Precision & Accuracy
- Every experimental measurement has some amount of uncertainty associated with it. Everyone wants the results to be precise and accurate. Precision and Accuracy are often referred to whenever we talk about measurement.
- Precision refers to the closeness of various measurements for the same quantity. It does not depend on true value.
- Accuracy is the agreement of a particular value to the true value of the result. It depends on the true value.
Significant figures
- The uncertainty in the experimental or the calculated values is indicated by mentioning the number of significant figures.
- Significant figures are meaningful digits which are known with certainty.
- There are certain rules for determining the number of significant figures. They are as follows:
- All non-zero digits are significant. For example: – There are 4 significant figures in 328.2 cm.
- Zeros preceding to first non-zero digit are not significant. For example: – 0.03 has one significant figure and 0.0052 has two significant figures.
- Zeros between two non-zero digits are significant. For example: – 2.005 have four significant figures.
- Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. For example: – 200 g has three significant figures. But, if otherwise, the zeros are not significant. For example, 100 have only one significant figure.
- Exact numbers have an infinite number of significant figures. For example:- In 2 balls or 20 eggs, there are infinite significant figures as these are exact numbers and can be represented by writing infinite number of zeros after placing a decimal i.e.,2 = 2.000000 or 20 = 20.000000.
Addition & Subtraction
- The result cannot have more digits to the right of the decimal point than either of the original numbers. For example: – (12.11 + 18.0+ 1.012) = 31.122. As 18.0 have only one digit after the decimal point therefore the result will be 31.1, one digit after the decimal point.
Multiplication & Division
- In these operations, the result must be reported with no more significant figures as are there in the measurement with the few significant figures. For example: – 2.5×1.25 = 3.125. Answer will be 3.1 because 2.5 have two significant figures and the result should not have more than two significant figures.
Dimensional Analysis
- When calculating, there is a need to convert units from one system to other.
- The method used to accomplish this is called factor label method or unit factor method or dimensional analysis. For example: – A jug contains 2L of milk. Calculate the volume of the milk in m3.
Problem:-
How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
Answer:-
(i) 0.0025
There are 2 significant figures.
(ii) 208
There are 3 significant figures.
(iii) 5005
There are 4 significant figures.
(iv) 126,000
There are 3 significant figures.
(v) 500.0
There are 4 significant figures.
(vi) 2.0034
There are 5 significant figures.
Problem:-
Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Answer:-
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810
Problem:-
A jug contains 2L of milk. Calculate the volume of the milk in m3.
Answer:-
Since 1 L = 1000 cm3 and 1m = 100 cm which gives
(1 m)/ (100 cm) = 1 = (100 cm/1m)
To get m3 from the above unit factors, the first unit factor is taken and it is cubed.
Therefore, (1m) 3 / (100cm) 3 =1 = (1m3)/ (106 cm3) (unit factor)
Now 2 L of milk = 2×1000 cm3
The above is multiplied by the unit factor
= 2 x1000 cm3 x (1 m3/106 cm3)
= 2 x 10-3 m3
Question 16. [NCERT Excercise]
What do you mean by significant figures?
Solution.
Significant figures : The significant figures in a number are all the certain digits plus one doubtful digit, e.g., 2005 has four significant figures.
Question 19.[NCERT Excersice]
How many significant figures are present in the following?
(i) 0.0025
(ii) 208
(iii) 5005
(iv) 126,000
(v) 500.0
(vi) 2.0034
Solution.
(i) 2
(ii) 3
(iii) 4
(iv) 3
(v) 4
(vi) 5
Question 20. [NCERT Excersice]
Round up the following upto three significant figures:
(i) 34.216
(ii) 10.4107
(iii) 0.04597
(iv) 2808
Solution.
(i) 34.2
(ii) 10.4
(iii) 0.0460
(iv) 2810 or 2.81 x 103
Q. Round off up to 3 significant figures (a) 1.235 (b) 1.225
Ans. (a) 1.24 (b) 1.22
Q. State the number of significant figures in each of the following
(i) 208.91 (ii) 0.00456 (iii) 453 (iv) 0.346
Ans.
(i) 208.91 has five significant figures.
(ii) 0.00456 has three significant.
(iii) 453 has three significant figures.
(iv) 0.346 has three significant figures.
Q. Express the results of the following calculations to the appropriate number of significant figures.
(i) (3.24 x 0.08666) / 5.006 (ii) (1.36 x 10-4 ) (0.5)/2.6
Ans.
(i) 0.05608 = 0.0561
(ii) 0.2615 x 10-4 = 0.3 x 10-4