Turkish Tubitak olympiads problems and solutions by Rafiq Abbasov

  • #1
Killua Rafiq
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Pdf is below.
 

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  • #2
kuruman
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Your solution to Problem 1 is incorrect. It should be $$\frac{l_1}{l_2}=\frac{\tan\theta-\mu_2}{\tan\theta-\mu_1}.$$
 
  • #3
Killua Rafiq
18
1
Homework Statement:: Hi guys,i want to share with you my project about advanced physics-Physics olimpiads.I find 5 hard problems and solved these.I think it will be useful for students who prepares to olimpiad's and interests in adv physics.Have fun.
Relevant Equations:: $P=\rho gl$,$GM=gR^2$(and etc.)

Pdf is below.
 
  • #4
Killua Rafiq
18
1
Your solution to Problem 1 is incorrect. It should be $$\frac{l_1}{l_2}=\frac{\tan\theta-\mu_2}{\tan\theta-\mu_1}.$$

Your solution to Problem 1 is incorrect. It should be $$\frac{l_1}{l_2}=\frac{\tan\theta-\mu_2}{\tan\theta-\mu_1}.$$
No.It was test and there wasn't variant you said.I explained this problem.There $\mu_2 > \mu_1$.Try to solve again
 
  • #5
Killua Rafiq
18
1
$\mu_2 >\mu_1$
 
  • #6
kuruman
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I messed up an overall negative sign and the correct solution should be $$\frac{l_1}{l_2}=\frac{\mu_2-\tan\theta}{\tan\theta-\mu_1}.$$You have $$\frac{l_1}{l_2}=\frac{\mu_2\tan\theta}{\tan\theta-\mu_1}.$$Did you omit the sign between ##\mu_2## and ##\tan\theta##?

Also, please familiarize how to use LaTeX here. You need two $ instead of one. Also you can use ## for inline equations.
 
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  • #7
Killua Rafiq
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I messed up an overall negative sign and the correct solution should be $$\frac{l_1}{l_2}=\frac{\mu_2-\tan\theta}{\tan\theta-\mu_1}.$$You have $$\frac{l_1}{l_2}=\frac{\mu_2\tan\theta}{\tan\theta-\mu_1}.$$Did you omit the sign between ##\mu_2## and ##\tan\theta##?

Also, please familiarize how to use LaTeX here. You need two $ instead of one. Also you can use ## for inline equations.
I wrote same,bruh
 
  • #8
Killua Rafiq
18
1
I wrote same,bruh
The answer should be $$\frac{\mu_2 tan\theta}{tan\theta -\mu_1}$$
 
  • #9
Killua Rafiq
18
1
You just try to solve again.Definetly you did a mistake and i explained problem in pdf.I think you should read solution again.Good luck
 
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  • #10
@kuruman is correct. The work-energy theorem spits out$$mg(l_1 + l_2)\sin{\theta} - \mu_1 l_1 mg \cos{\theta} - \mu_2 l_2 mg \cos{\theta} = \Delta T = 0$$Divide by ##mg l_1 \cos{\theta}##, and you'll end up with$$\left(1+ \frac{l_2}{l_1} \right) \tan{\theta} = \mu_1 + \mu_2 \frac{l_2}{l_1}$$Then it's a small matter of rearranging for ##l_2/l_1##, $$\frac{l_2}{l_1} = \frac{\mu_1 - \tan{\theta}}{\tan{\theta} - \mu_2}$$
 
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  • #11
Killua Rafiq
18
1
Check solution of 7th problem.Totally same that i write.
 

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  • #12
Killua Rafiq
18
1
I didn't even understand your first equation if it's equation.
 
  • #13
Killua Rafiq
18
1
I wrote same,bruh
@etotheipin @kruman i am so sorry.I forgot just add "-".
 
  • #14
kuruman
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@etotheipin @kruman i am so sorry.I forgot just add "-".
It's OK. We all make mistakes.
 
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  • #15
Killua Rafiq
18
1
Your solution to Problem 1 is incorrect. It should be $$\frac{l_1}{l_2}=\frac{\tan\theta-\mu_2}{\tan\theta-\mu_1}.$$
Homework Statement:: Hi guys,i want to share with you my project about advanced physics-Physics olimpiads.I find 5 hard problems and solved these.I think it will be useful for students who prepares to olimpiad's and interests in adv physics.Have fun.
Relevant Equations:: $P=\rho gl$,$GM=gR^2$(and etc.)

Pdf is below.
Edit:The answer must be in first problem :$$\frac{\mu_2-tan\theta}{tan\theta-\mu_1}$$
 
  • #16
kuruman
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While I have your attention, check your solution to Problem 5. The process A-B is adiabatic. Near the end of your solution you say that it is a isotherm. That is simply not true. I suggest that you correct this and post a corrected version of the pdf.
 
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  • #17
Killua Rafiq
18
1
It's OK. We all make mistakes.
While I have your attention, check your solution to Problem 5. The process A-B is adiabatic. Near the end of your solution you say that it is a isotherm. That is simply not true. I suggest that you correct this and post a corrected version of the pdf.
But in ishoterm "T" is const.I wanted to specify that ishotermal and aidabatic similar except adiabatic prosses gives eneregy only with work.And i am agree with you.I will fix problems
 
  • #18
kuruman
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The isothermal and the adiabatic processes are quite different and I would not call them similar. In an isothermal process the temperature does not change and the work done by the gas is equal to the heat that enters the gas. In an adiabatic process, no heat enters the gas and the temperature drops or increases depending on whether the gas is doing positive or negative work. The Carnot cycle is an example of the use of two isotherms and two adiabats to get the highest possible efficiency.
 
  • #19
Killua Rafiq
18
1
$$E=mc^2$$
 

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  • #20
Killua Rafiq
18
1
The isothermal and the adiabatic processes are quite different and I would not call them similar. In an isothermal process the temperature does not change and the work done by the gas is equal to the heat that enters the gas. In an adiabatic process, no heat enters the gas and the temperature drops or increases depending on whether the gas is doing positive or negative work. The Carnot cycle is an example of the use of two isotherms and two adiabats to get the highest possible efficiency.
Agreed Totally.Check it https://www.physicsforums.com/threads/turkish-tubitak-olympiads-problems-with-solutions-by-rafiq-abbasov-fi.992496/
 
  • #21
Killua Rafiq
18
1
$$E=mc^2$$
 

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  • #22
kuruman
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Homework Statement:: Again me.All bugs fixed
Relevant Equations:: &&P=\rho gl$$
$$E=mc^2$$
All bugs are not fixed. In the solution to Problem 5 (this version), you claim that in an adiabatic process the temperature does not change. That is not true. It could be true for a solid that does not change its volume when heat is added to it. A gas can expand (or contract) and do positive (or negative) work on the environment. As you know, according to the first law ##\Delta U=Q-W##, where ##W## is the work done by the gas, ##Q## is the heat that enters the gas and ##\Delta U## is the change in internal energy of the gas. In an adiabatic process, ##Q=0## so that ##\Delta U=-W##. This says that if the gas expands adiabatically and does positive work as is the case in Problem 5, the internal energy and hence the temperature must decrease.

Basic stuff like this must be clear in your mind before you attempt Olympiad-level problems.
 
  • #23
Killua Rafiq
18
1
All bugs are not fixed. In the solution to Problem 5 (this version), you claim that in an adiabatic process the temperature does not change. That is not true. It could be true for a solid that does not change its volume when heat is added to it. A gas can expand (or contract) and do positive (or negative) work on the environment. As you know, according to the first law ##\Delta U=Q-W##, where ##W## is the work done by the gas, ##Q## is the heat that enters the gas and ##\Delta U## is the change in internal energy of the gas. In an adiabatic process, ##Q=0## so that ##\Delta U=-W##. This says that if the gas expands adiabatically and does positive work as is the case in Problem 5, the internal energy and hence the temperature must decrease.

Basic stuff like this must be clear in your mind before you attempt Olympiad-level problems.
There is no reason to be rude like this.Dude,i remember i wrote T=constant in previous document?. There was ishoterm because of i confused and i was exhausted.So i forgot change this.I know kinda most part adv physics but it was my first article.I will fix this.
 
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  • #24
Killua Rafiq
18
1
There is no reason to be rude like this.Dude,remember i wrote T=constant in previous document?. There was ishoterm because of i confused and i was exhausted.So i forgot change this.I know kinda most part adv physics but it was my first article.I will fix this.
 
  • #25
Killua Rafiq
18
1
Summary:: I fixed all bugs.So you can enjoy now.I will try share with you new one.Have fun

$$E_k=mc^2(\gamma-1)$$
 

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  • #26
kuruman
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Your solution to Problem 5 is still incorrect, specifically your calculation of the work during the adiabatic expansion. There is no enjoyment in saying this repeatedly and, considering your reaction to constructive criticism, I will discontinue posting on this thread. Good luck with your solving challenging problems. Adios.
 
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  • #27
fresh_42
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The offered solutions unfortunately had to be corrected a bit too often. We usually discuss questions or certain papers, and do not deliver answers to unasked problems, even less wrong ones.

It also has derailed into its meta level, which is why I deleted this off topic part of the discussion.

This thread is closed.
 
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