Assessment of total uncertainty in the final result

 

To assess the total uncertainty or error, it is necessary to evaluate the likely uncertainties in all the factors involved in that calculation. The maximum possible uncertainty or error in the final result can be found as follows. 

1. For addition and subtraction 

Absolute uncertainties are added: 

For example, the distance x determined by the difference between two separate position measurements x1=10.5±0.1 cm  and x2 = 26.8 1±0.1 cm is recorded as

x=x2 - x1=16.3±0.2 cm.

2. For multiplication and division 

Percentage uncertainties are added. 

For example the maximum possible uncertainty in the value of resistance R of a conductor determined from the measurements of potential difference V and resulting current flow I by using

R = V/I is found as follows: 

V=5.2±0.1 V

I = 0.84±0.05A

The %age uncertainty for V is = 0.1V/5.2V × 100 = about 2%

The %age uncertainty for I is = 0.05A/0.84A × 100 = about 6%

Hence total uncertainty in the value of resistance R when V is divided by I is 8%. The result is thus quoted as 

R = 5.2V/0.84A = 6.19 V/A = 6.19 ohms with a % age uncertainty of 8%

that is R = 6.2 1±0.5 ohms.

The result is rounded off to two significant digits because both V and R have two significant figures and uncertainty, being an estimate only, is recorded by one significant figure.

4. For power factor

Multiply the percentage uncertainty by that power. 

For example, in the calculation of the volume of a sphere using

V = 4/3 π r3

%age uncertaintyin V= 3 x %age uncertainty in radius r.

As uncertainty is multiplied by power factor, it increases the precision demand of measurement. If the radius of a small sphere is measured as 2.25 cm by a vernier callipers with least count 0.01 cm, then the radius r is recorded as

r= 2.25±0.01 cm 

Absolute uncertainty = Least count = 0.01 cm

%age uncertainty in r= 0.01 cm / 2.25 cm ×100 = 0.4% 

Total percentage uncertainty in V = 3 × 0.4 = 1.2%

Thus volume  V= 4/3π r3

            = 4/3 × 3.14 × ( 2.25 cm)3

            = 47.689 (cm)3

 with 1.2% uncertainty

Thus the result should be recorded as

V = 47.71 (+0.6 or -0.6) cm3


4. For uncertainty in the average value of many measurements.

(i) Find the average value of measured values.

(ii) Find deviation of each measured value from the average value.

(iii) The mean deviation is the uncertainty in the average value. 

For example, the six readings of the micrometer screw gauge to measure the diameter of a wire in mm are:

1.20, 1.22, 1.23, 1.19, 1.22, 1.21 

Then Average:

             =(1.20+1.22+1.23+1.19+1.22+1.21)/6

         = 1.21 mm


The deviation of the readings, which are the difference without regards to the sign, between each reading and average value are 0.01, 0.01,-0.02, 0.02, 0.01, 0. 

Mean of deviations = (0.01 +0.01 +0.026+0.02 +0.01+ 0)/6 

                                    = 0.01 mm 


Thus, likely uncertainty in the mean diametre 1.21 mm is 0.01 mm recorded as 1.21±0.01 mm.

5. For the uncertainty in a timing experiment 

The uncertainty in the time period of a vibrating body is found by dividing the least count of timing device by the number of vibrations. 

For example, the time of 30 vibrations of a simple pendulum recorded by a stopwatch accurate upto one tenth of a second is 54.6 s, the period 

T = 54.6 s / 30 = 1.82 s with uncertainty 0.1/30 = 0.003 s

Thus, period T is quoted as T = 1.82±0.003 s.

Hence, it is advisable to count large number of swings to reduce timing uncertainty.

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