All-Russian Olympiad schoolchildren in physics consists of four stages ...
In the first of them, school, students can participate, starting from the seventh grade. It is held in September-October. As a rule, participants are asked to solve 4-5 problems.
This is followed by the municipal stage, which also takes place for schoolchildren in grades 7-11. And in the next, regional, only students of grades 9-11 can participate (for seventh and eighth graders, an analogue of the third and fourth stages is held - the Maxwell Olympiad). It is at this stage that an experimental one is added to the theoretical one.
About 300 schoolchildren take part in the final stage every year. In it, as well as in the regional one, there are two rounds. The winners and prize-winners of the final are enrolled in specialized universities without exams.
School Olympiads in physics have been held in Moscow since 1938. The first all-Union competition took place in 1962.
What's new
How to participate
- Let the school know about your desire to participate in the Olympiad, find out when and where the first stage will take place.
- Participate in the school stage.
- Wait for your results, ask the school for the passing score for the municipal stage and information about the conduct.
- Get ready and come to the municipal stage.
- Compare your verified work with the criteria, in case of disagreement with the points - ask the jury.
- Find out the passing scores on regional stage and information about it. For example, on the page of the All-Russian Olympiad in your region. Sites of organizers in the regions →
- Come to the regional stage. For a successful performance, one must participate in two rounds: theoretical and experimental.
- Wait for the results, review your proven work and criteria. If you find discrepancies, ask questions of the jury and file an appeal.
- Passing points for The final stage search on the Internet, they are published by the Ministry of Education and Science of Russia.
- All information about the trip to the final will be given to you by the person responsible for the All-Russian Olympiad in your region. Contacts of those responsible for the Olympiad in the regions →
What's Special
How to prepare
Solve problems from previous years Deal with difficult passages with the teacher. Ask questions. The school is interested in your success - this increases its prestige. Tasks and solutions →
Discussion of the Olympiad
Anna Solntseva, November 26, 2016 Hello. I have been looking for information on the Vseros website for a long time that the winners and prize-winners of last year can participate this year. But I don’t understand from what stage of the All-Russian Olympiad for schoolchildren. For example, if I am a prize winner in physics municipal stage in Moscow last year for class 8, can I immediately go to the regional stage this year for class 9? Or can I go to the municipal stage this year for 9 cells, but not to the region? I remember there was a link to an official document somewhere, but I can't find it. Help, who is in the subject, please! Another second point: is it absolutely a rule that the winners and prize-winners of the last year go to the next stage of this year (or the same stage of this year) automatically? Or does it depend on the number of points scored last year compared to the level of this year's participants? Example: last year there were sooo many winners in English in the municipal stage of grade 8. And only the winners were taken to the region. Well, what should the winner of last year do in this case, he did not know that the level would increase in this way, so he did not go to the same stage this year, he decided to use the right of prize-winning. And this year, his last year's prize-winning was not enough. It turns out that it is risky to sit and hope for last year's medals, because this year they can raise the requirements, and let only the winners go to the next round. So? Or to establish a new limit of points, with which the winner of last year will be in the span and will not be able to automatically pass on the basis of last year's merits to the next round. Could it be? Or, REGARDLESS of what levels and rules for announcing winners and winners of the municipal stage of this year, last year's winners and winners will automatically go to the region? Please comment. And if you know a link on this topic, please send.
Second (municipal) stage
All-Russian Olympiad for schoolchildren in physics
10.1. A thin hoop of mass lies flat on a smooth horizontal table. M. A light inextensible thread is wound around the perimeter of the hoop, we pull with force on the free end of the thread F directed tangentially to the hoop. With what acceleration does the end of the thread, for which we pull, move?
Decision
The hoop will slide on the table, and at the same time the thread will unwind from it. As a result, the hoop will make a complex movement, which can be represented as the sum of the translational movement of the hoop as a whole (in the absence of rotation) and the rotational movement of the hoop around its axis (with the center of the hoop stationary). Since the thread is inextensible, the desired acceleration of its end is equal to the tangential (tangential) acceleration of the hoop point where it touches the thread. In accordance with the rule of addition of accelerations, this acceleration is equal to the sum of the acceleration associated with the translational movement of the hoop and the tangential component of the acceleration of the points of the hoop associated with its rotational movement: a = a post + a rotation
Since the hoop makes forward movement under constant force F, then a post = F/M. Due to the fact that the hoop is thin and all its elements are at the same distance from the axis of rotation, the tangential component of the acceleration of the points of the hoop is also equal to a rotation = F/M. Therefore, the desired acceleration of the end of the thread is equal to a threads = a = 2F/M.
Criteria
Points | What are points for? |
Complete correct solution |
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Correctly found a post and a rotating, but then they are incorrectly folded or not folded at all. |
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Correctly found a post or a rotation (any one of the quantities). |
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10.2. At your disposal are 6 resistors with a resistance of 100 ohms. How should they be connected to get a resistor as close to 60 ohms as possible? It is not necessary to use all resistors!
Decision
Consider three electrical circuit diagrams:
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Let's calculate the resistance of these circuits:
100 ohm/2 = 50 ohm |
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≈ 66.7 Ohm
= 60 ohmConnecting resistors according to scheme 3 gives the best result, exactly 60 ohms.
Criteria
Points | What are points for? |
The scheme is given desired circuit and a calculation was made proving that its resistance is 60 ohms. |
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3 or more circuits of various circuits were considered and calculations of their resistances were made, but the circuits of the desired circuit (with a resistance of exactly 60 ohms) are not among them. |
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1 or 2 circuits of various circuits are considered and calculations of their resistances are made, but the circuits of the desired circuit (with a resistance of exactly 60 Ohms) are not among them. |
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1 circuit diagram was considered and a calculation of its resistance was made, but this circuit is not the desired one (with a resistance of exactly 60 ohms). |
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There are separate equations or drawings related to the essence of the problem, in the absence of a solution (or in case of an erroneous solution). |
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The solution is incorrect or missing. |
10.3. Two streams of liquids with different temperatures are fed into the vessel through two tubes. After mixing and establishing the temperature in the vessel, the excess liquid flows out. In the first experiment, the temperatures of the liquids were +50°C and +80°C, and the resulting temperature in the vessel was +60°C. In the second experiment, the flow rate of the first liquid was increased by 1.2 times, and its temperature was brought up to +60 °C. The flow rate of the second liquid and its temperature did not change. Find the steady temperature.
Decision
Let us write the heat balance equations for both experiments. Let us denote the flow rates of liquids by mass through M and a∙M, respectively, their specific heat capacity - through c, temperatures - through t 1 = +50 °С, t 2 = +80 °С, t 3 \u003d +60 ° С, and the desired temperature - through t.
Let's solve the resulting system of equations:
=>
=>
Criteria
Points | What are points for? |
Complete correct solution |
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The correct solution, in which there are minor flaws that generally do not affect the solution (misprints, errors in calculations, etc.). |
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The heat balance equations were written correctly for both experiments, but no solution was obtained. |
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The heat balance equation is correctly written for only one of the experiments. |
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There are separate equations related to the essence of the problem, in the absence of a solution (or in case of an erroneous solution). |
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The solution is incorrect or missing. |
10.4. On a smooth horizontal table there is a light rod, to the ends of which short, inextensible pieces of a light thread are tied. Weights are attached to the free ends of the pieces of thread M and 3 M lying on the table (see picture). The threads do not sag at first. A force is applied to the middle of the rod F, parallel to the pieces of thread and perpendicular to the rod. Find the acceleration of the middle of the rod. Count quickly before the rod turns!
Decision
Since the rod is light, the sum of the moments of the tension forces of the threads T 1 and T 2 and strength F, calculated relative to the axis passing through any point, must be equal to zero. Hence, T 1 = T 2 = F/2.
Since the threads are inextensible and do not sag, the accelerations of the ends of the rod are equal to the accelerations of the loads attached to them: 
for the left end of the rod and https://pandia.ru/text/78/452/images/image014_43.gif" width="123" height="42 src=">.
Criteria
Points | What are points for? |
Complete correct solution |
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The correct solution, which has minor flaws that generally do not affect the solution (for example, typos). |
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Accelerations of the ends of the rod (or weights) are correctly found, but the acceleration of the middle of the rod is not defined. |
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Correctly found the tension forces of the threads. |
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There are separate equations or drawings with explanations related to the essence of the problem, in the absence of a solution (or in case of an erroneous solution). |
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The solution is incorrect or missing. |