The Caliper and Micrometer

The Vernier Caliper and Micrometer
The activity involved the use of both vernier and micrometer calipers accurately on measuring provided materials. The objectives of the act were to familiarize the students with the use of the said scales and to demonstrate their differences. The dimensions of a metallic cube made of steel and a metallic washer were measured using a vernier caliper while that of a marble made of glass and the same metal cube were measured using the micrometer caliper. The density and percentage error of each material were then computed using the values obtained. Guide Questions:

1.) Differentiate the vernier and micrometer scales?
The Vernier caliper is an extremely precise measuring instrument; its accuracy is 0.05mm. It can show measurements up to two decimal places in millimeters. It has main scale, which shows whole numbers and the vernier scale which gives decimal values. The vernier is capable of measuring the outer and inner dimensions including the depth. A micrometer caliper uses a calibrated screw for measurement, rather than a slide which the vernier caliper uses. . It can show measurements up to three decimal places in millimeters. It also has a main scale like the vernier caliper with the same purpose with the micrometer scale showing decimal values. Its accuracy is equal to 0.01 mm making it more accurate than the vernier caliper. The micrometer can only measure the outer dimensions of an object.
2.) Draw the figure for micrometer readings below:
i. 3.685 mm
ii. 1.596
3.) State some of the errors the one might make in measuring length using both vernier and micrometer calipers.
The errors that one may make in measuring length using both the vernier and micrometer calipers are the incorrect reading of measurements. There can also be human misinterpretation, meaning that the person may have set the instrument too tight which may deform the object or too lose which leaves extra space for error. The device can also be broken promoting errors.
4.) Determine the percentage error for an observed value of 1.112×10-5 if the standard value is 1.117×10-5?
%error = actual value-observed value x100
actual value
%error= 1.117×10-5-1.112×10-5 x100
%error= 5.000×10-8 x100
%error= 4.476×10-3 x 100
%error= .4476%

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