Calculate Density of Standard Kilogram Cylinder: Quick Guide

[maej26]

This is simple, but I'm missing the last step. I've almost got the answer.

--The standard kilogram is a platinum-iridium cylinder 39.0 mm in height and 39.0 mm in diameter. What is the density of the material?

I know that the correct answer is 2.15 x 10^4 kg/m^3.

If someone could tell me how to get from my answer, 2.15 x 10^-5, TO 2.15 x 10^4 I would really appreciate it. Thanks.

 

[dav2008]

Check your units.

What are the units of the answer that you obtained? What are the units of the given answer?

That's the first thing you should always check when you obtain an answer that isn't consistant with a given answer.

 

[maej26]

I know that's the problem but I just don't know how to do it and would really like to understand it seeing as how it's so basic.

 

[dav2008]

Well the units of the given answer are kg/m³.

What are the units of the answer that you obtained?

 

[maej26]

See, that's what I'm bad at.

I think it's kg/m^3 because (19.5 m)^2 makes it m^2 and that's multiplied by (39.0 m) therefore making it m^3. I'm not too sure though.

 

[dav2008]

Right.

So if the desired units are kg/m³ and you used mm instead of m to solve for the density, what units could you use to obtain kg/m³ in your final answer?

Basically I'm saying that if you understand why your answer is in kg/mm³ then you should see what you could do to obtain an answer in kg/m³

 

[maej26]

Okay, I just confused myself.

I guess what the more important question is if I didn't already know the correct answer how would I know that the units are supposed to be kg/m^3?

 

[dav2008]

You don't, however typically you want your answers to be in standard mks units (everything in terms of meters, kilograms, seconds).

It's like asking if an answer of 1 hour is more correct than an answer of 60 minutes. They're equivalent.

 

[maej26]

lol

(1000^3) (2.15 x10^-5)

 

[maej26]

Okay, I get it now! Thanks! So, I basically had the correct answer all along. lol Do you think it would be marked correctly on an exam?

 

[dav2008]

That's one way to do it. Since there are 1000³ mm³ in 1 m³ you could multiply your answer that's in terms of kg/mm³ by 1000³ to obtain an answer in terms of kg/m³

$$\frac{2.15\cdot10^{-5} \ kg}{1 \ mm^3} \cdot \frac{1000^3 \ mm}{1 \ m^3}=\frac{2.15 \cdot 10^4 \ kg}{m^3}$$

Another way to have done it is to just use meters instead of millimeters in your initial calculations.

$$\frac{1 \ kg}{(\pi){(.0195 \ m)}^2(.039 \ m)}=2.15 \cdot 10^4 \frac{kg}{m^3}$$

 

[maej26]

I think I like the latter better. I think I'll try that. Thanks! :)

 

[dav2008]

Okay, I get it now! Thanks! So, I basically had the correct answer all along. lol Do you think it would be marked correctly on an exam?

It depends. Like I said usually answers are wanted in mks units unless otherwise specified. It can't hurt to ask your teacher/prof if they want certain units used.

 

[maej26]

Okay, I'll keep that in mind.