# Physics Lesson 1.3.2 - How to find the number of significant figures in a given numerical value?

Welcome to our Physics lesson on How to find the number of significant figures in a given numerical value?, this is the second lesson of our suite of physics lessons covering the topic of Significant Figures and their Importance, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

## How to find the number of significant figures in a given numerical value?

There are five basic rules to find the correct number of significant figures in a given numerical value obtained during a measurement. They are:

### 1. The zeroes at the end of a whole number are not counted

For example, in the value 7800 kg the significant numbers are only 2 (7 and 8). This is because zeroes here act simply as placeholders; they are not meaningful digits. They are written in the measurement result only to show that the measurement value contains 78 hundreds of kilogram, not 78 kilograms or 7.8 kilograms.

### 2. Zeroes at the beginning of a decimal number are not counted

For example, if during a measurement the result of speed is 0.000945 km/s, the number of significant figures in this number is 3 (9, 4 and 5) because the other digits (zeroes) act again as placeholders. We can write 0.000945 km/s = 0.945 m/s. The last number indicates that we have a speed of 945 thousandths of a metre per second. However, we can also say "the speed is equal to 945 millionths of a kilometre per second" and in both cases we are correct. Therefore, only 9, 4 and 5 are meaningful in this value.

### 3. All zeroes between two non-zero digits are always counted

For example, we are allowed to write the measurement value 0.003705 quintals as 0.3705 kilograms or 370.5 grams but we cannot write it as 0.00375 quintals, 0.375 kilograms or 375 grams because these values are not equal to the original one.

### 4. All zeroes at the end of a decimal are always counted

It is because zeroes at the end of a decimal indicate a higher degree of accuracy compared to cases in which we don't write them at all. For example, in physics the value 3.200 m is not the same as 3.2 m because the first shows that the measurement is taken using a device divided into mm while the other number (3.2 m) shows that the measurement is less accurate. Either it has been measured using an apparatus divided into dm or the result is rounded up to the nearest dm after the measurement was taken.

### 5. If the measurement value is written in standard form (scientific notation), only the numerical part before the powers of ten is counted. The exponential part is not considered

For example, if the result of the speed measurement taken during the observation of a meteorite shows the value 3.507 × 104 m/s, the number of significant figures in this result is 4 (3, 5, 0 and 7). This is because this result can also be written as 35 070 m/s and from the first rule of significant figures, it is obvious the zero at the end is not counted. It simply acts as a placeholder. On the other hand, all the other digits are counted because if you remove one of them, the result is not the same as the original anymore.

### Example 2

The dimensions of a room are 5.4 m × 4.27 m. Calculate the surface area of the room by expressing the result in the correct number of significant figures.

### Solution 2

The room sides are measured with different levels of accuracy. The smallest division of the apparatus when measuring the length was 0.1m (1 dm), while for the width it was 0.01 m (1 cm). Therefore, we must use the level of accuracy of the less accurate measurement to be sure. It is better rounding up a dimension to a less accurate value and be sure your result is close to the correct value than working in vain with numbers that probably are not correct. In other words, it is better to write the width as 4.3 m (by rounding up the value to the nearest tenth of metre) instead of 4.30 m. This is because we are not sure whether its exact value was 4.25 m, 4.26 m or 4.34 m as all these numbers become 4.3 m when rounded to the nearest tenth of metre.

Therefore, the surface area of the room is

Area = Length × Width = 5.7m × 4.3m = 24.51m2

The accuracy of the result is 0.01 m2 or 1dm2. Giving that length and width are expressed in tenths of metre, this is a value that complies with the desired order of accuracy because 1/10 × 1/10 = 1/100 = 0.01. Therefore, there are 4 significant figures in the result. The first two figures (2 and 4) show its whole part (square metres) and the other two figures (5 and 1) represent the tenths and hundredths of square metre respectively.

## More Significant Figures and their Importance Lessons and Learning Resources

Units and Measurements Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
1.3Significant Figures and their Importance
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
1.3.1What are significant figures?
1.3.2How to find the number of significant figures in a given numerical value?
1.3.3How to round up a number to the required number of significant figures?
1.3.4Why are significant figures so important in measurements?

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