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Physics lab on propagation of errors. In this experiment I investigated the propagation of errors while calculating the volume of two objects.

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                                                     , By Satish Ahuja

In this experiment I investigated the propagation of errors while calculating the volume of two objects. I came to the conclusion that a measuring instrument like a screw gauge has some problematic limitations. This, along with other human factors, introduces uncertainties and errors in the measurement. Careful procedure can minimize these errors but cannot remove them completely. The errors in individual measurements contribute to the result of calculations using the measured quantities. Various precautions need to be taken to minimize errors in measurements and study of how these errors propagate during the various calculations needs to be taken.

The aim of this experiment was to investigate the propagation of errors while calculating the volume of a Cylindrical and a Spherical object.

My experiment included calculating the volume of two objects by two different methods.

 Apparatus: Vernier calipers, given spherical object, given cylindrical object, a measuring cylinder, and string.

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2 = 11 cm3

200 cm3

211 cm3

3. V3 = 12 cm3

200 cm3

212 cm3

4. V4 = 10 cm3

200 cm3

210 cm3

Therefore, the final volume is (12+11+12+10)/ 4

 = 11.25 cm3.

The error in this case is + 2 cm3.

So volume for the first case is 11.25cm3 + 2 cm3





1. D1 = 2.46 cm

2.4 cm

0.06 cm

2. D2 = 2.46 cm

2.4 cm

0.06 cm

3. D3 = 2.48 cm

2.4 cm

0.08 cm

4. D4 = 2.48 cm

2.4 cm

0.08 cm

5. D5 = 2.46 cm

2.4 cm

0.06 cm

TRUE VALUE = 2.46 cm

Now with the diameter we can calculate the volume of the sphere using the formula for volume of a sphere:

V=4/3 π r3

V = 4/3 x πx (2.46/2)3

Volume = 12.9cm3

The error is equal to thrice the percentage error of the radius as it is cubed.

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3 = 310 cm3

100 cm

410 cm

4. V4 = 309 cm3

100 cm

409 cm

Thus, the final volume (average) is (311+310+310+309)/ 4 = 310 cm3.

The error is = + 5 cm.

Hence the Volume = 310cm3 +5cm.

For volume of the cylinder

We calculate the volume by the formula V = πr2h

Thus, Volume = πx (9.5/2)2 x 6.4

                        = 305.614 cm3

The error is twice the percent error of the radius. (Since the radius is squared)

Error = 2 (% error of radius)

Error = 2 x (0.01/6.4) x 100

          = 0.03125 % of the volume

Thus, error = (0.03125/100) x 305.614

                   = 0.0955 cm3

Volume of Cylinder = 305.60 +0.096 cm3

I noticed that there is a big difference between both the methods (with the sphere and the cylinder). E.g.: - volume of sphere with water displacement is 10 +1 cm2, but when we use the formula it comes out to be 9.525 + 0.108 cm2. This shows us that with the correct precautions and methods, we can get an answer that has the least possible error and is therefore the most accurate. My expectation regarding the difference was hence right. I found this to be an engaging and eye-opening experiment and I enjoyed doing it. I look forward to more experiments of this type in the future

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