{"kind":"task","effective_mode":"full","benchmark":{"kind":"benchmark","effective_mode":"full","slug":"frontierscience","formal_name":"FrontierScience","introduction":"専門的な科学課題を解く能力を評価するベンチマークです。公開データはolympiadとresearchに分かれ、競技問題と研究課題を区別して扱います。\n\nFrontierScience evaluates the ability to solve expert-level scientific tasks. Its public data separates olympiad and research problems so that competition and research tasks can be examined independently.","introduction_ja":"","introduction_en":"","category":"Category not supplied","task_count":null,"acquisition_status":"Acquisition status not supplied","official_url":"https://huggingface.co/datasets/openai/frontierscience","indexing_mode":"noindex"},"task_id":"b8c37597-b03d-5ac0-835d-48fdbae54f31","task_key":"olympiad--test--20fcf049~2db356~2d45cf~2d83d8~2d833f280ee107","task_revision_id":"1","upstream_id":"20fcf049-b356-45cf-83d8-833f280ee107","short_description":"In this question, we will explore the Brownian motion of particles in the…","config":"olympiad","split":"test","body":"{\"problem\":\"In this question, we will explore the Brownian motion of particles in the horizontal direction. Let `\\\\(m\\\\)` be the mass of the particle, `\\\\(x(t)\\\\)` be the coordinate of the particle at time `\\\\(t\\\\)`, and `\\\\(f(t)\\\\)` be the net force exerted on the particle by the medium molecules (i.e., water molecules). We divide `\\\\(f(t)\\\\)` into two parts: one is the viscous drag force `\\\\(-αv\\\\)`, and the other is the random force `\\\\(F(t)\\\\)`. Based on this, we can write the equation of motion for the particle:\\n\\n`\\\\(m\\\\frac{dv}{dt}=- \\\\alpha v + F(t)\\\\)`\\n\\nSince `\\\\(F(t)\\\\)` is random, its average value `\\\\(〈F(t)〉=0\\\\)` . We will use 〈.〉 to denote the ensemble average, which means if we replicate this system many times and let each of them evolve over time (the evolution process can be regarded as a probability distribution), then we average the values obtained by measuring these systems in the same way.\\n\\nLet us denote the average time interval between two consecutive collisions of water molecules as `\\\\(\\\\tau^{*} \\\\ll \\\\frac{m}{\\\\alpha}\\\\)`. Then `\\\\(K(s)=\\\\langle F(t+s)F(t)\\\\rangle\\\\)` can be considered as 0 when `\\\\( s \\\\gg \\\\tau^{*}\\\\)`, because during this time interval, the particle has been hit multiple times by different water molecules, and `\\\\(F(t+s)\\\\)` and `\\\\(F(t)\\\\)` can be considered independent. However, if \\\\\\\\(s\\\\\\\\) is in the order of `\\\\(\\\\tau^{*}\\\\)`, then `\\\\(K(s)>0\\\\)`. When answering the following question, account for appropriate approximations using the properties above. To facilitate the estimation of the interaction strength between the particle and water molecules, we define the following integral:\\n\\n`\\\\(J=\\\\int\\\\limits_{0}^{\\\\infty}K(s)ds\\\\)`\\n\\nDerive an expression for `\\\\(\\\\alpha\\\\) as the particle reaches thermal equilibrium with the water molecules at temperature \\\\(T\\\\)`. Express the answer in terms of `\\\\(J, k_B, T, m\\\\)`, where `\\\\(k_B\\\\)` is the Boltzmann constant.\\n\\nThink step by step and solve the problem below. At the end of your response, write your final answer on a new line starting with “FINAL ANSWER”. It should be an answer to the question such as providing a number, mathematical expression, formula, or entity name, without any extra commentary or providing multiple answer attempts.\",\"subject\":\"physics\"}","display_format":"text","language":"","answer_status":"published","assets":[],"source_url":"https://huggingface.co/datasets/openai/frontierscience","history":"initial import","indexing_mode":"noindex","subproblems":[],"grids":[]}