The result of computation with approximate numbers, which contain more than one uncertain digit, should be rounded off. The rules for rounding off numbers to the appropriate significant figures are obvious in most cases. A number 2.746 rounded off to three significant figures is 2.75, while the number 2.743 would be 2.74.
The rule by convention is that the preceding digit is raised by 1 if the insignificant digit to be dropped (the underlined digit in this case) is more than 5, and is left unchanged if the latter is less than 5. But what if the number is 2.745 in which the insignificant digit is 5. Here, the convention is that if the preceding digit is even, the insignificant digit is simply dropped and, if it is odd, the preceding digit is raised by 1.
Then, the number 2.745 rounded off to three significant figures becomes 2.74. On the other hand, the number 2.735 rounded off to three significant figures becomes 2.74 since the preceding digit is odd.
In any involved or complex multi-step calculation, you should retain, in intermediate steps, one digit more than the significant digits and round off to proper significant figures at the end of the calculation. Similarly, a number known to be within many significant figures, such as in 2.99792458 × $10^8$ m/s for the speed of light in vacuum, is rounded off to an approximate value 3 × $10^8$ m/s, which is often employed in computations.
Finally, remember that exact numbers that appear in formulae like $T=2\pi \sqrt{\frac{L}{g}}$ , have a large (infinite) number of significant figures. The value of π = 3.1415926.... is known to a large number of significant figures. You may take the value as 3.142 or 3.14 for π, with limited number of significant figures as required in specific cases.
Example 1
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?
Answer
The number of significant figures in the measured length is 4. The calculated area and the volume should therefore be rounded off to 4 significant figures.
Surface area of the cube = 6(7.203)$^2 \ m^2$
= 311.299254 $m^2$ = 311.3 $m^2$
Volume of the cube = (7.203)$^3 \ m^3$ = 373.714754 $m^3$ = 373.7 $m^3$
Example 2
5.74 g of a substance occupies 1.2 $cm^3$. Express its density by keeping the significant figures in view.
Answer
There are 3 significant figures in the measured mass whereas there are only 2 significant figures in the measured volume. Hence the density should be expressed to only 2 significant figures.
Density = $\frac{5.74}{1.2} \ g.cm^{-3}$
= 4.8 $g.cm^{-3}$
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