Please check solutions to Error Uncertainty calculations

AI Thread Summary
The discussion focuses on verifying the accuracy of error calculations related to measuring the density of a steel ball. The volume calculation is based on the formula for a sphere, yielding a volume of approximately 2.85 x 10^-6 m^3 with an uncertainty of 2.77 x 10^-13 m^3. The calculated density is 7719 kg/m^3, with an uncertainty of 479 kg/m^3, which is compared to the standard value of 7860 kg/m^3. Participants suggest using the percentage difference criterion and checking if the accepted value falls within the calculated range to assess consistency. The conversation emphasizes the importance of proper error propagation methods in scientific calculations.
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Please check if my solutions are correct or if I am totally wrong. Thank you!

1. Homework Statement [/b]
In an experiment to measure the density of a steel ball. The diameter of the ball is d= 1.76 +- 0.02 cm and the mass to be m= 22 +- 1g. (include error calculations and report values with uncertainties.)

a) The volume of a sphere is given by V = 4/3 x pie x r^3. What is the volume of the ball in m^3?

Attempt Solution: d= 1.76 +- 0.02 cm --> 0.0176 +- 0.0002 m
r= 0.00088 +- 0.0001 m

V = 4/3 x pie x r^3
= 4/3 x pie x (0.00088)^3
= 2.85 x 10^-6

Uncertainty:
V = V(3xr)
= (2.85 x 10^-6 )((3)(0.00010/0.0176))
= 4.86 x 10^-8

Final: V= 2.85 x 10^-6 +- 2.77 x 10^-13 m^3


b) Density is given by p= m/v. What is the density of the ball in kg/m^3/

0.022 +- 0.001 kg

p=(0.022/2.85x10^-6)
= 7719.3 kg/m^3

Uncertainty: P = P(m/m + V/V)
= 7719.3 ((0.001/0.022) + (4.86x10^-8)/(2.85x10^-6))
= 478.98

Final: P= 7719 +- 479 kg/m^3

c) Is your value consistent with a tabulated value of 7860 kg/m^3

How do I answer this scientifically?
 
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I don't know if the error propagation method you use is satifactory enough but I strongly suggest to use the least squares method.

Anyway, to answer (c) you can use the percentage difference criterion. You can also answer it by checking if the accepted value falls within the range of the calculated value.
 
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