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Density of a platinum-iridium cylinder

 
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Sep2-08, 07:49 PM   #1
 

Density of a platinum-iridium cylinder


1. The problem statement, all variables and given/known data

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?

2. Relevant equations

now I know that to find volume you have to use v=h(pi)r^2
and I also know you have to convert it to meters. When I do I get this
(0.039)(pi)(0.0195)^2
which gives me 4.6589033654573236278091385713189e-5
so this is what I get over all
1 / (39.0 * pi * (19.5^2)) = 2.14642786 × 10-5
yet the answer is 2.15*10^4k/m^3
can someone tell me what im doing wrong, and if im suppose to convert the 2.14

3. The attempt at a solution
 
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Sep2-08, 08:07 PM   #2
 
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Well, 2.1464 is rounded up to 2.15 if one is using three significant digits in the final answer.

One's solution is correct, the density is about 21464 or approximately 21500 kg/m3
 
Aug28-09, 02:15 PM   #3
 
That answer is correct but be careful with units. 21464 is in millimeters. so it would be 21.464 kg/m^3
 
Aug28-09, 08:19 PM   #4
 
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Density of a platinum-iridium cylinder


1 / (39.0 * pi * (19.5^2)) = 2.14642786 × 10-5

This step is wrong. It should be
1 / (0.0390 * pi * (0.195^2)) = 2.14642786 × 10^4
 
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