In this experiment, we will try to find the density of an object by using its dimensions with the usage of vernier caliper, micrometer and weighing scale.

INTRODUCTION

        In this experiment, we will use some of the measurement technics by using vernier scale and micrometer as distance and dimensions, weighing scale to find the weight of the object. The objective of the experiment is to find the density of an object by using some formulas which are includes the total area and the weight of the object.

INSTRUMENTATION and PROCEDURE

        

        

                                        THE SPECIMEN

        Vernier scale : A vernier scale is an additional scale which allows a distance or angle measurement to be read more precisely than directly reading a uniformly-divided straight or circular measurement scale. It is a sliding secondary scale that is used to indicate where the measurement lies when it is in between two of the marks on the main scale.

Verniers are common on sextants used in navigation, scientific instruments used to conduct experiments, machinists' measuring tools  used to work materials to fine tolera and on theodolites used in surveying.

        When a measurement is taken by mechanical means using one of the above mentioned instruments, the measure is read off a finely marked data scale. The measure taken will usually be between two of the smallest graduations on this scale. The indicating scale  is used to provide an even finer additional level of precision without resorting to estimation.

        

        Micrometer : A micrometer, sometimes known as a micrometer screw gauge, is a device used widely in mechanical engineering and machining for precision measurement, along with other metrological instruments such as dial calipers and vernier calipers. Micrometers are often, but not always, in the form of calipers.The additional digit of  micrometers is obtained by finding the line on the sleeve vernier scale which exactly coincides with one on the thimble. The number of this coinciding vernier line represents the additional digit.

        Digital Scale : An electronic measuring device that uses fiber optics to detect and transmit its position to a digital or computer readout for display.

        Procedure : We used vernier scaler to measure a, c, d and micrometer for L and b. Then we used digital scale to find mass of the specimen. After we got final results we used error analysis for vernier as 0.02 , micrometer as 0.01 and digital scale as 0.5 g.

RESULTS and DISCUSION

Calculations :

        

        v = a x (L x b – 4 x c x d) = 7.57 cm3

        

v = [(dv/da x wa)^2+(dv/db x wb)^2 + (dv/dL x wL)^2 + (-4dv/dc x wc)^2+(-4dv/dd)^2]^1/2

      [4.45x10^-3 + 0.12 + 1.24 x 10^-4 + 0.08 + 5.38 x 10^-4]

       = 0.45

        v = 7.57 ± 0.45 cm3

        m = 20 ± 0.5 g

        density = m / v = 20 / 7.57 = 2.64  g/cm3

        density = [ (1/ v x wm)^2 + (- m/v^2 x wv)^2]^1/2

        density = (4.36 x 10^-3 + 1.41)= 1.19

        density = 2.64  ± 1.19 g/cm3

ERROR ANALYSIS

        The reason of the errors might be the measurement mistakes. The calculations shows that when we find the density of the aluminium part is 2.64 (the correct density is 2.70) we did a little mistake when we find it. However when we find the error discusion the answer is plus or minus 1.19 which is so huge number for an error. This error might be beacuse of the measurement mistakes which we did it by two different machine; vernier scale and micrometer..both have different error scale between each other so this might affect the answer of the error analysis.

CONCLUSION

        

        In this experiment,we find the density of an object by using the measurements of the dimensions and the weight of the object with the usage of vernier caliper, micrometer and weighing scale. While we find the dimensions of the parts we have some error analysis because of the usage of different materials which are have different error scales. When try to find the density of the try to find the weight of the object to use it in the Formulas. When we use the formulas and find the density we got some error calculation. How ever when we discuss the reasons of the errors we find some facts that are cause these errors. For the error analysis of the density we find the mass 4.36x10^-3 and the volume is 1.14. That shows us the error of the v is more than the mass. Which we can understand that to find a better answer for the error analysis we have to change the materials that we use to find the v. For that reason to find better answers we need more spectacular materials to measure the dimensions.