Determining the Density of a sphere.

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Homework Help Overview

The problem involves determining the density of a solid sphere given its radius and mass, along with their respective uncertainties. The calculations include finding the volume of the sphere and subsequently the density, while also addressing the propagation of uncertainty in these measurements.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the density and its uncertainty using the measured values for radius and mass, but expresses confusion regarding the correctness of their final answer. Some participants question the method used for calculating the uncertainty in density, suggesting a different approach involving relative uncertainties.

Discussion Status

Participants are actively discussing the calculations related to uncertainty, with some providing guidance on how to properly propagate errors when dividing quantities. There is an ongoing exploration of the relationship between the exponent in the volume formula and the multiplication of uncertainty in radius.

Contextual Notes

There is a noted concern regarding the response time to posts, indicating a desire for quicker feedback. The original poster's calculations are based on specific significant figures and error propagation rules, which are under scrutiny.

teamsoulglo
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Homework Statement



The radius of a sphere is measured to be (6.90 0.25 ) cm, and its mass is measured to be (1.75 0.05) kg. The sphere is solid. Determine its density in kilograms per cubic meter and the uncertainty in the density. (Use the correct number of significant figures.)

Homework Equations



Volume = (4/3) (pi)(radius)^3

= (4/3) (3.1416)(6.9/100)^3 = 1.376 X 10^-3 m^3

Density = mass / vol = 1.75 / 1.376 X 10^-3 = 1271.8 kg / m^3

= 1.27 X 10^3 kg /m^3


The Attempt at a Solution



% error in radius = (0.25/6.9)X100 = 3.6%
therefore % error in volume = 3 X 3.6% = 10.8%
% error in mass = (0.05/1.75) X 100 = 2.86%

So % error in the density = 10.8 + 2.86 = 13.66%

Therefore undertainty in density = 13.66% of 127 X 10^3 kg/m^3

= 173 kg/m^3

So density = (1.27±0.173) X 10^3 kg/m^3

According to webassign, my answer is incorrect. I can't figure out what I did wrong. Any help would be appreciated.
 
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My first time using this board, and no response for over 4 hours. Yet others who post after me are getting help asap. Thanks for the help guys!
 
Welcome to the forums teamsoulglo,

Patients is a virtue, and besides, I'd be impressed if on average questions were answered quicker on any other boards. Now, to your question, you've done everything right up until when you calculate the uncertainty in the density. Suppose we have two quantities with their appropriate errors;

(A\pm\delta A)\hspace{1cm}(B\pm\delta B)

Now, suppose we want to find X = A/B, then the uncertainty in this case is;

\frac{\delta X}{X} = \sqrt{\left(\frac{\delta A}{A}\right)^2 + \left(\frac{\delta B}{B}\right)^2}

Also note that when you divide a value by a constant, you must also divide the error by that same constant. Hope this helps.
 
I was a little unclear why the radius uncertainty is multiplied my three. Is there a relation between the exponent and the amt its multiplied by?? Truly wondering.
 
denverdoc said:
I was a little unclear why the radius uncertainty is multiplied my three. Is there a relation between the exponent and the amt its multiplied by?? Truly wondering.
Indeed there is (as far as I know). If we again have some quantity (A\pm\delta A) and we wish to find Z = A^n then the error in this case is;

\frac{\delta Z}{Z} = n\cdot\frac{\delta A}{A}
 
Thanks hoot. Learn something every day on this forum.
 
denverdoc said:
Thanks hoot.
No problem :smile:
denverdoc said:
Learn something every day on this forum.
Oh how true that is, more often that not its me doing the learning!
 

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