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:
■ 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
• 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
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
(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
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
(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
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
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 |
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
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
Three important points to remember:
• 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
Now we will see some solved examples
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
In the next section, we will see a few more solved examples on significant figures
No comments:
Post a Comment