Question 4: Briefly explain how do we indicate uncertainties? With the help of examples explain how uncertainties are measured in final result in different cases?

To indicate uncertainty in a measurement, consider the following example.

Suppose the length of a block is measured with the help of a meter rod. Let the length be 22.5 cm. The least count of the meter rod is 0.1 cm. Hence, the total uncertainty in the measurement is ±0.1 cm. Therefore, we can conclude these end values are maximum values of uncertainties.

When we measure the length of the block, we take two readings, one on the first end of the block and the other on the second end of the block. Let one end of the block coincides with, say, 10.5 cm mark and the other coincides with 33.0 cm mark. This means ±0.05 cm uncertainty develops at one end and ±0.05 cm on the other end. Therefore, (33.0 ±0.05) cm – (10.5 ± 0.05) cm = (22.5 ± 0.1) cm is the length of the block along with the uncertainty. Therefore, length of the rod is sure in the range somewhere from 22.4 cm to 22.6 cm.

## Uncertainty in the final results

Let x, y and z are three different measurements such that x is obtained when y and z are mathematically operated (i-e, added, subtracted, divided, multiplied etc). Let ±Δ x, ±Δ y and ±Δ z are their respective uncertainties.

### (1) Rule for addition or subtraction

Let in the above case, x is the final result obtained by the addition or subtraction of y and z. Then the uncertainty in the final value of x will be the sum of the uncertainties in the measurements of y and z.

Therefore, ±Δ x = ± (Δ y + Δ z)

This expression represents the total uncertainty in the final result and valid for both addition and subtraction.

Example

Let y and z represents the lengths of two lines to be added or subtracted and x is the final result. Further suppose y = 6.3±0.1 cm and z = 4.3 ±0.1 cm. Then
x = y + z = 6.3±0.1+4.3 ±0.1 = (6.3+4.3) ± (0.1+0.1) = 10.6 ± 0.2 cm
And in case x = y-z = (6.3±0.1)-(4.3±0.1) = (6.3 – 4.3) ± (0.1+0.1) = 2.0 ±0.2 cm

### (2) Rule for product or quotient (multiplication or division)

Let x be the final result obtained by the multiplication or division of y and z. In such cases, we find the percent uncertainties of the measurements operated to find percent uncertainty in the product/quotient. So,

Percentage uncertainty in x = ± (percent uncertainty in y + percent uncertainty in z)

Once %age uncertainty in x is determined, it is always easy to find the uncertainty in the actual value.

Example

Take the above example once again. Percent uncertainty in the length of first line is

0.1/6.3×100=2 cm approximately —– (1)

And percent uncertainty in the length of second line is

0.1/4.3×100=2 cm approximately —- – (2)

Therefore, in the case of multiplication, x = yz,

Percentage uncertainty in x = percentage uncertainty in y + percentage uncertainty in z

Percentage uncertainty in x = (2 +2) = 4%

[Note: This is percent uncertainty, that is, uncertainty in 100 cm.]

If we are interested in the actual value of x with the uncertainty included, then

x=yz±(actual uncertainty in yz)

=(6.3×4.3)±{4/100 (6.3×4.3) }

=27.09±(0.04×27.09)

=27.09±1.08 cm

This equation gives the value of x = yz along with total uncertainty of the product.

Now if we are concerned with the quotient (division) of the lines, then we apply the quotient rule.

Percentage uncertainty in x = ± (percentage uncertainty in y + percentage uncertainty in z)

Percentage uncertainty in x = ± (2+2) = 4 cm.

The actual value of the quotient could be calculated as follow.

x±∆x=y/z±(actual uncertainty in y/z)

=1.5±{4/100×1.5}

=1.5±(0.04×1.5)

=1.5±0.06 cm

This is called uncertainty in the quotient (division).

### (3) Uncertainty in power

In the final result of a power factor, the percentage uncertainty is multiplied with the power factor.

Let us calculate the uncertainty in the measurement of the volume of a sphere. Let radius of the sphere is with the range of uncertainty as r=2.25±0.01 cm

Volume of the sphere=V=4/3 πr3 —— (1)

Put the value of r in the above equation

V=4/3×3.14×2.25×2.25×2.25=47.68 cm3 — – (2)

Now consider the radius measurement of the sphere; there is an uncertainty of ±0.01 in the total length of 2.25 cm of the radius. Therefore, the percent uncertainty is,

= 0.01/2.25×100=0.0044×100=0.4

Now by equation (1) the power of ‘r’ is 3 while calculating the volume. Therefore,
Percent uncertainty in volume = 0.4 *3= 1.2 (percentage uncertainty in radius multiplied by power factor).

This means that in the volume of the sphere of 100 cm3, the uncertainty is 1.2 cm. So in the present case when the volume is 47.68 cm3

Therefore, uncertainty in the case in hand will be,

1.2/100×47.68=0.57

Therefore, the volume of the sphere will be

Volume=V=47.68±0.57 cm

### (4)Uncertainty measurement in average value

Follow the following steps for finding uncertainty or error in the average value when you make a number of measurements of a certain quantity.

1. Find the average value of all measurements.
2. Find the deviations of the individual values from the average value. Here you have to ignore the sign, if the deviation is negative.
3. Find the mean value of the deviations.
4. Mean deviation is the average value of all deviations.

Example

Let 1.50 cm, 1.51 cm and 1.52 cm are the measured values of the diameter of a wire.

1. The average value of all measurements = (1.50+1.51+1.52)/3=4.53/3=1.51
2. Deviation of first measurement from the average value = 1.50 – 1.51 = 0.01 cm. (The negative sign is ignored). Deviation of second measurement from the average value = 1.51 -1.51 = 0 cm.Deviation of third measurement from the average value = 1.52 – 1.51 = 0.01 cm.
3. Mean deviation = (0.01+0+0.01)/3=0.02/3=0.007

This gives the possible error or uncertainty in the measurement of the diameter. Therefore,

Diameter of the wire = 1.51±0.007 cm

### (5) Uncertainty in time period

Let us calculate the uncertainty in the measurement of the time period of a pendulum. Let the time of 30 vibrations recorded by a stopwatch is 54.6 s and the least count of the stopwatch is 0.1 s. Then,

Time period=(Total time )/(number of vibrations)=54.6/30=1.82 s

Uncertainty in time period is calculated as;

uncertainty in time period=(least count)/(no.of vibrations)=0.1/30=0.003

Therefore, time period of the pendulum = 1.82 ± 0.003 s

Note: Since least count is constant, therefore, if we take a large number of vibrations in one reading, then the amount of uncertainty or error will be less.