physics labreport

7/28/2019 Physics Labreport

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where (m) is the mass and (v) is the

volume;

Finally the percent error is calculated withthe formula:

3. MethodologyFor this experiment, the group used avernier caliper, a micrometer caliper, a footrule, a sphere, and an electronic grambalance.

In order to measure the diameter of thesphere using the foot rule, different pointsalong its circumference were used.

In the case of the vernier caliper, thesphere was inserted between its jaws and

the screw clamp was tightened to hold the jaws in place. The diameter was readthrough the use of the main scale and thevernier scale.

In the case of the micrometer caliper, thesphere was positioned by the anvil whilesimultaneously rotating the thimble and theratchet knob until a click is heard;effectively locking the sphere into placebetween the anvil and the spindle. The

diameter was read through the use of thethimble and the sleeve. The sleevedisplayed the millimeters by ones while thesleeve divides the millimeters starting withthe ones in the decimals place.

4. Results and Discussion

Table 1 shows the measured diameters of the sphere using the respective instruments.Ten trials were made in order to get the

precise measurements of the sphere.

Table 1. Measured diameters of thesphere with its mean diameter

Diameter of sphere (cm)

Trial Foot

rule

Vernier

Caliper

Micrometer

Caliper

1 1.72 1.655 1.66

2 1.72 1.660 1.66

3 1.72 1.660 1.66

4 1.72 1.660 1.66

5 1.72 1.655 1.66

6 1.72 1.670 1.65

7 1.72 1.655 1.65

8 1.72 1.660 1.65

9 1.72 1.660 1.66

10 1.72 1.655 1.66

MeanDiameter

1.72 1.662 1.657

Table 2. Computed Average Deviations (a.d.),Average deviations of the mean (A.D.),volume, experimental value of density (EVD),Accepted value of density (AVD), % error for Density

Footrule

Vernier caliper

Micrometer caliper

a.d. 0 3.4 x 10- 1.328 x 10-

A.D. 0 1.07 x 10-1.32 x 10-3

Volume(cm3)

2.66 2.40 2.38

Mass (g) 18.80 18.80 18.80

EVD(g/cm3)

7.05 7.82 7.89

AVD(g/cm3)

7.87 7.87 7.87

% error (%)

10.45 0.63 0.25

The instruments that were used producedprecise but not accurate measurements.Precision is how close the measurementsare with each other while accuracy entailshow close the measurements are to thetrue value.

According to the results, the instrumentthat gave the highest percentage of error



was the foot rule by having 10.45% of error.Next to the foot rule is the vernier caliper which had a substantial difference with thepercentage error of the foot rule. Thevernier caliper only had a 0.63% of error.

However, compared to the vernier caliper,the micrometer caliper only had a 0.25% of error. Therefore, it has the most accuratemeasurement because its measurement isthe one that’s nearest to the acceptedvalue.

The probable sources of error may be theuncalibrated instruments used or themisreading of an instrument. Human error is also a factor that could lead to the

misinterpretation of data.

Table 3. Width of thumb (in)

Groupmember

1 2 3 4

Width of thumb

(in)

0.60 0.65 0.65 0.62

Table 3 shows the widths of the thumbs of

the members in the group. Member nos. 2and 3 had the widest thumbs in the groupfollowed by member no.4, and then lastlymember no. 1.

5. Conclusion

Although instruments are accurate andprecise, the measurements are stilldependent on the errors encountered bythe experimenters, may they be systematic

or random.

After 10 trials of measuring the sphereusing the foot rule, the vernier caliper, andthe micrometer caliper, the experimentersgot the values of 1.72, 1.662, and 1.657 astheir means respectively. Their averagedeviations were 0, 0.003, and 0.001respectively.

At the end of the experiment, the

experimenters solved for the density of thesphere using the respective instruments.

The instrument that gave the leastpercentage of error was the micrometer caliper having only 0.25% error. It wasfollowed by the vernier caliper, having only0.63% error, and then by the foot rule

having a rather drastic 10.45% of error.

From the comparison of the results of theexperimental measurements, it could besaid that the micrometer caliper gives theleast percentage of error and is thereforethe most accurate instrument.

6. Applications1. Which among the three measuring

devices give you the least % error?

Is the accuracy of a measurementaffected by the least count of themeasuring device?

- The device that gave us the least %error was the micrometer caliper. It onlygave 0.25% error while the foot rulegave us 10.45% error. The least countis the smallest reading that can bemeasured by the instrument. Therefore,the least count is indirectly proportionalto the accuracy of the instrument.

2. What do you mean by error? Whatare the types of error? What are theerrors you encountered in thisexperiment?

- Error refers to the disagreementbetween a measurement and the trueaccepted value. The 2 types of error are the random and systematic errors.

The random errors are unpredictableerrors brought about by things out of the experimenter’s control. Systematicerrors are errors usually brought aboutby the measuring equipment. Since aperson does not always get the samemeasurements because of human error,our group encountered a random error.

3. Sketch a) a vernier caliper that reads3.48 cm b) a micrometer caliper that

reads 7.05 mm



a) Vernier caliper

b) Micrometer caliper

4. A student weighs himself using a

bathroom scale calibrated inkilograms. He reported his weight inpounds. What is the percentageerror in his reported weight if heuses this conversion: 1kg = 2.2pounds?

- Suppose the student weighed 75 kg.With his reported weight, using theconversion 1 kg = 2.2 pounds, he would

weigh 165 pounds. On the other hand,if we were to use the standard kilogram

which is equal to 2.2046 pounds, hisweight would be 165.345. To calculatethe percentage error, the formula is

Substituting:

%error =

% error = 0.20865

Therefore, there is a 0.2087% error.

5. In an experiment on determination of mass of a sample, your groupconsisting of 5 students obtainedthe following results: 14.34 g, 14.32 g,14.33 g, 14.30 g and 14.23 g. Find themean, a.d. and A.D. Suppose thatyour group is required to make onlyfour determinations for the mass of the sample. If you are the leader of the group, which data will you omit?Recalculate the mean, a.d., and A.D.

without this data. Which results willyou prefer?

- Given the first set of data:Mean = (14.34 + 14.32 + 14.33 + 14.30+ 14.23) / 5= 14.304

Average deviation:14.34 – 0.036

14.32 – 0.01614.33 – 0.02614.30 – 0.00414.23 – 0.074= (0.036 + 0.016 + 0.026 + 0.004 +0.074) = 0.156/5a.d. = 0.0312

A.D. = 0.0312/ √ A.D. = 0. 0139

Ommitting 14.23:



Mean = (14.34 + 14.32 + 14.33 +14.30) / 4= 14.322

Average deviation:

14.34 – 0.01814.32 – 0.00214.33 – 0.00814.30 – 0.022= (0.018 + 0.002 + 0.008 + 0.022) =0.05/4a.d. = 0.0125

A.D. = 0.0125/ √ A.D. = 0.0062

We prefer the 2nd results because itproduces a lower A.D. which means thedata are more precise and accurate.

6. Give other parts of the body that arebeing used for measurement eventoday. What are some properties tobe a standard of measurement?

- The fingers are used to measure or count to ten. The hand was previouslyused to measure three inches, but it isnow currently used to measure 4 inches. A measurement should have clear definitions to be useful. They must alsobe the same or consistent no matter thesituation or condition.

7. Reference:

[1] Dealing with error and uncertainty inmeasured values. Retrieved on June 24,2013 fromhttp://www.chem1.com/acad/webtext/pre/mm2.html

[2] Errors and uncertainties by JohnKennedy. Retrieved on June 24, 2013 fromhttp://www.slideshare.net/thephysicsteacher/12b-errors-and-uncertainties

[3] Edmonds, Dean S.Jr, Cioffari’sExperiments in College Physics,

Massachussets: D.C. Heath and Company,1988.

http://www.chem1.com/acad/webtext/pre/mm2.html



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