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Precision of measuring tools and significant figures

An important factor in the precision of measurements involves the precision of the measuring tool. In general, a precise measuring tool is one that can measure values in very small increments. For example, a standard ruler can measure length to the nearest millimeter whereas a caliper can measure length to the nearest 0.01 mm. The caliper is a more precise measuring tool because it can measure extremely small differences in length. The more precise the measuring tool, the more precise the measurements.

When we express measured values, we can only list as many digits as we measured initially with our measuring tool. For example, if we use a standard ruler to measure the length of a stick, we may measure it to be 36.7 cm. We can’t express this value as 36.71 cm because our measuring tool is not precise enough to measure a hundredth of a centimeter. It should be noted that the last digit in a measured value has been estimated in some way by the person performing the measurement. For example, the person measuring the length of a stick with a ruler notices the stick length seems to be somewhere in between 36.6 cm and 36.7 cm, and he or she must estimate the value of the last digit. Using the method of significant figures    , the rule is that the last digit written down in a measurement is the first digit with some uncertainty . To determine the number of significant digits in a value, start with the first measured value at the left and count the number of digits through the last digit written on the right. For example, the measured value 36.7 cm has three digits, or three significant figures. Significant figures indicate the precision of the measuring tool used to measure a value.

Zeros

Special consideration is given to zeros when counting significant figures. The zeros in 0.053 are not significant because they are placeholders that locate the decimal point. There are two significant figures in 0.053. The zeros in 10.053 are not placeholders; they are significant. This number has five significant figures. The zeros in 1300 may or may not be significant, depending on the style of writing numbers. They could mean the number is known to the last digit or they could be placeholders. So 1300 could have two, three, or four significant figures. To avoid this ambiguity, we should write 1300 in scientific notation as 1.3 × 10 3 , 1.30 × 10 3 , or 1.300 × 10 3 , depending on whether it has two, three, or four significant figures. Zeros are significant except when they serve only as placeholders .

Significant figures in calculations

When combining measurements with different degrees of precision, the number of significant digits in the final answer can be no greater than the number of significant digits in the least-precise measured value . There are two different rules, one for multiplication and division and the other for addition and subtraction.

  1. For multiplication and division, the result should have the same number of significant figures as the quantity with the least number of significant figures entering into the calculation . For example, the area of a circle can be calculated from its radius using A = πr 2 . Let’s see how many significant figures the area has if the radius has only two—say, r = 1.2 m. Using a calculator with an eight-digit output, we would calculate
    A = π r 2 = ( 3.1415927 ) × ( 1.2 m ) 2 = 4.5238934 m 2 .

    But because the radius has only two significant figures, it limits the calculated quantity to two significant figures, or
    A = 4.5 m 2 ,

    although π is good to at least eight digits.
  2. For addition and subtraction, the answer can contain no more decimal places than the least-precise measurement. Suppose we buy 7.56 kg of potatoes in a grocery store as measured with a scale with precision 0.01 kg, then we drop off 6.052 kg of potatoes at your laboratory as measured by a scale with precision 0.001 kg. Then, we go home and add 13.7 kg of potatoes as measured by a bathroom scale with precision 0.1 kg. How many kilograms of potatoes do we now have and how many significant figures are appropriate in the answer? The mass is found by simple addition and subtraction:
    7.56 kg 6.052 kg + 13.7 kg 15.208 kg = 15.2 kg .

    Next, we identify the least-precise measurement: 13.7 kg. This measurement is expressed to the 0.1 decimal place, so our final answer must also be expressed to the 0.1 decimal place. Thus, the answer is rounded to the tenths place, giving us 15.2 kg.

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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