Chapter 2.10 - Significant figures - Solved examples

In the previous section, we saw the 'rules for arithmetic operations with significant figures'. We saw some solved examples also. In this section, we will see a few more solved examples.

Solved example 2.15
Fill in the blanks
(Note: In stating numerical answers, take care of significant figures)
(a) The volume of a cube of side 1 cm is equal to .....m³
(b) The surface area of a solid cylinder of radius 2.0 cm and height 10.0 cm is equal to  ...(mm)²
(c) A vehicle moving with a speed of 18 km h^-1 covers ....m in 1 s
(d) The relative density of lead is 11.3. Its density is ....g cm^-3 or ....kg m^-3
Solution:
Part (a):
1. We have: Volume  = s × s × s
• Where 's' is the length of side
2. The side is given in cm. But we want the volume in m³
• 1 cm = 0.01 m
    ♦ 1 cm have 1 significant figure
    ♦ 0.01 m also have 1 significant figure
3. So volume in m³ = (0.01)³ = 10^-6 m³  
Part (b):
1. We have: Total surface area of a solid cylinder = 2𝞹r² + 2𝞹rh = 2𝞹r(r + h)
See details here
2. First we evaluate the quantity within the brackets
• r = 2.0 cm
• h = 10.0 cm
(r + h) = 2.0 + 10.0 = 12.0 cm
(Since both values have the same number of digits after the decimal point, we do not need to draw the magenta line that we saw in fig.2.18 above)
3. So total surface area = 2𝞹r × 12.0 = (2 × 𝞹 × 2.0 × 12.0) cm²
• 2 is a factor
It has infinite number of significant figures
• 2.0 has 2 significant figures
• 10.0 has 3 significant figures
• So the final result must have only 2 significant figures
• We will take the value of 𝞹 in such a way that, it has 3 (one more than what is required in the final result) significant figures
• So 𝞹 = 3.14
4. Thus total surface area = (2 × 3.14 × 2.0 × 12.0) cm² = 150.72 cm²
5. one cm = 10 mm
• So 1 cm² = 100 mm² 
• Here 100 is a factor. It has infinite number of significant figures
6. We can now convert the 'area in cm²' to 'area in m²'
• Total surface area in m² = 150.72 × 100
• In scientific notation, this value is 1.5072 × 10⁴ m²
7. But in (2) we saw that, the final result must have only 2 significant digits. This is valid even if units change
• So total surface area = 1.5 × 10⁴ m².
Part (c):
1. Distance traveled in 1 h = 18 km
• '18' has 2 significant figures
• 1 h = (60 × 60) = 3600 s
• So distance traveled in 1 s = (¹⁸⁄₃₆₀₀) km
2. 3600 is a factor. It has infinite number of significant figures
• So the result of (¹⁸⁄₃₆₀₀) must have only 2 significant figures
3. One km = 1000 m
• So (¹⁸⁄₃₆₀₀) km = [(¹⁸⁄₃₆₀₀) × 1000] m
4. 1000 is a factor. It has infinite number of significant figures
• So the result of [(¹⁸⁄₃₆₀₀) × 1000] must have only 2 significant figures 
5. We have: [(¹⁸⁄₃₆₀₀) × 1000] = 5 
• If '5' is to have two significant figures, we must write it as 5.0
• So we get: 
A vehicle moving with a speed of 18 km h^-1 covers 5.0 m in 1 s
Part (d):
• Relative density of a material is it's density expressed in relation to the 'density of another material' (usually water)
• Relative density of a material A is given by: ^{Density of A}⁄_{Density of water}
■ Numerator and denominator must be in the same units: g cm^-3  OR  kg m^-3
• So relative density does not have a unit. It is just a ratio
• Now we can write the steps:
1. When the unit is g cm^-3: 
• From data book, density of water is 1.000 g cm^-3
• 1.000 has 4 significant figures
2. We can write: ^{Density of lead}⁄_{Density of water} = 11.3
⇒ ^{Density of lead}⁄_1.000 = 11.3
⇒ Density of lead = (11.3 × 1.000) g cm^-3
• 11.3 has 3 significant figures
• 1.000 has 4 significant figures
• So the result (11.3 × 1.000) must have only 3 significant figures
3. In scientific notation, we can write (11.3 × 1.000) as 1.13 × 10¹
• (1.13 × 10¹)  = 11.3
It has only 3 significant figures
• So we get: Density of lead = 11.3 g cm^-3  
4. When the unit is kg m^-3: 
• From the data book, density of water is 1000 kg m^-3
• 1000 has 4 significant figures
6. We can write: ^{Density of lead}⁄_{Density of water} = 11.3
⇒ ^{Density of lead}⁄₁₀₀₀ = 11.3
⇒ Density of lead = (11.3 × 1000) kg m^-3
• 11.3 has 3 significant figures
• 1000 has 4 significant figures
• So the result (11.3 × 1000) must have only 3 significant figures
7. In scientific notation, we can write (11.3 × 1000) as 1.13 × 10⁴
• It has only 3 significant figures
• So we get: Density of lead = 1.13 × 10⁴ kg m^-3

Solved example 2.16
The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
Solution:
1. Area of top and bottom faces = length × breadth = 4.234 × 1.005 = 4.25517 m².
• 4.234 has 4 significant figures
• 1.005 has 4 significant figures
• So the result 4.25517 should have only 4 significant figures
• Rounding off, we get: 4.255
• So area of the top and bottom faces = 4.255 m².
2. Area of front and rear faces = length × thickness
• Thickness is given to us in cm. We have to convert it into m
• 1 cm = 0.01 m
• So we have to multiply each cm by 10^-2 
• So 2.01 cm = 2.01 × 10^-2 m
3. Now we can find the area:
• length × thickness = 4.234 × (2.01×10^-2) = 8.51034 ×10^-2 m³.
• 4.234 has 4 significant figures
• (2.01×10^-2) has 3 significant figures
• So the result (8.51034 ×10^-2) should have only 3 significant figures
• Rounding off, we get: (8.51 ×10^-2)
• So area of the front and rear faces = (8.51 ×10^-2) m².
4. Area of left and right faces = breadth × thickness
= 1.005 × (2.01×10^-2) = 2.02005 ×10^-2 m³.
• 1.005 has 4 significant figures
• (2.01×10^-2) has 3 significant figures
• So the result (2.02005 ×10^-2) should have only 3 significant figures
• Rounding off, we get: (2.02 ×10^-2)
• So area of the left and right faces = (2.02 ×10^-2) m²    
5. So total area = 2 [4.255 + (8.51 ×10^-2) + (2.02 ×10^-2)]
= 2 [4.255 + 0.0851 + 0.0202]
6. The addition inside the square brackets is shown in fig.2.19 below:

Fig.2.19

• The digits on the right side of the magenta line should be rounded off
• After rounding off, we get: 4.360
7. Thus the total area becomes: 2 × [4.360] = 8.72
• 2 is a factor. It has infinite number of significant figures
• 4.360 has 4 significant figures
• So the result 8.72 must also have 4 significant figures
• Thus the total area becomes: 8.720 m².
• We can write:
Total surface area of the metal sheet = 8.720 m².
8. Calculation of volume:
• We have: Volume = area of top face × thickness
= 4.255 × (2.01×10^-2) = 8.55255 ×10^-2
• 4.255 has 4 significant figures
• (2.01×10^-2) has 3 significant figures
• So the result (8.55255 ×10^-2) should have only 3 significant figures
• Rounding off, we get: (8.55 ×10^-2)
• So volume = (8.55 ×10^-2) m³.

Solved example 2.17
The mass of a box measured by a grocer’s balance is 2.3 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to the box. What is (a) the total mass of the box, (b) the difference in the masses of the pieces to correct significant figures?
Solution:
Part (a):
1. Mass of the box = 2.3 kg
• Mass of gold piece 1 = 20.15 g
• Mass of gold piece 2 = 20.17 g
2. We will convert g into kg
• In this way, we will get measurements with decimal places
• If we convert kg into g, the decimal point will vanish
(2.3 kg = 2300 g)
• For problems involving addition/subtraction, we need to have numbers with decimal places. This will enable us to apply the rule more easily
3. 20.15 g = (20.15 ×10^-3) kg = 0.02015 kg
    ♦ There is no change in the number of significant figures
20.17 g = (20.17 ×10^-3) kg = 0.02017 kg
    ♦ There is no change in the number of significant figures
4. Sum of all the weights is shown in fig.2.20 (a) below:

Fig.2.20

• The digits on the right side of the magenta line should be rounded off
• After rounding off, we get: 2.3 kg
■ This result does not show the effects of adding the gold pieces to the box. This is because, the box is weighed with a device of low precision
Part (b):
• The difference in weights is shown in fig.2.20(b) above
• There are no digits on the right side of the magenta line. So there is no rounding off to be done
• We get: The difference in the masses of the pieces = 0.02 g

Solved example 2.18
The radius of a circle is 2.12 m. What is it's area according to rules of significant rules?
Solution:
1. We have: Area of circle = 𝞹r² 
2. So in our present case, area = (𝞹 × 2.12 × 2.12) m²
• 2.12 has 3 significant figures
• So the final result must have 3 significant figures
• We will take the value of 𝞹 in such a way that, it has 4 (one more than what is required in the final result) significant figures
• So 𝞹 = 3.141
3. Substituting the value, we get:
• Area = (3.141 × 2.12 × 2.12) = 14.1169104 m²
• Rounding off to 3 significant figures, we get:
Area = 14.1 m².

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In the next section, we will learn about errors

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Chapter 2.9 - Arithmetic operations with Significant figures

In the previous section, we saw how to find the number of significant figures in a given measurement. In this section, we will see the 'rules for arithmetic operations with significant figures'. We will first write about the basics: 

1. Consider an experiment in which we want to find the 'density of the material' of an object
• The mass of that object is measured to be 4.237 g
• It’s volume is measured to be 2.51 cm³
2. We know that density = ^mass⁄_volume
• In our present case, we get: density of the material of the object = ^4.237⁄_2.51 
3. So we have to perform a division
• If we use a calculator, we will get the density as: 1.68804780876 g cm^-3
4. This result has 12 significant figures
• When a person reads this result at a later stage, what will he think?
Answer: He will think that, during the experiment, the measurements were made using instruments of very high precision.
• That means, he will be getting all wrong information about the procedure of the experiment
• We cannot let such wrong information to be conveyed
5. Scientists have given clear rules that we must follow in such situations. The rules are based on the following facts:
• The measurements will be done using various instruments
• Each instrument will be having it’s own precision
• The result of the experiment may be obtained by the multiplication/division or addition/subtraction of those measurements 
• Now consider two items:
(i) Precision of the result of the experiment
• Remember that, the result is obtained by the multiplication/division or addition/subtraction of the measurements
(ii) Precision of the lowest precise measurement
■ Item (i) cannot be expected to be higher than item (ii) 
6. Now we will see the two rules
Rule 1:
• This rule is to be followed while doing multiplication or division
• The steps are as follows:
(i) Note down the number of significant figures in each of the numbers which are being multiplied or divided
(ii) Note down the smallest among those numbers
(iii) The number of significant figures in the product/quotient must be same as the number in (ii) above
Let us apply this rule to our present case:
Step (i):
• 4.237 has 4 significant figures
• 2.51 has 3 significant figures
Step (ii): '3' is the smallest number of significant figures
Step (iii): The result '1.68804780876' should not have more than 3 significant figures
• So in this case, there can only be two digits after the decimal place
■ What happens to the digits coming after those two digits?
We will see the answer when we learn 'rounding off significant figures' which is discussed further down below

Rule 2:
This rule is to be followed while doing addition or subtraction
The steps are as follows:
(i) Write the numbers to be added or subtracted, one below the other
• The decimal points should all come in a vertical line
(ii) Find the number with the ‘least number of decimal places’
(iii) Draw a vertical line just to the right of that number
(iv) The ‘number of decimal places in the result' should not go beyond that vertical line


Two examples are shown in the fig.2.18 below:

Fig.2.18

■ What happens to the digits coming after the magenta line?
We will see the answer in the topic 'rounding off significant figures' which is discussed below

Rules for rounding off significant figures

• We have seen that, the result obtained by multiplication/division or addition/subtraction should not contain more significant figures than required
• We have seen the methods to determine the ‘position of the last significant figure’ in the result
Examples:
• In ‘1.68804780876’ that we obtained as density in the example for rule 1, the ‘position of the last significant figure’ is the (¹⁄₁₀₀)^th place
• In fig.2.18(a), the ‘position of the last significant figure’ in the result is the (¹⁄₁₀₀)^th place
• In fig.2.18(b), the ‘position of the last significant figure’ in the result is the (¹⁄₁₀)^th place
‘position of the last significant figure’ comes very often in our discussions. So we will shorten it as: PLS. Note that, PLS indicates a 'position'. It is not a digit.
So we can write the above examples as:
• In ‘1.68804780876’ that we obtained as density in the example for rule 1, the PLS is the (¹⁄₁₀₀)^th place
• In fig.2.18(a), the PLS in the result is the (¹⁄₁₀₀)^th place
• In fig.2.18(b), the PLS in the result is the (¹⁄₁₀)^th place

• Now, 3 questions arise:

(i) What happens to the digits beyond the PLS?
(ii) What happens to the digits before the PLS?
(iii) What happens to the digit at the PLS?
• The ‘rules for rounding off significant figure’ will give the answers to these questions
Let us see the rules:
■ First, always consider the digit just to the right of the PLS
Based on the value of this digit, we must select from among 3 rules:
Rule 1: If this digit is greater than 5, add 1 to the digit at the PLS
• After that, discard all digits to the right of the PLS
Examples:
    ♦ If 5.23781 should have only 3 significant figures, it will be rounded off to 5.24
Explanation: Here PLS is the (¹⁄₁₀₀)^th place. The digit '7' which is just to the right of PLS is greater than 5 
    ♦ If 0.0037681 should have only 2 significant figures, it will be rounded off to 0.0038
Explanation: Here PLS is the (¹⁄₁₀₀₀₀)^th place. The digit '6' which is just to the right of PLS is greater than 5 

Rule 2: If this digit is less than 5, add nothing to the digit at the PLS

• After that, discard all digits to the right of the position of the last significant figure     
Examples:
    ♦ If 5.23481 should have only 3 significant figures, it will be rounded off to 5.23
Explanation: Here PLS is the (¹⁄₁₀₀)^th place. The digit '4' which is just to the right of PLS is less than 5
    ♦ If 0.0037281 should have only 2 significant figures, it will be rounded off to 0.0037
Explanation: Here PLS is the (¹⁄₁₀₀₀₀)^th place. The digit '2' which is just to the right of PLS is less than 5
Rule 3: If this digit is exactly equal to 5, two sub rules have to be considered. They are given as (a) and (b) below:
Sub rule (a): If digit at PLS is an even number, add nothing to it
• After that, discard all digits to the right of the PLS
Examples:
    ♦ If 5.26581 should have only 3 significant figures, it will be rounded off to 5.26
Explanation: Here PLS is the (¹⁄₁₀₀)^th place. The digit which is just to the right of PLS is 5. The digit '6' at the PLS is even 
    ♦ If 0.0038581 should have only 2 significant figures, it will be rounded off to 0.0038
Explanation: Here PLS is the (¹⁄₁₀₀₀₀)^th place. The digit which is just to the right of PLS is 5. The digit '8' at the PLS is even 
Sub rule (b): If digit at PLS is an odd number, add 1 to it
• After that, discard all digits to the right of the PLS
Examples:
    ♦ If 5.23581 should have only 3 significant figures, it will be rounded off to 5.24
Explanation: Here PLS is the (¹⁄₁₀₀)^th place. The digit which is just to the right of PLS is 5. The digit '3' at the PLS is odd
    ♦ If 0.0037581 should have only 2 significant figures, it will be rounded off to 0.0038
Explanation: Here PLS is the (¹⁄₁₀₀₀₀)^th place. The digit which is just to the right of PLS is 5. The digit '7' at the PLS is odd

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Three important points to remember:

(1) Complex problems often involves many steps. Suppose that, a problem involves 5 steps
• The result obtained in the step 5 is to be reported as the final answer
• We will get results in the intermediate steps also. Each of those results may be used in the succeeding step
• We must not round off those intermediate results. If we do, there may not be enough digits for rounding off in the final step 5
• Rounding off should be done only on the result obtained in the last step
(2) Speed of light has been determined with a very high degree of precision
• It's value is 2.99792458 × 10⁸  ms^-1 in vacuum
• In some problems we may need to take this value as such
• In some problems, we may need to take only upto the (¹⁄₁₀₀₀₀)^th place, which is 2.9979 × 10⁸  ms^-1  
• In some problems, we may need to take only 3 × 10⁸  ms^-1
• The value to be taken depends on the number of significant figures required in the final result
(3) The value of 𝞹 has been determined with a very high degree of precision
• It's value is 3.1415926 . . .
• In some problems we may need to take a large number of decimal places
• In some problems, we may need to take only a small number of decimal places
• The value to be taken depends on the number of significant figures required in the final result

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Now we will see some solved examples

Solved example 2.13
Each side of a cube is measured to be 7.203 m. What are the total surface area and the volume of the
cube to appropriate significant figures?
Solution:
• Length of one edge of the cube (s) = 7.203 m
1. We have: Total surface area of cube = 6 × s × s = 6  × 7.203 × 7.203 = 311.299254 m²
2. To find the number of significant figures allowable, we will apply the rule for multiplication/division
Step (i):
• 6 is a factor. It has infinite number of significant figures
• 7.230 has 4 significant figures
Step (ii): '4' is the smallest number of significant figures
Step (iii): The result '311.299254' should not have more than 4 significant figures
• So in this case, there can only be one digit after the decimal place
3. So we must round off 311.299254
• If 311.299254 should have only 4 significant figures, it will be rounded off to 311.3
Explanation: Here PLS is the (¹⁄₁₀)^th place. The digit '9' which is just to the right of PLS is greater than 5
■ So the total surface area with appropriate significant figures is 311.3 m² 
4. To find volume:
We have: Volume = s × s × s = 7.203  × 7.203 × 7.203 = 373.714754427 m³
5. To find the number of significant figures allowable, we will apply the rule for multiplication/division
Step (i):
• 7.230 has 4 significant figures
Step (ii): '4' is the smallest number of significant figures
Step (iii): The result '373.714754427' should not have more than 4 significant figures
• So in this case, there can only be one digit after the decimal place
6. So we must round off 373.714754427
• If 373.714754427 should have only 4 significant figures, it will be rounded off to 373.7
Explanation: Here PLS is the (¹⁄₁₀)^th place. The digit '1' which is just to the right of PLS is less than 5
■ So the volume with appropriate significant figures is 373.7 m³

Solved example 2.14
5.74 g of a substance occupies 1.2 cm³. Express its density by keeping the significant figures in view.
Solution:
• Mass = 5.74 g
• Volume = 1.2 cm³ 
1. We have: Density = ^mass⁄_volume
• So in our present case, density = ^5.74⁄_1.2 = 4.78333. . . g cm^-3 
2. To find the number of significant figures allowable, we will apply the rule for multiplication/division
Step (i):
• 5.74 has 3 significant figures
• 1.2 has 2 significant figures
Step (ii): '2' is the smallest number of significant figures
Step (iii): The result '4.78333. . .' should not have more than 2 significant figures
• So in this case, there can only be one digit after the decimal place
3. So we must round off 4.78333. . .
• If 4.78333. . . should have only 2 significant figures, it will be rounded off to 4.8
Explanation: Here PLS is the (¹⁄₁₀)^th place. The digit '8' which is just to the right of PLS is greater than 5
■ So the density with appropriate significant figures is 4.8 g cm^-3

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In the next section, we will see a few more solved examples on significant figures

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Chapter 2.8 - Number of Significant figures

In the previous section, we saw the basics about significant figures. In this section, we will see the rules for finding the number of 'significant figures' in a given measurement.

• Before learning the rules, we must first find the answer to an important question:
Do 'changing of units' have any effect on significant figures?
• We will try to find the answer in steps:
1. Consider the measurement 2.308 cm
• Based on what we saw in the previous section, we get the following information:
(i) The last digit is uncertain
So 0.008 cm is the uncertain quantity
(ii) 2.30 cm is the reliable quantity
• The yellow arrow was stuck just beyond the 2.30 cm mark
(iii) Consider the two points:
    ♦ The 2.30 cm mark
    ♦ The tip of the yellow arrow
• The length between those two points is found by judgement. It is 0.008 cm
(iv) So 8 is the last significant figure
• 2, 3 and 0 are the other significant figures
• In total, there are 4 significant figures 
2. Now, let 2.308 cm be written as 0.02308 m
• Do we get the same 3 information that we wrote in (1)?
Let us try:
Consider the measurement 0.02308 m
(i) The last digit is uncertain
• So 0.00008 m is the uncertain quantity
• But 0.00008 m = 0.008 cm
• That means, 0.008 cm is the uncertain quantity
• This is the same result that we obtained in 1(i) above
(ii) 0.0230 m is a reliable quantity
    ♦ The yellow arrow was stuck beyond the 0.0230 m mark
• But 0.0230 m = 2.30 cm
• That means 2.30 cm is the reliable quantity
    ♦ The yellow arrow was stuck beyond the 2.30 cm mark
• This is the same result that we obtained in 1(ii) above
(iii) Consider the two points:
    ♦ The 0.0230 m mark
    ♦ The tip of the yellow arrow
• The length between those two points is found by judgement
    ♦ This length is 0.00008 m
• But 0.00008 m = 0.008 cm
    ♦ So 0.008 cm is obtained by judgement
• This is the same result that we obtained in 1(iii) above
(iv) So 8 is the last significant figure
• 2, 3 and 0 are the other significant figures
• In total, there are 4 significant figures
3. Let 2.308 cm be written as 23.08 mm
• Do we get the same 3 information that we wrote in (1)?
Let us try:
Consider the measurement 23.08 mm
(i) The last digit is uncertain
• So 0.08 mm is an uncertain quantity
• But 0.08 mm = 0.008 cm
• That means, 0.008 cm is an uncertain quantity
• This is the same result that we obtained in 1(i) above
(ii) 23.0 mm is a reliable quantity
    ♦ The yellow arrow was stuck beyond the 23.0 mm mark
• But 23.0 mm = 2.30 cm
• That means 2.30 cm is a reliable quantity
    ♦ The yellow arrow was stuck beyond the 2.30 cm mark
• This is the same result that we obtained in 1(ii) above
(iii) Consider the two points:
    ♦ The 23.0 mm mark
    ♦ The tip of the yellow arrow
• The length between those two points is found by judgement
    ♦ This length is 0.08 mm
• But 0.08 mm = 0.008 cm
    ♦ So 0.008 cm is obtained by judgement
• This is the same result that we obtained in 1(iii) above 
(iv) So 8 is the last significant figure
• 2, 3 and 0 are the other significant figures
• In total, there are 4 significant figures
■ Let us write an inference based on (1), (2) and (3):
The units of 2.308 cm can be changed and written as 0.02308 m or 23.08 mm. But the reliable quantity and uncertain quantity will not change. So the number of significant figures do not change

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Now we can write the rules for 'finding the number of significant figures'

Rule 1: All non-zero digits are significant
Explanation:
• In a measurement, we may see any of the 9 digits: 1, 2, 3, 4, 5, 6, 7, 8, and 9
• We must count them as significant

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• Once we understand Rule 1, there should be no doubts regarding non-zero digits

• The rest of the rules are all related to zeros

Rule 2: All captive zeros are significant

Explanation: 
• Captive zeros are those zeroes which are in between two non-zero digits
• Such zeros are significant. We need not worry whether there is a decimal point or not
Examples:
    ♦ The under lined zero in 2037 is significant
    ♦ The under lined zeros in 5.70021 are significant
Rule 3: IF the measurement is less than 1 AND there is one or more zeros just to the right side of the decimal point, those zeros are NOT significant
Explanation:
• If the measurement is less than 1, there will not be any non-zero digits on the left side of the decimal point. There will be only zeros
• On the right side of the decimal point, there can be both zeros and non-zero digits
• Consider two items:
(i) the decimal point
(ii) the first non-zero digit coming after the decimal point
• If there are any zeros between the two items, none of those zeros are significant
Examples:
    ♦ The under lined zero in 0.035 is not significant
    ♦ The under lined zero in 0.0402 is not significant
    ♦ The under lined zeros in 0.00295 are not significant
Rule 4:
• This rule is related to trailing zeros. There can be two cases:
Case 1: The measurement has a decimal point
Case 2: The measurement does not have a decimal point
We will consider each case separately
Case 1:
• In the previous section, we saw this:
If length is exactly 2.3 cm (on a ruler which can measure both cm and mm), we must write it as 2.30 cm
• Another example:
A tape measure is graduated in m and cm. The yellow arrow is exactly at 4 m and 20 cm. Then the reliable measurement is 4.20 m. We must add one more zero to denote the uncertain part. Thus the measurement becomes: 4.200 m
■ Based on the two examples, we can write: All trailing zeros are significant
Case 2:
• In this case, the trailing zeros are not significant
Example:
    ♦ The underlined zeros in 87300 are not significant

Ambiguity in Rule 4 (case 2):
We will write about the ambiguity in steps:
1. Consider the measurement 4.700 m
• Based on Rule 4 (case 1), we know that the zeros after 7 are significant
• So the number of significant figures in 4.700 m is 4
2. Let us write this measurement in different units
• We have: 4.700 m = 470.0 cm = 4700 mm = 0.004700 km
3. Consider the third measurement: 4700 mm
    ♦ Based on Rule 4 (case 2), the number of significant figures will be 2
    ♦ When it is written as 4.700 m, the number of significant figures is 4
■ But the number of significant figures cannot change by mere 'changing of units'
4. To remove such ambiguities in determining the 'number of significant figures', the best way is to record every measurement in scientific notation
• The details about this notation can be written in 3 steps as follows:
(i) In this method, every number is expressed as a × 10^b
• Where a has to satisfy two conditions:
    ♦ Condition 1: a should be greater than or equal to 1
    ♦ Example: 1, 2.5, 3.48, 7, 8.0569, 9 etc.,
    ♦ Condition 2: a should be less than 10
    ♦ Example: 7, 8, 9.21, 9.999 etc.,
• The two conditions can be written together mathematically as: 1 ≤ a <10
(ii) b is the positive or negative power of 10
• If b is positive and large, the number itself will be very large
    ♦ Example: 3.79 × 10⁸
• If b is negative and large (numerically), the number itself will be very small  
    ♦ Example: 2.341 × 10^-15
(iii) It is often customary to put the decimal point after the first digit
■ If we follow the above 3 steps, the confusion can be avoided
5. Consider our example again. We have:
4.700 × 10^-3 km = 4.700 m = 4.700 × 10² cm = 4.700 × 10³ mm
• We see that, in all measurements, the base a = 4.700
• For the base, Rule 4 (case 1) can be easily applied
6. Scientific notation is ideal for reporting measurement. But if the measurement given to us is not in that notation, we can directly apply Rule 4 (case 1 OR case 2, which ever is applicable)
Rule 5:
• If a number is less than 1, we usually put a zero in front of the decimal point
• Such a zero is not significant
    ♦ Example: The under lined zero in 0.325 is not significant
Rule 6:
• We must not try to find the significant figures in 'multiplying or dividing factors'
• They are exact
• Their precision cannot be increased or decreased
• We can assume that, they have an infinite number of significant figures
Example: 
    ♦ Radius is obtained by dividing diameter by the factor 2
    ♦ Circumference of a circle is obtained by multiplying 𝞹r by a factor 2

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Now we will see some solved examples:

Solved example 2.12
Find the number of significant figures in the following measurements:
(a) 0.0002801 cm (b) 13.5 g (c) 4100 mL (d) 7.400 × 10³ m
Solution:
(a) 0.0002801 cm
(i) Rule 1 says that all non-zero digits are significant
• So 2, 8 and 1 are significant
(ii) The zero between 8 and 1 is a captive zero
• Rule 2 says that it is significant
(iii) There are 3 zeros just after the decimal point
• Rule 3 says that they are not significant
(iv) There is just a zero in front of the decimal point 
• Rule 5 says that it is not significant
(v) So the total number of significant figures is: 4
(b) 13.5 g
• Here we apply Rule 1
• All digits are non-zero digits
• So the number of significant figures = 3
(c) 4100 mL
• This measurement should have been given in scientific notation
• But the person who recorded it, does not give importance for presenting it in that notation
• So we can apply Rule 4 (case 2)
• We get: Number of significant figures = 2
(d) 7.400 × 10³ m
• This measurement is in scientific notation
• We need to consider the base only
• The base is 7.400
• So the number of significant figures = 4

Order of magnitude

We will explain this in steps:
1. We have seen the details about scientific notation
• If a number is given to us in that notation, we can get an approximate idea about the 'size' of that number by rounding off a
2. We know that a will be greater than or equal to 1
• If a is less than or equal to 5, it will become 1
3. We know that a will also be less than 10
• If a is greater than 5, it will become 10
4. When a is rounded off in this manner, b gets a new name: order of magnitude 
5. Let us see some examples:
Example 1:
• The diameter of the earth is 1.28 × 10⁷ m
• To get an approximate idea about the diameter, we round of 1.28
• Based on (2) above, when 1.28 is rounded off, we get 1
• After rounding off 1.28 to 1, we cannot say this:
The diameter of the earth is 10⁷ m
• We must say this:
Diameter of the earth is of the order of  10⁷ m with the order of magnitude 7
• 10⁷ is very large: 1 followed by seven zeroes
Example 2:
• The actual diameter of hydrogen atom is 1.06 × 10^-10 m 
• To get an approximate idea about the diameter, we round off 1.06
• Based on (2) above, when 1.06 is rounded off, we get 1
• After rounding off 1.06 to 1, we cannot say this:
The diameter of the hydrogen atom is 10^-10 m
• We must say this:
Diameter of the hydrogen atom is of the order of  10^-10 m with the order of magnitude -10

• 10^-10 is very small:
    ♦ There is only a zero in front of the decimal point
    ♦ Just after the decimal point, there will be 9 zeroes
    ♦ After those 9 zeroes, there will be just a '1' 
6. What is the difference between the following two items:
(i) Order of magnitude of diameter of hydrogen atom
(ii) Order of magnitude of diameter of earth
• Obviously, the difference is: [7 - (-10)] = 17
• Thus we can write:
Diameter of earth is 17 orders of magnitude larger than the diameter of hydrogen atom

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We have seen the basics about significant figures. In the next section, we will see the rules for arithmetic operations with significant figures

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