Density of a platinum-iridium cylinder

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The discussion focuses on calculating the density of a platinum-iridium cylinder, specifically the standard kilogram. The volume is calculated using the formula V = h(pi)r^2, with dimensions converted to meters. The initial calculations led to a misunderstanding of unit conversions, resulting in an incorrect density value. The correct density, after proper unit adjustments, is approximately 21,464 kg/m³ or 21.464 kg/m³ when rounded to three significant figures. Accurate unit conversion is essential for obtaining the correct density measurement.
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Homework Statement



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?

Homework 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 I am doing wrong, and if I am suppose to convert the 2.14

The Attempt at a Solution

 
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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
 
That answer is correct but be careful with units. 21464 is in millimeters. so it would be 21.464 kg/m^3
 
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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