Problems - Significant Figures

1. State the number of significant figures in each of the following:

(i) 208.91 (ii) 0.00456 (iii) 453 (iv) 2.945 x 10⁴ (v) 0.346

Solution

(i) 208.91 has five significant figures.

(ii) 0.00456 has three significant figures.

(iii) 453 has three significant figures.

(iv) 2.945 x 10⁴ has four significant figures.

(v) 0.346 has three significant figures.

2. Express 0.00000345 in scientific notation and calculate the number of significant figures.

Solution

0.00000345 can be expressed in scientific notation as 3.45 x 10⁵. The significant figures are three.

Calculations Involving Significant Figures

Rule I

When addition or subtraction is to be carried out in the numbers having different precisions, the final result should be reported up-to the same number of decimal places as are present in the term having the least number of decimal places.

• Addition of numbers

In the addition of 42.2 + 4.22 + 0.422, all the three numbers have three significant figures, but 42.2 has the least number of decimal place, namely one. The answer is therefore limited to one decimal place.

• Subtraction of numbers

The subtraction of numbers is carried out in the same way as addition. In the difference of 5.2748 and 5.2722 the answer (0.0026) has four decimal places. The numbers have five significant figures but the answer has only two significant figures.

Rule II

In the multiplication or division of numbers, the final result should be reported up to the same number of significant figures as are present in the term with the least number of significant figures.

Multiplication of numbers

In the multiplication of 5.1028 (five significant figures) with 1.30 (three significant figures) the value is 6.63364, but the correct answer is 6.63.

Division of numbers

In the division of 5.2765 (five significant figures) by 1.25 (three significant figures) the result is 4.2212 but the correct answer is 4.22.

Retention of Significant Figures - Rounding off Figures

The rounding off procedure is applied to retain the required number of significant figures.

• If the digit coming after the desired number of significant figures happens to be more than 5, the preceding significant figure is increased by one, 4.317 is rounded off to 4.32.

• If the digit involved is less than 5, it is neglected and the preceding significant figure remains unchanged, 4.312 is rounded off to 4.31.

• If the digit happens to be 5, the last mentioned or preceding significant figure is increased by one only in case it happens to be odd. In case of even figure, the preceding digit remains unchanged. 8.375 is rounded off to 8.38 while 8.365 is rounded off to 8.36.

The rounding off is done only in the final answer when the problem involves more than one step.

Problems

3. Express the results of the following calculations to the appropriate number of significant figures.

Solution

= 0.05608 = 0.0561

The number of significant figures in the term 3.24 is 3, therefore the result should have 3 significant figures. Therefore, the correct answer is 0.0561. The number after zero is 8 and therefore it is rounded off to one.

= 0.2615 x 10^-4 = 0.3 x 10^-4

The answer should have one significant figure because 0.5 has one significant figure.

4. Calculate to proper significant figures:

(i) 12.6 x 11.2 (ii) 172.8/15

Solution

(i) 12.6 x 11.2 = 141.12

Correct answer = 141 (up to three significant figures)

(ii) 172.8/15 = 11.52

Correct answer = 12 (up to two significant figures)

Dimensional Analysis

It is often necessary to convert one set of units to another. This can be done in a systematic way by the method called dimensional analysis or the conversion factor method. Conversion factor is equal to the ratio of the magnitudes of the same quantity in the two units. To obtain a conversion factor the magnitude of the physical quantity in the unit to be converted is placed in the denominator, while that in the desired unit is placed in the numerator.

For example, to convert 25.6 metres to centimetres, we know that

100 cm/1m is called the conversion factor because it converts a quantity expressed in one unit to a quantity expressed in another unit. Multiply the given quantity by the conversion factor and retain the units of the physical quantity and that of the conversion factor in such a way that all units cancel out leaving behind only the required units.

Similarly to convert 10 minutes to seconds, one proceeds as follows:

1min = 60 s

10 min = 10 min x 1

The advantage of this method is that if the equation is written correctly, all the units except the desired one get cancelled. If the conversion factor is written incorrectly, then the desired unit is not obtained.

To convert the units of a physical quantity involving many terms, it is convenient to place the units of each physical quantity directly in the steps of the calculations along with the number, cancel out the common units and finally convert the units by using proper conversion factor.

Problems

5. Express each of the following in SI units:

(1) 93 million miles (distance between earth and sun)

(2) 5 feet and 2 inches (height of an Indian female)

(3) 100 miles per hour (speed of Rajdhani Express)

(4) 14 pounds per square inch (atmospheric pressure)

(5) 0.4 Å (bond length of hydrogen molecule)

Solution

(1) 93 million miles = 93 x 10⁶ miles

1 mile = 1.60934 km = 1.60934 x 10³ m

= 1.5 x 10¹¹ m

(2) 5 feet and 2 inch = (5 x 12 + 2) inch = 62 inch

1 inch = 2.54 x 10^-2 m

then, 5 feet and 2 inch = 62 inch

1 mile = 1.60934 x 10³ m, 1 min 60 x 60 s

1 pound force = 4.448 N, 1 inch = 2.54 x 10^-2 m

= 9.66 x 10⁴ N m^-2

= 0.74 x 10^-10 m

7.4 x 10^-11 m

6. The density of ethanol is 0.79 g cm^-3 at 25^o C. What is its value in SI units?

Solution

1 cm³ of ethanol weighs 0.79 g or 0.79 x 10^-3 kg

1 m³ i.e., 10⁶ cm³ weighs = 0.79 x 10^-3 x 10⁶ kg

= 0.79 x 10³ kg

0.79 g cm^-3 = 0.79 x 10³ kg m^-3