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By plotting diameter and circumference data sets, determine an experimental value for pi. The diameters of...

By plotting diameter and circumference data sets, determine an experimental value for pi.

The diameters of various cylinders were measured using a Vernier caliper. The corresponding circumference of each cylinder was measured using the method described in the next paragraph.

A thin strip of paper was carefully wrapped around the cylinder, and the strip of paper was marked with a pencil at the point where the strip overlapped. The thin strip was unwound, laid flat, and the distance from the edge of the strip to the pencil mark was carefully measured to determine the circumference.

Both the diameters and circumferences were measured in millimeters.

Using the (diameter, circumference) data pairs, create a scatter plot of the data pairs using Excel and determine an experimental value for pi to three (3) decimal places from the fit equation. If the fit equation is not displaying at least three decimal places, the decimal places of the fit equation may need to be adjusted accordingly. The data pair (0,0) should be used in addition to the data pairs given below. In addition to appropriate axis labels and titles, the fit and fit equation should be displayed on the plot.

In addition to the scatter plot, include a data table with the diameter/circumference pairs that were used to create the scatter plot.

Finally, type the experimental value of pi to three decimal places in a cell. Highlight the experimental value of pi with a border and yellow background.

If the plot, table, and highlighted experimental value for pi are all constricted to fit within Rows 1-44 and Columns A-I, everything should fit on one page.

1st data Pair: ( 32.5, 102.3), 2nd data pair: (41.9, 131.9), 3rd data pair: ( 62.5, 195.6), 4th data pair: (81.9, 258.8), 5th data pair: (101.3, 316.1)

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Answer #1
Diameter Circumference
32.5 mm 102.3 mm
41.9 mm 131.9 mm
62.5 mm 195.6 mm
81.9 mm 258.8 mm
101.3 mm 316.1 mm

We plot circumference on y - axis and diameter on x-axis.

The slope will be the value of

therefore,

= 3.13

as there must be four total significant figures ( three decimal places)

= 3.130 ( this is very close to known value of with error of only 0.32 % )

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