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# Significant Figures

In Physics, measurement plays an important role. The smallest magnitude of a quantity that can be measured by an instrument is called least count of that instrument (For eg: the least count of metre scale is 0.1 cm). Since there is a least count associated with each instrument, the results of measurements made with it, always have limited accuracy.

Let a quantity measured by an instrument be given by 87.2. The quantity is written up to the first decimal place. This indicates that the least count of the measuring instrument is 0.1. The digits 8 and 7 are reliable and certain while the last digit 2 in the given number (87.2) is uncertain by an amount ± 1. So, the given number has three significant figures. Thus, the significant figures are the number of digits required to report the result of an experiment or a calculation accurately. It is that number of digits in a quantity that is known reliably, plus one that is uncertain. As the number of significant figures increases, the accuracy also increases.

For example, if the period of oscillation of a simple pendulum is measured as 1.45 seconds, the digits 1 and 4 are reliable and certain, while the digit 5 is uncertain. Thus, the measured value has three significant figures.

Consider an iron rod whose length is measured as 5.6 cm using a metre scale. The value is observed as 5.61 cm when measured with a vernier caliper. In the first case, the length is measured with an accuracy of 1/10th of a centimetre while in the second case it is 1/100th of a centimetre. The number of significant figures is 2 with the metre scale while it is 3 that with the vernier calipers. Hence the most accurate value of the length is 5.61 cm with significant figures.

If the same measured quantity is represented in other units, there is no change for the significant figures. i.e., 5.61 cm = 5.61 × 10-2 cm. Here also the number of significant figures is 3.

## Rules for determining significant figures

The rules for determining the number of significant figures are:

1. All non zero digits are significant.
Example: 326.84 has five significant figures.
2. All zeroes occurring in between two non zero digits are significant.
Example: 804.003 has six significant figures.
3. All the zeros after decimal but before a non zero digit are not significant.
Example: 0.00942 has only three significant figures. Here the three zeros are not significant.
4. All the zeros right side of the decimal and right side of a non zero digit are significant.
Example: 0.35000 has five significant figures.
5. If the given number does not contain a decimal point, the final zeroes are ambiguous and are not significant.
Example: 346000 has only three significant figures.
In such cases the number is written as a power of 10, putting the decimal point after the first digit, i.e.,  346000 = 3.46 × 105 .
But if the number obtained is on the basis of an actual measurement, all zeroes to the right of the last non zero digit are significant.
Example: The distance between two places is 4030 metre, then 4030 m contains 4 significant figures.
6. If there is a decimal at the end of the whole number, all zeros at the right extreme just before the decimal are significant.
Example: 21300. has five significant figures.
7. If the number has an integral part and a decimal part, all zeros in the number are significant.
Example: 28.20 has four significant figures.

Below table gives you significant figures for different numbers.

 Numbers Significant Figures 2456 4 24.56 4 24.05 4 0.2456 4 0.02456 4 0.24560 5 0.024560 5 0.035 2 0.007 1 2456.0 5 2456.20 6 6300 2 63.00 4 6.300 4 6.320 4 6.032 4 0.00006032 4 4.86 ×105 3 4.860 × 105 4 4.8 × 10-4 2 0.4 × 10-4 1 2.64 × 1025 3

Table 1: Different numbers and their significant figures

## Rules for rounding significant figures

Rounding off is made in correcting a physical quantity with least variation from its original value after dropping insignificant figures.

The rules for rounding off numbers to the appropriate significant figures are given below:

1. If the digit after the last significant digit is greater than 5, then the last significant digit is raised by 1.
2. If the digit after the last significant digit is less than 5, then the last significant digit is left unchanged.
3. If the digit after the last significant digit is equal to 5, then the last significant digit is not changed if it is even and is raised by 1, if it is odd.

For example, the number 8.26 rounded off to two significant figures is 8.3, while the number 8.24 would be 8.2.
Similarly, the number 8.25 rounded off to two significant figures is 8.2, while on the other hand the number 8.35 rounded off to two significant figures becomes 8.4 since the preceding digit is odd.

For any complex multi-step calculation, one digit more than the significant digits should retain in intermediate steps, and round off to proper significant figures at the end of the calculation.

Also, learn the order of magnitude of a physical quantity.

## Rules for Arithmetic operations with Significant figures

When an experiment Is performed, a number of observations are made and the result is obtained by computing (adding, subtracting, multiplying, dividing) different data. The computed result cannot be more accurate than the original measurement which has the fewest significant figures.

### Rules for Multiplication and Division with significant figures

In multiplication and division, the computed result should retain the significant digits, equal to those present in the least significant number involved in the calculation.

Examples:

• 63 × 3.041 = 191.583
Since 63 has the least significant figures(i.e., 2) among the quantities multiplied, the final result must be rounded to 2 significant figures. Also, the third significant figure in the answer is less than 5(i.e., 1), so the third figure can be dropped. So the answer after rounding off is 1.9 X 102
• 4800 ÷ 10.5 = 457.1428
Since 4800 has the least significant figures(i.e., 2) among the quantities divided, the final result must be rounded to 2 significant figures. Also, the third significant figure in the answer is greater than 5(i.e., 7), so 1 should be added to the last digit retained. So the answer after rounding off is 4.6 X 102

### Rules for Addition and Subtraction with significant figures

In addition or subtraction of given numbers, the same number of decimal places is retained in the result as are present in the number with the least number of decimal places.

Examples:

• 84.426 + 14.4 = 98.826
Since 14.4 has least decimal places(i.e., 1) among the quantities added, the final result must be rounded to 1 decimal place. Also, the fourth significant figure in the answer is less than 5(i.e., 2), so the fourth figure can be dropped. So the answer after rounding off is 98.8
• 15.32 + 62.6184 = 77.9384
Since 15.32 has least decimal places(i.e., 2) among the quantities added, the final result must be rounded to 2 decimal places. Also, the fifth significant figure in the answer is greater than 5(i.e., 8), so 1 should be added to the last digit retained. So the answer after rounding off is 77.94
• 21.2 + 33.58 + 17.5864 = 72.3664
Since 21.2 has least decimal places(i.e., 1) among the quantities added, the final result must be rounded to 1 decimal place. Also, the fourth significant figure in the answer is greater than 5(i.e., 6), so 1 should be added to the last digit retained. So the answer after rounding off is 72.4
• 7.735 – 2.4837 = 5.2513
Since 7.735 has least decimal places(i.e., 3) among the quantities added, the final result must be rounded to 3 decimal places. Also, the fifth significant figure in the answer is less than 5(i.e., 3), so the fifth figure can be dropped. So the answer after rounding off is 5.251
• 5.4 ×10-4 – 2.2 × 10-5 =  5.4 ×10-4 – 0.22 × 10-4  =  5.18 × 10-4
Since 5.4 × 10-4 has least decimal places (ie., 1) among the quantities added, the final result must be rounded to 1 decimal place. Also, the third significant figure in the answer is greater than 5(i.e., 8), so 1 should be added to the last digit retained. So the answer after rounding off is 5.2 × 10-4

NOTE:
• 326.32 + 127.2 + 0.301 =  453.821
Here, 127.2 has the least number of decimal places(i.e., 2) among the quantities added and therefore the final result must be rounded to 1 decimal place. So the answer after rounding off is 453.8. Note that we should not use the rule applicable for multiplication and division which is based on significant digits. For addition and subtraction, the rule is in terms of decimal places.

## Significant figures solved examples

1. The result of an experimental calculation corrected up to seven significant figures is 7.363573. Round it off to six, five, four, three and two significant figures.

The given quantity P = 7.363573
When rounding off to 6 significant figures, P6 = 7.36357 ( the dropped digit is less than 5, 6th significant figure remains the same).
When rounding off to 5 significant figures, P5 = 7.3636 (5th significant figure is added by 1, as the dropped digit is greater than 5).
Similarly, When rounding off to 4 significant figures, P4 = 7.364
When rounding off to 3 significant figures, P3 = 7.36
When rounding off to 2 significant figures, P2= 7.4

2. The length, breadth and thickness of a rectangular sheet of metal are 2.324m, 2.005m and 1.01 cm respectively. Find the surface area and volume of the sheet to the correct significant figures.

Solution:

Given, Length, l      = 2.324 m
thickness, t             = 1.01 cm = 0.0101 m LARGE begin{aligned} Total;area& =2(ltimes b+ltimes t+btimes t)\ &= 2(2.324times 2.005+2.324times 0.0101+2.005times 0.0101)\ &= 9.4066858;m^{2} end{aligned}

As thickness has least number of significant figures (i.e., 3 ), the area should be rounded off to three significant figures. Also, the fourth significant figure in the answer is greater than 5(i.e., 6), so 1 should be added to the last digit retained.
Area corrected to significant figres = 9.41 m2 large begin{aligned} Similarly,;Volume& =ltimes btimes t\ &= 2.324times 2.005times 0.0101\ &= 0.047062162;m^{3} end{aligned}
Similarly, the volume should be rounded off to three significant figures. Also, the fourth significant figure in the answer is greater than 5(i.e., 6), so 1 should be added to the last digit retained.
Volume corrected to significant figures  = 0.0471 m3

3. 6.84 g of a substance occupies 1.3cc. Express its density by keeping the significant figures in view.

Ans:

Given, Mass, m     = 6.84 g
Volume, v              = 1.3 cc large begin{aligned} Density,; rho & =frac{Mass}{Volume} \ &= frac{6.84}{1.3}\ &= 5.26153846;g/cm^{3} end{aligned}
Since the volume has minimum number of significant figures(i.e., 2), the result must be rounded off to two significant figures. Also, the third significant figure in the answer is greater than 5(i.e., 6), so 1 should be added to the last digit retained.
Density corrected to significant figures  = 5.3 g/cc

4. The mass of a box measured by a grocer’s balance is 3.1 kg. Two gold piece of masses 22.35 g and 22.39 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 the correct significant figures?

Solution:

Given, mass of box, m1      = 3.1 kg
Mass of gold piece 1, m = 22.35 g = 0.02235 kg
Mass of gold piece 2 , m3 = 22.39 g = 0.02239 kg
(a) Total mass of the box  = m1 + m2 + m3
= 3.1 + 0.02235 + 0.02239
= 3.14474 kg
During addition, the final result should retain same number of decimal places as are present in the number with least decimal places. Here mass m1 has least decimal places(i.e., 1) among the quantities added and therfore the final result must be rounded to 1 decimal place. Also, the third significant figure in the answer is less than 5(i.e., 4), so the third figure can be dropped.
the total mass of the box after correcting to significant figures = 3.1 kg.

(b) Difference in the masses of gold pieces = 22.39 – 22.35
= 0.04 g
Similarly during subtraction, the final result should retain same number of decimal places as are present in the number with least decimal places. Here both the masses have two decimal places and therfore the final result must be rounded to two decimal places. here, the result is already having only two decimal places and therefore difference in masses of gold pieces corrected to significant figures remains the same as 0.04 g.

5. The radius and length of thin wire are measured to be 0.54mm and 33.5 cm respectively. Find the volume of the wire up to the appropriate number of significant figures.

Ans:

Given, length of the wire, L   = 33.5 cm
Radius of wire, r                    = 0.54 mm = 0.054 cm

Volume of the wire ,          V  = πr2L
= 3.14 × 0.0542 × 33.5
= 0.306734 cm3

Since the radius of wire has minimum number of significant figures(i.e., 2), the result must be rounded off to two significant figures. Also, the third significant figure in the answer is greater than 5(i.e., 6), so 1 should be added to the last digit retained.
Volume of the wire corrected to significant figures  = 0.31 cm3

6. Each side of a cube is measured to be 8.405 m. Calculate the total surface area and the volume of the cube to appropriate significant figures.

Ans:

Given, length of a side of the cube, a = 8.405 m
The number of significant figures in the measured length is four. Therefore the calculated area and volume should be rounded off to four significant figures.
Total surface area of the cube   = 6a2
= 6 × 8.4052
= 423.86415 m2
i.e., Total surface area of the cube with appropriate significant figures = 423.9 m2

Volume of the cube   = a3
= 8.4053
= 593.76303 m3
i.e., Volume of the cube with appropriate significant figures = 593.8 m3

The significant figures and their different rules are discussed in this article.