Note : (i) The uncertainty in any value is indicated by mentioning the number of signification figures.

(ii) Significant figures are meaningful digits which have certainty.

(iii) Uncertainty is indicated by writing the certain digits and the last uncertain digit e.g. if there is a result written as `11.2` `mL`.

Formal Definition : The total number of digits in a number including the last digit whose value is uncertain is called

the number of significant figures.


`"Example-"` If we say the period of oscillation of a simple pendulum is 1.62 s, the digits 1 and 6 are reliable and certain, while the digit 2 is uncertain. Thus, the measured value has three significant figures.

• A choice of change of different units does not change the number of significant digits or figures in a measurement.

`"Example-"` The length 2.308 cm has four significant figures. But in different units, the same value can be written as `0.02308 m` or `23.08mm` or `23080 μm`. All these numbers have the same number of significant figures (digits 2, 3, 0, 8), namely four.

Defination : (1). A leading zero is any 0 digit that comes before the first nonzero digit

(2). trailing zeros are a sequence of 0's after which no other digit follows.

(i) All non-zero digits are significant.

(ii) Zeros preceding to first non-zero digit are not significant. These zeros only indicates the position of decimal point only.

(iii) Zeros between two non-zero digit are significant.

(iv) Zero at the end or right of a number are significant if they are on the right side of the decimal point.

e.g. `0.200g` has `3` significant figures.

If there is no decimal point then the terminal zeros are not significant.

e.g. `tt((text{Number} , text{no. of significant figures} ) , (100 , 1) , (100. , 3) , (100.0 , 4))`


(v) Objects which can be counted have infinite significant figure as these are exact numbers. And can be represented as writing infinite number of zero after placing a decimal.

e.g. `2 = 2.000000` or `20 = 20.00000`

(i) If the right most digit to be removed `> 5`, then preceding number is increased by one.

e.g. `1.386`

If we remove `6`, then it is rounded off to `1.39`.

(ii) If the right digit to be removed `< 5`, then the preceding number is not changed.

e.g. `4.334`

If we remove `4`, then it is rounded off to `4.33`.

(iii) If the right most digit to be removed `= 5`

Case I : If preceding number is even, then it is not changed.

Case II : If preceding number is odd, then it is increased by one.

The result obtained can not have more digits to the right of the decimal point than either of the original numbers.

e .g.

`12.11`
`18.0`
`1.012`
` underline overline(31.122)`

Multiplication and Division of Significant Figures

(Since 18.0 has only one digit after decimal point . So, the answer should be reported as `31.1`.)

The result obtained should have the significant figures equal to the number which has minimum number of significant figures.

e.g. `2.5xx1.25 = 3.125`

So, answer should be reported as `3.1`.

• If length `l=16.2cm` and breadth `b=10.1cm` then `l` may be written as `l = 16.2 ± 0.1 cm = 16.2 cm ± 0.6 %` and `b` may be written as `b = 10.1 ± 0.1 cm = 10.1 cm ± 1 %`.

Then, the error of the product of two (or more) experimental values, using the combination of errors rule, will be `l b = 163.62 cm^2 + 1.6%
= 163.62 + 2.6 cm^2`

This leads us to quote the final result as `l b = 164 pm 3 cm^2`. Here `3 cm^2` is the uncertainty or error in the estimation of area of rectangular sheet.

• If a set of experimental data is specified to `n` significant figures, a result obtained by combining the data will also be valid to `n` significant figures.
However, if data are subtracted, the number of significant figures can be reduced.

• The relative error of a value of number specified to significant figures depends not only on n but also on the number itself.
`"Example-"` The accuracy in measurement of mass 1.02 g is ± 0.01 g whereas another measurement 9.89 g is also accurate to ± 0.01 g.
The relative error in 1.02 g is = (± 0.01/1.02) × 100 % = ± 1%
Similarly, the relative error in 9.89 g is = (± 0.01/9.89) × 100 % = ± 0.1 %