What's inside this article?

No measurement is perfect. The uncertainty in measurement is the important factor in determining the accuracy and precision. You'll understand what the accuracy and precision mean if you continue to read this article.

And that's where comes the significant figures. The significant figures rules come in handy in significant figures calculation and you'll also see the examples of significant figures calculation in this article. The uncertainty in measurement is something that you'll always care in Physics measurements and significant figures calculation is always required.

Including everything, without scientific notation you can not think of doing calculation in Physics. The significant figures calculation is always required everywhere in Physics as well as in scientific notion.

## Accuracy and Precision

Accuracy and precision represent the same meaning in everyday language but there is a little bit difference between them in technical language.

For example, an old watch gives the value of time in hour, minute and second which is not the correct time, then the old watch is **precise** as that is able to give the value of even the second but a new watch gives the value of time in hour and minute not the second which is the correct time but in this case the watch is **accurate** not precise.

Accuracy always wants to be the nearest of the exact or true value but precision is a different thing as it has nothing to do with accuracy and focuses on how much deeper the measurement is.

## Uncertainty in Measurement

When you measure the length of your pencil with your hand ruler and found the answer to be $5.4\text{cm}$, your answer has an uncertainty of $\pm 0.1\text{cm}$. It means your answer lies somewhere between $5.3\text{cm}$ and $5.5\text{cm}$.

Because you can not be exactly certain that your answer is $5.4\text{cm}$, it can be plus $0.1\text{cm}$ or minus $0.1\text{cm}$ and that is called uncertainty in measurement. This uncertainty depends on how much deep your scale can measure.

In this case your measurement has an **uncertainty** of $\pm 0.1\text{cm}$ which is the maximum difference between the measured value and the actual value. You can not exactly say that your answer is $5.4\text{cm}$ due to how much accurate and precise your measuring device is.

The uncertainty depends on the measuring device you use and how much deeper or precisely the measuring device actually measures.

For example, if you measured the thickness of your note book using a more accurate and precise vernier calliper which can measure up to $0.01\text{cm}$ scale and found the value to be $4.34\text{cm}$, your value has an uncertainty of $\pm 0.01\text{cm}$; it means your value lies somewhere between $4.33\text{cm}$ and $4.35\text{cm}$.

Your value with your uncertainty can be written as $4.34\pm 0.01\text{cm}$. If your measuring device can measure up to $0.001\text{cm}$ scale, your uncertainty will be $\pm 0.001\text{cm}$. Greater the accuracy and precision of the measuring device, lesser the uncertainty.

## Significant Figures

The significant figures are digits that are significant or meaningful in a numerical value. For example, you know the distance between two cities to be $144\text{km}$ within the uncertainty of $\pm 1\text{km}$ then all the digits in $144$ are meaningful or significant because you are sure up to the final digit within an uncertainty of $\pm 1$ and there are three **significant figures**.

Consider $45892\pm 1$. The final digit has uncertainty of $\pm 1$. It means all the digits are meaningful and there are 5 significant figures. In $0.0034$ there are two significant figures because the zeros in the left hand side are only the place holders for the decimal point and they are not significant and this number obviously has the uncertainty of $\pm0.0001$.

Now what happens when you add or subtract the numbers having uncertainties? When you add two or more numbers having uncertainties the result will have the same uncertainty as in the number with fewest digits to the right of the decimal point.

For example $3.54+3.3=6.8$ not $6.84$. The number $3.54$ has the uncertainty of $\pm 0.01$ and that of $3.3$ is $\pm 0.1$. Now the result has the same uncertainty as there is in $3.3$ due to only one digit to the right of the decimal point. And the same rule applies for the subtraction as well, for example $4.43-2.6=1.8$ not $1.83$.

When you add or subtract numbers having uncertainties the computed number will also have the uncertainty equal to the uncertainty with the number having fewest digits to the right of the decimal point.

The general rule of the multiplication and division of the numbers with uncertainties generally in the lack of complete analysis of uncertainties is that the result always has the same number of significant figures as there is in the number with least number of significant figures.

For example, multiply $7.86$ and $3. 112$, the result of multiplication has the same number of significant figures as there is in the number with fewest number of significant figures. So the value of multiplication of $7.86$ and $3. 112$ is 24.5 not 24.53892.

When you multiply and divide the numbers having uncertainties the result has the same number of significant figures as in the number with fewest significant figures.

Example 1| Find the total number of significant figures in $\frac{(2.59)(3.2)}{2.69}$.

As we have discussed before the total number of significant figures in the result of multiplication and division is equal to the number of significant figures in the number with least number of significant figures in the calculation. So the answer is 3.1 not 3.081040892 as your calculator gives. And note that do not forget to round your final answer.

Example 2| Find the total number of significant figures in $5.3-3.21$.

As we have already known that for the addition and subtraction the location of decimal point matters not the number of significant figures, that is the result will have the same uncertainty as there is in the number with fewest digits to the right of decimal point. The answer is $2.1$ not $2.09$. You should not forget to round your answer.