Please check solutions to Error Uncertainty calculations

In summary, the conversation discusses an experiment to measure the density of a steel ball using the given diameter and mass. The volume and density of the ball are calculated with uncertainties using the formulas provided. The final values for volume and density are 2.85 x 10^-6 +- 2.77 x 10^-13 m^3 and 7719 +- 479 kg/m^3, respectively. The question of whether the calculated value is consistent with the tabulated value of 7860 kg/m^3 is answered using the percentage difference criterion or by checking if the accepted value falls within the range of the calculated value.
  • #1
madeeeeee
87
0
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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  • #2
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.
 

Related to Please check solutions to Error Uncertainty calculations

1. What are error and uncertainty calculations?

Error and uncertainty calculations are methods used to determine the margin of error or uncertainty in a measurement or experimental data. They involve analyzing the precision and accuracy of the measurement or data and taking into account any potential sources of error.

2. Why is it important to check solutions to error and uncertainty calculations?

It is important to check solutions to error and uncertainty calculations to ensure the accuracy and reliability of the data or measurement. By identifying and accounting for sources of error, the results can be more precise and trustworthy.

3. What are some common sources of error in scientific experiments?

Some common sources of error in scientific experiments include human error, equipment limitations, environmental conditions, and systematic errors in measurement or data collection.

4. How do you calculate error and uncertainty?

Error and uncertainty can be calculated using various statistical methods, such as standard deviation, confidence intervals, and propagation of error. These methods take into account the precision and accuracy of the measurement or data, as well as any sources of error.

5. Can error and uncertainty be completely eliminated?

While error and uncertainty can never be completely eliminated, they can be minimized through careful experimental design, precise measurement techniques, and thorough analysis of results. It is important to acknowledge and account for any remaining error or uncertainty in the data.

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