ObjectMass / g

± 0.05 g *Length / cm

± 0.003 cm **Width / cm

± 0.003 cm **Height / cm

± 0.003 cm **

T1T2T3T4T5T1T2T3T4T5T1T2T3T4T5

A44.95.9755.9805.9805.9805.9751.5001.5001.5001.5001.5000.6000.6000.6050.6000.600 B16.54.4004.3954.4004.3954.3951.5001.5001.4951.5001.4950.3000.3050.3050.3000.300 C22.53.0002.9953.0003.0003.0001.5001.5051.4951.5001.5000.6000.5950.6000.5950.595 D22.66.0056.0006.0106.0056.0001.5001.5001.4951.5001.4950.2950.3000.3000.3050.300 E11.43.0403.0453.0403.0753.0701.5001.4951.5001.5001.4950.3000.3000.3050.3000.300 F5.801.5501.5501.5401.5501.5501.4951.5001.5001.4951.5000.3000.3000.3000.3000.300 Table 1: Raw data table

Half the smallest division: 0.1 ÷ 2 = 0.05 g

** Half the smallest division: 0.005 ÷ 2 = 0.0025 ≃ 0.003 cm Inaccurate set of data due to random error. (See Evaluation)

Table 2: Data processing 1

ObjectMass / g

± 0.05 gAverage L / cmAbs. Unc. L / cmAverage W / cmAbs. Unc. W / cmAverage H / cmAbs. Unc. H / cm A44.95.980.0061.500.0030.600.007

B16.54.400.0061.500.0060.300.006

C22.53.000.0071.500.0080.600.006

D22.66.000.0091.500.0060.300.008

E11.43.060.03*1.500.0060.300.007

F5.801.550.02*1.500.0060.300.003

Since the uncertainties can only be written with 1 s.f, the number of decimal places cannot be the same as the others.

Calculating the average of length, width, and height and their uncertainties: In order to find the average all the numbers should be added up, and then divided by how many numbers there are. Average: (∑X)/N

Average length of object A:

(5.975+5.980+5.980+5.980+5.975)/5 = 5.978 ≃ 5.98 cm

Average width of object A:

(1.500+1.500+1.500+1.500+1.500)/5 = 1.500≃ 1.50 cm

Average height of object A:

(0.600+0.600+0.605+0.600+0.600)/5 = 0.601 ≃ 0.60 cm

After finding the average, the uncertainties should be calculated. The uncertainties are calculated as follows: |Average – Highest value |

|Average – Lowest value |

In this case however, since there is already an uncertainty of 0.003 cm in the vernier calliper, the number calculated above should be added to 0.003. Calculating the uncertainties of the averages for the length, width and height of object A are shown below: Length:

|5.978 – 5.980| = 0.002 cm

|5.978 – 5.975| = 0.003 cm

0.003 is the bigger value; therefore, this number is used. 0.003 cm is then added to the initial uncertainty which is also 0.003 cm in this case. 0.003 + 0.003 = 0.006 cm

Therefore the length and its uncertainty is: 5.98± 0.006 cm Width:

|1.500 – 1.500| = 0.00 cm

|1.500 – 1.500| = 0.00 cm

The values are both 0 since all the readings were the same; therefore, the uncertainty is only 0.003 cm which is the initial uncertainty. Therefore the length and its uncertainty is: 1.50 ± 0.003 cm Height:

Highest value=|0.601 – 0.605| = 0.004 cm

Lowest value=|0.601 – 0.600| = 0.001 cm

0.004 is the bigger value; therefore, this number is used. 0.004 cm is then added to the initial uncertainty which is 0.003 cm. 0.004 + 0.003 = 0.007 cm

Therefore the length and its uncertainty is: 0.60 ± 0.007 cm

Table 3: Data processing 2

ObjectMass / g

± 0.05 gVolume / 〖cm〗^3Abs. Unc. Volume / 〖cm〗^3

A44.95.390.08

B16.51.990.06

C22.52.690.05

D22.62.700.09

E11.41.380.06

F5.800.700.02

Calculating the volume and its uncertainty:

To calculate the volume of an object 3 variables are required: length, width and height of the object. By multiplying these three variables, the volume of an object can be determined. Object A’s volume can be calculated like so: Δ Length × Δ width × Δ height = Volume

5.978 × 1.500 × 0.601 = 5.389 ≃ 5.39 〖cm〗^3

To calculate the uncertainty of the volume the following steps should be taken: Step 1: Add up the relative uncertainties of length, width, and height. To do this, the...