The Caliper and Micrometer

The vernier scale Caliper and Micrometer Experiment 1AbstractThe activity involved the use of both vernier and micrometer gauge 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 gold-bearing cube made of steel and a metallic washer were mensural 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 faulting of each material were then computed using the values obtained. Guide Questions1.) Differentiate the vernier and micrometer scales?The Vernier caliper is an extremely precise measuring cats-paw its accuracy is 0.05mm. It can show measurements up to two decimal places in millimeters. It has main scale, which shows whole amount 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, kind of 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 exchangeable 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 belowi. 3.685 mmii. 1.5963.) 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 small promoting errors.4.) Determine the percentage error for an observed value of 1.11210-5 if the standard value is 1.11710-5?%error = actual value-observed value x100 actual value%error= 1.11710-5-1.11210-5 x100 1.11710-5 %error= 5.00010-8 x100 1.11710-5 %error= 4.47610-3 x 100 %error= .4476%