Determining the Density of Turpentine: Archimedes' Principle Explained

  • Thread starter Thread starter frozen7
  • Start date Start date
  • Tags Tags
    Archimedes
AI Thread Summary
The discussion centers on calculating the density of turpentine using Archimedes' Principle with a solid aluminum cylinder. The first method calculates the density as 887 kg/m³, while the lecturer's method yields a density of 2699.4 kg/m³, suggesting a nearly equal density to aluminum. It is noted that if the densities were indeed similar, the aluminum would displace its weight, resulting in a near-zero reading when immersed. The consensus is that the initial calculation is correct, as the professor's approach appears flawed. The correct density of turpentine is confirmed to be 887 kg/m³.
frozen7
Messages
163
Reaction score
0
A solid aluminium cylinder with density of 2700 has a measured mass of 67g in air and 45g when immersed in turpentine. Determine the density of turpentine.

I do it in this way:
W = ( 0.067 - 0.045 ) (9.8)
= 0.2156

Volume of aluminium = 0.067 / 2700 = 2.48 x 10-5

0.2156 = (density of turpentine ) (2.48 x 10-5) (9.8)
density of turpentine = 887

However, my lecturer do it in this way:

W = (density of aluminium) (g)(volume of solid)
volume of solid = xxx

W = (density of turpentine) ( g ) ( volume of solid)
density of turpentine = 2699.4

Which solution is correct actually?
 
Physics news on Phys.org
If you think about it ...

By Archimedes' Principle, the upthrust on an object immersed in a fluid is equal to the weight of the fluid displaced.
According to your professor, turps and aluminium have approximately the same density (2699.4 vs 2700)
This would mean that the aluminium would displace approximately its own weight and so would register a weight of almost zero instead of 45 g.

You working out looks fine.

I wonder if your professor equated the weight of the aluminium cylinder with the weight of the turps displaced ?
 
Your solution is correct.

ehild
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top