draw several arcs with the center at R,0. It is obvious that the geometric length of each arc between two lines is proportional to the distance to the point R,0. It follows from the above that the optical path a product of geometric length and refractive index along each arc between the two lines close to each other is the same for all the arcs.

According to the Huygens principle, the secondary sources on this line emit secondary waves. The wave fronts of secondary waves, shown in Fig. At the beginning the wave front of the light coincided with the x axis, it means that inside the plate the light will move along the circle with center at the point R,0. Correct description of refraction at points A and B up to 2 points 3. Calculation of B x up to 1 point 4.

Calculation of d up to 5 points Problem 3 A scientific expedition stayed on an uninhabited island. The members of the expedition had had some sources of energy, but after some time these sources exhausted. Then they decided to construct an alternative energy source. Unfortunately, the island was very quiet: there were no winds, clouds uniformly covered the sky, the air pressure was constant and the temperatures of air and water in the sea were the same during day and night.

Fortunately, they found a source of chemically neutral gas outgoing very slowly from a cavity. The pressure and temperature of the gas are exactly the same as the pressure and temperature of the atmosphere. The expedition had, however, certain membranes in its equipment. One of them was ideally transparent for gas and ideally non-transparent for air. Another one had an opposite property: it was ideally transparent for air and ideally non-transparent for gas. The members of the expedition had materials and tools that allowed them to make different mechanical devices such as cylinders with pistons, valves etc.

They decided to construct an engine by using the gas from the cavity. Show that there is no theoretical limit on the power of an ideal engine that uses the gas and the membranes considered above. Solution Let us construct the device shown in Fig. B 1 denotes the membrane transparent for the gas from the cavity, but non-transparent for the air, while B 2 denotes the membrane with opposite property: it is transparent for the air but non-transparent for the gas.

Initially the valve Z 1 is open and the valve Z 2 is closed. Let 0 V denotes an initial volume of the gas at pressure 0 p. Now we close the valve Z 1 and allow the gas in the cylinder to expand. During movement of the piston in the downwards direction we obtain certain work performed by excess pressure inside the cylinder with respect to the atmospheric pressure 0 p. Due to the membrane B 2 the partial pressure of the air in the cylinder all the time is 0 p and balances the air pressure outside the cylinder.

It means that only the gas from the cavity effectively performs the work. It is obvious that the amount of work performed by the gas during isothermal expansion from 0 V to k V is represented by the area under the curve shown in the graph from 0 V to k V. Of course, the work is proportional to 0 V. We shall prove that for large enough k V the work can be arbitrarily large. It means that during isothermal expansion of a given portion of the gas we may obtain arbitrarily large work at the cost of the heat taken from sthe urrounding it is enough to take k V large enough.

After reaching k V we open the valve Z 2 and move the piston to its initial position without performing any work. The cycle can be repeated as many times as we want. In the above considerations we focused our attention on the work obtained during one cycle only. We entirely neglected dynamics of the process, while each cycle lasts some time.

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One may think that - in principle - the length of the cycle increases very rapidly with the effective work we obtain. This would limit the power of the device we consider. Take, however, into account that, by proper choice of various parameters of the device, the time taken by one cycle can be made small and the initial volume of the gas 0 V can be made arbitrarily large we consider only theoretical possibilities we neglect practical difficulties entirely. In our analysis we neglected all losses, friction, etc. One should remark that there are no theoretical limits for them. These losses, friction etc.

Does this contradict the second law of thermodynamics? It is true that there is no temperature difference in the system, but the work of the device makes irreversible changes in the system mixing of the gas from the cavity and the air. Model of an engine and its description up to 4 points 2. Proof that there is no theoretical limit for power up to 4 points 3. By using instruments provided by the organizers find the resistance of the resistor.

Remark: One may assume that the diode conducts current in one direction only. List of instruments: two universal volt-ammeters without ohmmeters , battery, wires with endings, graph paper, resistor with regulated resistance. Solution At the beginning we perform preliminary measurements by using the circuit shown in Fig. In this way we find that: 1. The black box conducts current in both directions; 2.

There is an asymmetry with respect to the sign of the voltage; 3. In both directions current is a nonlinear function of voltage. It is: Fig. It follows from the above that it is enough to take characteristics of the black box in both directions: by subtraction of the corresponding points graphically we obtain a straight line example is shown in Fig.

The solutions were marked according to the following scheme draft : B B A Theoretical part: 1. Proper circuit and method allowing determination of connections the elements in the black box up to 6 points 2. Determination of R principle up to 2 points 3. Remark that measurements at the same voltage in both directions make the error smaller up to 1 point 4.

Role of number of measurements affect on errors up to 1 point Experimental part: 1. Proper use of regulated resistor as potentiometer up to 2 points 2. Practical determination of R including error up to 4 points 3. Proper use of measuring instruments up to 2 points 4. Taking into account that temperature of diodes increases during measurements up to 1 point 5.

Taking class of measuring instruments into account up to 1 point Fig. Jan Mostowski for reading the text and for many valuable comments and remarks that allowed improving the final version. Altogether, 9 countries with 45 pupils participated. The entire event took place in the pedagogic academy of Gstrow. Pupils and leaders were accommodated inside the university academy complex.

On the schedule there was the competition and receptions as well as excursions to Schwerin, Rostock, and Berlin were offered. The delegation of the FRG reported of a very good organisation of the olympiad. The problems and solutions of the 8th International Physics Olympiad were created by a commission of university physics professors and lecturers. The same commission set marking schemes and conducted the correction of the tests. The correction was carried out very quickly and was considered as righteous and, in cases of doubt, as very generous.

The main competition consisted of a 5 hour test in theory and a 4. The time for the theoretical part was rather short and for the experimental part rather long. The problems originated from central areas of classical physics. The theoretical problems were relatively difficult, although solvable with good physics knowledge taught at school. The level of difficulty of the experimental problem was adequate. There were no additional devices necessary for the solution of the problems. Only basic formula knowledge was requested, and could be demanded from all pupils.

Critics were only uttered concerning the second theoretical problem thick lens. This problem requested relatively little physical understanding, but tested the mathematical skills and the routine in approaching problems e. However, it is also difficult to find substantial physics problems in the area of geometrical optics. Gorzkowski, in order to close one of the last few gaps in the IPhO-report collection. Contact: Dr. The best contestant came from the USSR and had 43 points. The first prize gold medal was awarded with 39 points, the second prize silver medal with 34 points, the third prize bronze medal with 28 points and the fourth prize honourable mention with 22 points.

Among the 45 contestants, 7 I. The following problem descriptions and solution are based mainly on a translation of the original German version from Because the original drafts are not well preserved, some new sketches were drawn. We also gave the problems headlines and the solutions are in more detail. Theoretical problem 1: Rotating rod A rod revolves with a constant angular velocity around a vertical axis A.

A body of mass m can glide along the rod. The angle is called friction angle. The angle does not change during rotation. Find the condition for the body to remain at rest relative to the rod. According to equations 2 and 4 this is equivalent to tan tan. Case 1: Z. That means, , tan 1 2 2 g r. For some special applications it is required, that the focal length is independent from the wavelength.

Sketch possible shapes of lenses and mark the central curvature points M 1 and M 2. Take into account the physical and the geometrical circumstances. Solution of problem 2: a The refractive index n is a function of the wavelength , i. According to the given formula for the focal length f see above which for a given f yields to an equation quadratic in n there are at most two different wavelengths indices of refraction for the same focal length. If the values of the radii r 1, r 2 and the thickness satisfy this condition the focal length will be the same for two wavelengths indices of refraction.

The parameters in this equation are subject to some physical restrictions: The indices of refraction are greater than 1 and the thickness of the lens is greater than 0 m. In the following it is assumed that r 1 is infinite and r 2 is finite. Equation 5 has only one physical correct solution, if Only the positve solution makes sense from the physical point of view. The ions were accelerated by a voltage U. They are deflected in a uniform magnetic field B that is perpendicular to the plane of paper.

The trajectories of the ions are symmetric to the middle perpendicular on QA. On a moving ion charge e and velocity v in a homogenous magnetic field B acts a Lorentz force F. Under the given conditions the velocity is always perpendicular to the magnetic field. Therefore, the paths of the ions are circular with Radius R.

Lorentz force and centrifugal force are of the same amount: 2 m v e v B R. Leaving the magnetic field they fly in a straight line along the last tangent. The centres of curvature of the ion paths lie on the middle perpendicular on QA since the magnetic field is assumed to be symmetric to the middle perpendicular on QA. The paths of the focussed ions are above QAdue to the direction of the magnetic field. Ions starting at an angle steeper than the tangent at Q, do not arrive in A. The boundaries of the magnetic field are given by: 1 sin cos a R r a. It is not allowed to use the multimeters as ohmmeters.

Write down your data in tabular form and plot your data. Before your measurements consider how an overload of the semiconductor element can surely be avoided and note down your thoughts. Sketch the circuit diagram of the chosen setup and discuss the systematic errors of the circuit. The input voltage U 1 varies between 0 V and 9 V. The semiconductor element is to be placed in the circuit in such a manner, that U 2 is as high as possible.

Describe the entire circuit diagram in the protocol and discuss the results of the measurements. What is a practical application of the circuit shown above? Hints: The multimeters can be used as voltmeter or as ammeter. The precision class of these instruments is 2. Therefore the measurements have to be processed in a way, that the product U I is always smaller than mW.

The figure shows two different circuit diagram that can be used in this experiment: The complete current-voltage- characteristics look like this: The systematic error is produced by the measuring instruments. Concerning the circuit diagram on the left Stromfehlerschaltung , the ammeter also measures the current running through the voltmeter. The current must therefore be corrected. Concerning the circuit diagram on the right Spannungsfehlerschaltung the voltmeter also measures the voltage across the ammeter. This error must also be corrected.

To this end, the given internal resistances of the measuring instruments can be used. Another systematic error is produced by the uncontrolled temperature increase of the semiconductor element, whereby the electric conductivity rises. The dynamic resistance is different for the two directions of the current.

By requesting that the semiconductor element has to be placed in such a way, that the output voltage U 2 is as high as possible, a backward direction should be used. Comment: After exceeding a specific input voltage U 1 the output voltage increases only a little, because with the alteration of U 1 the current I increases breakdown of the diode and therefore also the voltage drop at the resistance.

Comment: The circuit is a voltage divider circuit. Its special behaviour results from the different resistances. The resistance of the semiconductor element is much smaller than the resistance. It changes nonlinear with the voltage across the element. The circuit diagram can be used for stabilisation of voltages. If a multimeter is destroyed five points are deducted.

Solution a The block moves along a horizontal circle of radius sin R. The net force acting on the block is pointed to the centre of this circle Fig. The vector sum of the normal force exerted by the wall N, the frictional force S and the weight mg is equal to the resultant: sin 2 R m. The solution of the system of equations:. In this case there must be at least this friction to prevent slipping, i.

Condition for the minimal coefficient of friction is Fig. If increases, the block remains in equilibrium, if decreases it slips downwards. In case b : if the block slips upwards it stays there; if the block slips downwards it returns. If increases the block climbs upwards - , if decreases the block remains in equilibrium. Problem 2 The walls of a cylinder of base 1 dm 2 , the piston and the inner dividing wall are 90 a 0.

The valve in the dividing wall opens if the pressure on the right side is greater than on the left side. Initially there is 12 g helium in the left side and 2 g helium in the right side. The lengths of both sides are Outside we have a pressure of kPa. The piston is pushed slowly towards the dividing wall. When the valve opens we stop then continue pushing slowly until the wall is reached. Find the work done on the piston by us. Solution The volume of 4 g helium at C 0 temperature and a pressure of kPa is It follows that initially the pressure on the left hand side is kPa, on the right hand side kPa.

Therefore the valve is closed. Next the valve opens, thepiston is arrested. During this phase there is no change in the energy, no work done on the piston. An adiabatic compression follows from Problem 3 Somewhere in a glass sphere there is an air bubble. Describe methods how to determine the diameter of the bubble without damaging the sphere. Solution We can not rely on any value about the density of the glass. It is quite uncertain. The index of refraction can be determined using a light beam which does not touch the bubble.

Another method consists of immersing the sphere into a liquid of same refraction index: its surface becomes invisible. A great number of methods can be found. We can start by determining the axis, the line which joins the centers of the sphere and the bubble. The easiest way is to use the tumbler-over method. If the sphere is placed on a horizontal plane the axis takes up a vertical position. The image of the bubble, seen from both directions along the axis, is a circle. If the sphere is immersed in a liquid of same index of refraction the spherical bubble is practically inside a parallel plate Fig.

Its boundaries can be determined either by a micrometer or using parallel light beams. Along the axis we have a lens system consisting , of two thick negative lenses. The diameter of the bubble can be determined by several measurements and complicated calculations. If the index of refraction of the glass is known we can fit a plano-concave lens of same index of refraction to the sphere at the end of the axis Fig.

As ABCD forms a parallel plate the diameter of the bubble can be measured using parallel light beams. Focusing a light beam on point A of the surface of the sphere Fig. The rays strike the surface at the other side and illuminate a cap. Measuring the spherical cap we get angle. Angle can be obtained in a similar way at point B. The diameter of the bubble can be determined also by the help of X-rays. X-rays are not refracted by glass.

They will cast shadows indicating the structure of the body, in our case the position and diameter of the bubble. We can also determine the moment of inertia with respect to the axis and thus the diameter of the bubble. The X material is insoluble in the liquid. Examine the thermal properties of the X crystal material between room temperature and 70 C.

Determine the thermal data of the X material.

Description:

Tabulate and plot the measured data. You can use only the devices and materials prepared on the table. The damaged devices and the used up materials are not replaceable. Solution Heating first the liquid then the liquid and the crystalline substance together two time-temperature graphs can be plotted. From the graphs specific heat, melting point and heat of fusion can be easily obtained.

Literature [1] W. Compression follows an adiabatic process from point 1 to point 2, see Fig. The pressure in the cylinder is doubled during the mixture ignition The hot exhaust gas expands adiabatically to the volume V 2 pushing the piston downwards Then the exhaust valve opens and the pressure gets back to the initial value of 1 atm.

All processes in the cylinder are supposed to be ideal. The Poisson constant i. The compression ratio is the ratio of the volume of the cylinder when the piston is at the bottom to the volume when the piston is at the top. Are they realistic? Solution: a The description of the processes between particular points is the following: 01 : intake stroke isobaric and isothermal process 12 : compression of the mixture adiabatic process 23 : mixture ignition isochoric process 34 : expansion of the exhaust gas adiabatic process 41 : exhaust isochoric process 10 : exhaust isobaric process Let us denote the initial volume of the cylinder before induction at the point 0 by V 1 , after induction at the point 1 by V 2 and the temperatures at the particular points by T 0 , T 1 , T 2 , T 3 and T 4.

Hence: p 1 V. Denoting the temperature by T. We have thus obtained the correct result T. The exhaust gas does work on the piston during the expansion 34, on the other hand, the work is done on the mixture during the compression The heat is supplied to the gas during the process Since the gas is not ideal, the real eciency would be lower than the calculated one.

Dipping the frame in a soap solution, the soap forms a rectangle lm of length b and height h. White light falls on the lm at an angle measured with respect to the normal direction. The reected light displays a green color of wavelength 0. Derive the related equations. The mass of the thinnest lm thus cannot be determined by given laboratory scales. It would appear dark. An electron gun T emits electrons accelerated by a potential dierence U in a vacuum in the direction of the line a as shown in Fig. The target M is placed at a distance d from the electron gun in such a way that the line segment connecting the points T and M and the line a subtend the angle as shown in Fig.

Find the magnetic induction B of the uniform magnetic eld 5 T M a a electron gun d Figure 2: a perpendicular to the plane determined by the line a and the point M b parallel to the segment TM in order that the electrons hit the target M. Solution: a If a uniform magnetic eld is perpendicular to the initial direc- tion of motion of an electron beam, the electrons will be deected by a force that is always perpendicular to their velocity and to the magnetic eld.

Con- sequently, the beam will be deected into a circular trajectory. Namely, the motion of electrons will be composed of an uniform motion on a circle in the plane perpendicular to the magnetic eld and of an uniform rectilinear motion in the direction of the magnetic eld. The component v 1 of the initial velocity v, which is perpendicular to the magnetic eld see Fig.

The component v 2 parallel to the magnetic eld will remain 7 T M d a a v v v 1 2 electron gun Figure 4: constant during the motion, it will be the velocity of the uniform rectilinear motion. Kozel and Prof. Belonuchkin, I. Slobodetsky, S. The problem for the experimental competition has been worked out by O. Kabardin from the Academy of Pedagogical Sciences.

The engine is activated for a short time to pass at the lunar landing orbit. What amount of fuel should be spent so that when activating the braking engine at point A of the trajectory, the rocket would land on the Moon at point B Fig. In the second scenario of landing, at point A the rocket is given an impulse directed towards the center of the Moon, to put the rocket to the orbit meeting the Moon surface at point C Fig. What amount of fuel is needed in this case? Brass weights are used to weigh an aluminum-made sample on an analytical balance. The reflected light was then collected by the same telescope and focused at the photodetector.

Experi mental Probl em Define the electrical circuit scheme in a black box and determine the parameters of its elements. Li st of i nstruments: A DC source with tension 4. The rocket should then move along the elliptical trajectory with the focus in the Moons center. Thus the rocket velocity change v at point A must be. Using the momentum conservation law we obtain kg 2 2.

The difference in the scale indication F is determined by the change of difference of these forces. From expressions 2 and 3 we obtain. With the general pressure being equal, in the second case, some part of dry air is replaced by vapor: V m V m v a a a. When we wrote down expression 3 , we considered the sample mass be equal to the weights mass, at the same time allowing for a small error.

One may choose another way of solving this problem. Let us calculate the change of Archimedes force by the change of the air average molar mass. So the number of photons N getting into the pupil of the eye is equal e K h E N. The solid angle in which one can see the telescope mirror from the Moon, constitutes. The result obtained is only evaluative as the light flux is unevenly distributed inside the angle of diffraction. A transformer is built-in in a black box.

The black box has 4 terminals. To be able to determine the equivalent circuit and the parameters of its elements one may first carry out measurements of the direct current. The most expedient is to mount the circuit according to the layout in Fig. This enables one to make sure rightway that there were no e. The tests allowed for some estimations of values R and R The ammeter did not register any current between the other terminals.

This means that between these terminals there might be some other resistors with resistances larger than R L : ohm 10 Probably there might be some capacitors between terminals , , , Fig. Then, one can carry out analogous measurements of an alternative current. The taken volt- ampere characteristics enabled one to find full resistances on the alternative current of sections and Z 1 and Z 2 and to compare them to the values R 1 and R 2.

The character of the found dependences enabled one to draw a conclusion about the presence of ferromagnetic cores in the coils. J udging by the results of the measurements on the alternative current one could identify the upper limit of capacitance of the capacitors which could be placed between terminals , , , 6 9 min max 1 max 5 10 A 5 10 F 5nF 2 2 3. The plot of dependence of voltage U versus voltage U Fig. To do that one had to find the presence of induction coupling between terminals and , that is the appearance of e.

When comparing the direction of the pointers rejection of the voltmeters connected to terminals and one could identify directions of the transformers windings. Acknowledgement The authors would like to express their thanks and gratitude to Professor Waldemar Gorzkowski and Professor Ivo Volf for supplying the materials for the XI IphO in the Polish, Hungarian and Czech languages that have been of great help to the authors in their work at the problems of the Olympiad.

Gorzkowski, A. Theoretical Problem 1 A static container of mass M and cylindrical shape is placed in vacuum. One of its ends is closed. A fixed piston of mass m and negligible width separates the volume of the container into two equal parts. The closed part contains n moles of monoatomic perfect gas with molar mass M 0 and temperature T.

After releasing of the piston, it leaves the container without friction. After that the gas also leaves the container. What is the final velocity of the container? The gas constant is R. The momentum of the gas up to the leaving of the piston can be neglected. There is no heat exchange between the gas, container and the piston. The change of the temperature of the gas, when it leaves the container, can be neglected.

Do not account for the gravitation of the Earth. The nominal voltage of the lamp is ensured as the lamp is connected potentiometrically to the accumulator using a rheostat with resistance R. What is the maximal possible efficiency of the system and how the lamp can be connected to the rheostat in this case? The registered signal is proportional to the intensity of the detected waves.

The detector registers waves with electric vector, vibrating in a direction parallel to the sea surface. Determine the angle between the star and the horizont in the moment when the detector registers maxima and minima in general form. Does the signal decrease or increase just after the rise of the star? Determine the signal ratio of the first maximum to the next minimum.

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At reflection of the electromagnetic wave on the water surface, the ratio of the intensities of the electric field of the reflected E r and incident E i wave follows the low:. Does the ratio of the intensities of consecutive maxima and minima increase or decrease with rising of the star? Assume that the sea surface is flat. Solution of the Theoretical Problem 1 Up to the moment when the piston leaves the container, the system can be considered as a closed one. This temperature can be determined by the law of the adiabatic process:.

By this reason we find the first derivative. It means that the efficiency will be minimal when the rheostat resistance is minimal. The maximal current I max can be calculated using eq. The result is: I max mA. In this case the total current I will be minimal and equal to R U 0. Therefore the maximal efficiency is The used part of the rheostat is R 1 :. Solution of the Theoretical Problem 3 1 The signal, registered by the detector A, is result of the interference of two rays: the ray 1, incident directly from the star and the ray 2, reflected from the sea surface see the figure.

The phase of the second ray is shifted by due to the reflection by a medium of larger refractive index. The condition for an interference minimum is: 2 1 2 sin 2 2 max.

By this reason just after the rise of the star, the signal will increase. From the figure it is seen that max max 2. So, with the rising of the star the ratio of the intensities of the consecutive maxima and minima decreases. The mass of the rubber cord can be neglected. Make the following study: 1. Load the rubber cord with weights in the range 15 g to g. Put the data obtained into a table. Make a graph using suitable scale with the experimentally obtained dependence of the prolongation of the cord on the stress force F.

Using the experimental results, obtained in p. Do the calculations consequently for each two adjacent values of the loading in this range. Write down the formulas you have used for the calculations. Make an analytical proposition about the dependence of the volume on the loading. Determine the volume of the rubber cord, using the chronometer, at mass of the weight equal to 60 g. Write the formulas used. Solution of the Experimental Problem 1. The measurements of the cord length l n at different loadings m n must be at least The results are shown in Table I.

Table 1. The volume of the rubber cord at fixed loading can be determined investigating the small vibrations of the cord. The result for the volume V This article contains the competition problems, their solutions and a grading scheme. The organisational guidelines were laid down by the work group Olympiads for pupils of the conference of ministers of education of the German federal states.

A commission of professors, whose chairman was appointed by the German Physical Society, were concerned with the formulation of the competition problems. All other members of the commission came from physics department of the university of Kiel or from the college of education at Kiel.

The problems as usual covered different fields of classical physics. In the pupils had to deal with three theoretical and two experimental problems, whereas at the previous Olympiads only one experimental task was given. However, it seemed to be reasonable to put more stress on experimental work. The degree of difficulty was well balanced.

One of the theoretical problems could be considered as quite simple problem 3: hot-air balloon. Another theoretical problem problem 1: fluorescent lamp had a mean degree of difficulty and the distribution of the points was a normal distribution with only a few 1 Contact: Leibniz-Institute for Science Education IPN at the University of Kiel Olshausenstrasse 62, Kiel, Germany ipho ipn. The third problem problem 2: oscillation coat hanger turned out to be the most difficult problem. This problem was generally considered as quite interesting because different ways of solving were possible.

About one third of the pupils did not find an adequate start to the problem, but nearly one third of the pupils was able to solve the substantial part of the problem. That means, this problem polarized between the pupils. The two experimental tasks were quite different in respect of the input for the experimental setup and the time required for dealing with the problems, whereas they were quite similar in the degree of difficulty. Both required demandingly theoretical considerations and experimental skills.

Both experimental problems turned out to be rather difficult. The tasks were composed in a way that on the one hand almost every pupil had the possibility to come to certain partial results and that there were some difficulties on the other hand which could only be solved by very few pupils. The difficulty in the second experimental problem problem5: motion of a rolling cylinder was the explanation of the experimental results, which were initially quite surprising. The difficulty in the other task problem 4: lens experiment" was the revealing of an observation method with a high accuracy parallax.

The five hours provided for solving the two experimental problems were slightly too short. According to that, in both experiments only a few pupils came up with excellent solutions. In problem 5 nobody got the full points. The problems presented here are based on the original German and English versions of the competition problems.

The solutions are complete but in some parts condensed to the essentials. Almost all of the original hand-made figures are published here. Theoretical Problems Problem 1: Fluorescent lamp An alternating voltage of 50 Hz frequency is applied to the fluorescent lamp shown in the accompanying circuit diagram. Name and explain this function! Hint: The starter includes a contact which closes shortly after switching on the lamp, opens up again and stays open.

How does this affect the operation of the lamp and to what intent is this possibility provided for? Explain the differences between the two spectra. You may walk up to the lamp and you may keep the spectroscope as a souvenir. S 4 Solution of problem 1: a The total resistance of the apparatus is Hence the total ohmic resistance is R Therefore the inductance of the series reactor is: 2 2 L Z R This yields 1 Thus o This voltage is sufficient to ignite the lamp.

The main voltage itself, however, is smaller than the ignition voltage of the fluorescent tube. The two reactances subtract and there remains a reactance of The total resistance of the arrangement is now 2 2 Z' Thus the lamp has the same operating qualities, ignites the same way, and a difference is found only in the impedance angle , which is opposite to the angle calculated in b : 1 L C Such additional capacitors are used for compensation of reactive currents in buildings with a high number of fluorescent lamps, frequently they are prescribed by the electricity supply companies.

That is, a high portion of reactive current is unwelcome, because the power generators have to be layed out much bigger than would be really necessary and transport losses also have to be added which are not payed for by the customer, if pure active current meters are used.

The continuous spectrum results from the ultraviolet part of the mercury light, which is absorbed by the fluorescence and re-emitted with smaller frequency energy loss of the photons or larger wavelength respectively. Problem 2: Oscillating coat hanger A suitably made wire coat hanger can perform small amplitude oscillations in the plane of the figure around the equilibrium positions shown. In positions a and b the long side is 6 horizontal. The other two sides have equal length. The period of oscillation is the same in all cases. What is the location of the center of mass, and how long is the period?

Nothing is known, e. Solution of problem 2 First method: The motions of a rigid body in a plane correspond to the motion of two equal point masses connected by a rigid massless rod. The moment of inertia then determines their distance. Because of the equilibrium position a the center of mass is on the perpendicular bipartition of the long side of the coat hanger. If one imagines the equivalent masses and the supporting point P being arranged in a straight line in each case, only two positions of P yield the same period of oscillation see sketch.

One can understand this by considering the limiting cases: 1. Between these extremes the period of oscillation grows continuously. The supporting point placed in the corner of the long side c has the largest distance from the center of mass, and therefore this point lies outside the two point masses. The two other supporting points a , b then have to be placed symmetrically to the center of mass between the two point masses, i.

One knows of the reversible pendulum that for every supporting point of the physical pendulum it generally has a second supporting point of the pendulum rotated by o , with the same period of oscillation but at a different distance from the center of mass. The 7 section between the two supporting points equals the length of the corresponding mathematical pendulum. Second method: Let s denote the distance between the supporting point and the center of mass, m the mass itself and the moment of inertia referring to the supporting point. Then we have the period of oscillation T : T 2 mg s.

This quadratic equation has only two different solutions at most. Therefore at least two of the three distances are equal. Third method: This solution is identical to the previous one up to equation 2. The balloon is fastened with a rope. Calculate the force on the rope.

Which height h can be gained by the balloon under these conditions? Find out by qualitative reasoning what kind of motion it is going to perform! Therefore the driving force is proportional to the elongation out of the equilibrium position. This is the condition in which harmonic oscillations result, which of course are damped by the air resistance.

Only these parts may be used in the experiment. Solution of problem 4: a For the determination of f L , place the lens on the mirror and with the clamp fix the pencil to the supporting base. Lens and mirror are then moved around until the vertically downward looking eye sees the pencil and its image side by side. In order to have object and image in focus at the same time, they must be placed at an equal distance to the eye. In this case object distance and image distance are the same and the magnification factor is 1.

It may be proved quite accurately, whether magnification 1 has in fact been obtained, if one concentrates on parallatical shifts between object and image when moving the eye: only when the distances are equal do the pencil-tips point at each other all the time. The light rays pass the lens twice because they are reflected by the mirror. Thus we find for magnification 1: L L 2 2 g b and i. Therefore these rays have to hit the mirror at right angle and so the object distance g equals the focal length f L of the lens in this case. This is accomplished either by averaging several measurements or by stating an uncertainty interval, which is found through the appearance of parallaxe.

Half the thickness of the lens has to be subtracted from the distance between pencil- point and mirror. However, the actual focal length of the single lenses spread considerably. Each lens was measured separately, so the individual result of the student can be compared with the exact value. This focal length is to be determined by a mapping of magnification 1 as above. Problem 5: Motion of a rolling cylinder The rolling motion of a cylinder may be decomposed into rotation about its axis and horizontal translation of the center of gravity. In the present experiment only the translatory acceleration and the forces causing it are determined directly.

After letting the cylinder go, it rolls with constant acceleration. Discuss the results. Before starting the measurements, adjust the plane board horizontally. For present purposes it suffices to realize the horizontal position with an uncertainty of 1 mm of height difference on 1 m of length; this corresponds to the distance between adjacent markings on the level. What would be the result of a not horizontal position of the plane board? Describe the determination of auxiliary quantities and possible further adjustments; indicate the extent to which misadjustments would influence the results.

By means of knots, the strings are put into slots at the cylinder. They should be inserted as deeply as possible. You may use the attached paper clip to help in this job. The stop watch starts running as soon as the contact at A is opened, and it stops when the contact at B is closed.

The purpose of the transistor circuit is to keep the relay position after closing of the contact at B, even if this contact is opened afterwards for a few milliseconds by a jump or chatter of the cylinder. Solution of problem 5: Theoretical considerations: a The acceleration of the center of mass of the cylinder is 2 2 s a t.

For small values of r the torque produced by the string-tensions is not large enough to provide the angular acceleration required to prevent slipping. The interaction force between cylinder and plane acts into the direction opposite to the motion of the center of mass and thereby delivers an additional torque.

For large values of r the torque produced by string-tension is too large and the interaction force has such a direction that an opposed torque is produced. With 3 and 5 you may eliminate T and a from this equation. Adjustment mentioned of strings a horizontally and b in direction of motion 0.

At the origin of Ox axis is placed a perfectly reflecting wall. Find the length of the path of the particle before it comes to a final stop b. Sketch the potential energy x U of the particle in the force field x F. Draw qualitatively the dependence of the particle speed as function of his coordinatex. The switch K being closed the circuit is coupled with a source of alternating current. The current furnished by the source has constant intensity while the frequency of the current may be varied.

The switch K is now open. Calculate the frequency of electromagnetic oscillation in 2 2 1 1 L C C L circuit; c. Determine the intensity of the electric current in the AB conductor; d. Calculate the amplitude of the oscillation of the intensity of electric current in the coil 1 L. Neglect the mutual induction of the coils, and the electric resistance of the conductors. Neglect the fast transition phenomena occurring when the switch is closed or opened. The dependences of refraction indexes of the prisms on the wavelength are given by the relations 2 1 1 1. Determine the wavelength 0 of the incident radiation that pass through the prisms without refraction on AC face at any incident angle; determine the corresponding refraction indexes of the prisms.

Draw the ray path in the system of prisms for three different radiations iolet red v , , 0 incident on the system at the same angle. Determine the minimum deviation angle in the system for a ray having the wavelength 0. Calculate the wavelength of the ray that penetrates and exits the system along directions parallel to DC.

Atomics - Problem IV 7 points Compton scattering A photon of wavelength i is scattered by a moving, free electron. Find the de Broglie wavelength for the first electron before the interaction. To characterize the photons the following notation are used: Table 4.

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The figure presents the phenomenon of the curving of the trajectory of a jet of fluid around the shape of a cylindrical surface. The trajectory of fluid is not like the expected dashed line but as the circular solid line. Qualitatively explain this phenomenon first observed by Romanian engineer Henry Coanda in This problemwill be not considered in the general score of the Olympiad.

The best solution will be awarded a special prize. In the origin of the Ox axis is placed a perfectly reflecting wall.

Description:

Find the length of the path of the particle until its final stop b. Plot the potential energy x U of the particle in the force field x F. Qualitatively plot the dependence of the particles speed as function of its xcoordinate.