CHAPTER TWO: MEASUREMENT – PHYSICS S1-S2 NOTES

2.1  Measurement:  –is essential in physics. Before any measurement is taken, the quantity
to be measured and its unit must be specified.

(a)  Physical Quantities of Matter

These are the measurable properties of matter E.g. length, mass, speed etc. They are expressed in terms of a numerical value and a unit. There are so many quantities that can be measured in physics. However, some of the quantities are defined by using basic quantities.

(b)  Basic Quantities

These are quantities of matter that are used to define other quantities of matter. The basic quantities of matter
are:  – Length,

• Mass and
• Time.

Since these quantities are used to define other units, they are referred to as fundamental
quantities.

1. S.I system of Units
   

The S.I system of units is an International System of units based on the MKS (metre-kilogram-second) system. It is abbreviated as SI in all languages and is derived from French Le Système International d’Unités.

The basic quantities and their S.I units are shown in the table 2.1 below.

Table  2.1

NB:  When writing the names or symbols of the units, note that:

– Symbols of units have no plural forms.

e.g. we write: 2 kg  not  2 kgs

5 m  not  5 ms

(d)  Prefixes

A prefix is a word or letter placed before another. Examples of prefixes are:

micro (m),  milli (m),  centi (c),  deci (d),  kilo (k),  mega (M), giga (G) and tera (T).

Prefixes can be used with the units for measuring quantities.

E.g.  kilo with gram  forming  kilogram (kg),

kilo with meter  forming  kilometer (km),

milli with meter  forming  millimeter (mm),

kilo with byte forming  kilobyte (kB),

mega with byte  forming  megabyte (MB),

giga with byte forming  gigabyte (GB) etc.

(e)  Multiples and Sub-multiples

Multiples and sub-multiples are shown in tables 2.2 and table 2.3 below.

Table 2.2 Multiples   Table 2. 3 Sub-multiples

2.11  Instruments used for measurement

The devices used for measuring quantities are called instruments. Those used to measure the basic quantities are given in the table 2.4 below.

Table 2.4  Instruments for measuring the basic quantities of matter

2.12  Measurement of Length

Length is measured by using the
Tape measure, Metre
rule, Vernier calipers and the Micrometer screw gauge.

The choice of the instrument to be used depends on:

• The size of the distance to be measured (how long and short the distance is) and
• The accuracy to which the measurement is needed.

(i)  Measurement of Long distance

Lang distances are measured using, the metre rule and the tape measure. Both instruments are calibrated (graduated) in metres (m), centimeters (cm) and millimeters (mm).

1 m  = 100 cm and 1 cm  = 10 mm

 \  1 m  = 100 x 10 = 1 000 mm

The larger unit is kilometer (km).  1 km  = 1 000 m

Note:  Due to wear at the edges, when using the metre rule, it is advisable to start measuring from the 10 cm mark and deducting 10 cm from the total measurement.

(ii)  Measurement of Small and very small distances

For small distances, the Vernier calipers or the engineers calipers is used. While for very small distances such as the diameter of copper wire, thickness of paper, the micrometer screw gauge is used. Both the Vernier calipers and the micrometer screw gauge give readings with reasonable accuracy.

1. The Vernier Caliper
   

   A Caliper is a mechanical device used to determine small lengths with reasonable accuracy.
   

Types of Vernier Callipers

There are two types of Vernier Callipers

1. Simple caliper (commonly called Engineer’s caliper) and
2. Complex vernier

(i)  Simple caliper

Simple calipers have two movable parts/legs of some desired shape and length to meet the surfaces whose separation is to be measured. The adjusted width between the leg tips is then placed against some length scale. E.g metre rule and the reading taken.

Figure 2.1  Engineer’s caliper  Figure 2.2 The mechanical Vernier calipers

For measuring the internal diameter of a pipe, the caliper is turned the other way round. This way, the tips of the jaws point outward.

(ii)  Complex vernier

The more complex vernier caliper has two types of scale that allow direct reading of the adjusted width between the jaws. The types of the complex vernier are mechanical and digital. Figure 2.2 below shows the mechanical vernier.

Figure 2.3  Showing a digital vernier being used to measure length of a bird

How to use the Vernier calipers

The movable jaw is adjusted until it grips the object to be measured and then the reading is taken as described below.

How to read the Vernier

(i)  The Digital Vernier calipers

Once the vernier is adjusted to the width of the object to be measured, a reading is taken directly from a small screen engraved on it.

 E.g. in figure 2.3, the length of the bird’s bick is 2.20 cm.

(ii)  The Mechanical Vernie

 For the mechanical vernier, the reading is taken in three steps:

Step I  Read and record the main scale reading at the zero mark of the vernier scale to an accuracy of one millimeter.

E.g.  2.1 cm.

Step II  Read and record the vernier scale reading at the position on the vernier where a mark on it is coincident (i.e. coincides) with a mark (division) on the main scale in tenths of millimeters.

E.g. let the 6^th vernier division coincide with a mark on the main scale. In tenths, 6 becomes  = 0.6 mm  = 0.06 cm

Step III  Get the sum of the two readings (i.e add the main scale reading and the vernier scale reading to get the total reading).

Main scale reading = 2.10 cm

Vernier scale reading = + 0.06 cm

Total reading = 2.16 cm

Example 1

The S.1 students of TLA measured the thickness of a desk top during physics lesson and found that the main scale reading before the zero mark of the vernier scale to be 4.4 cm. Find the thickness of the desk top if the 4^th vernier mark coincides with one of the marks on the main scale.

Solution  Main scale reading = 4.40 cm

Vernier scale reading 4^th =  = 0.04  cm  

Total reading  = 4.44 cm

Note:  Instead of dividing the vernier reading by 10 to get the answer in mm and then by 10 to change to cm, we can divide the value directly by 100 to get the answer once in cm.

Example 2

Find the readings of the verniers shown figures 2.4 (i) and (ii) below.

Solution:  (i)  Main scale reading = 5.20 cm

Vernier scale reading 4^th =  = + 0.04 cm

Total reading  = 5.24 cm

 (ii)  Main scale reading = 0.80 cm

Vernier scale reading 8^th =  = + 0.08 cm

Total reading  = 0.88 cm

1. The Micrometer screw gauge
   

The Micrometer screw gauge is an instrument used to measure very small distances.

The Micrometer screw gauge is calibrated in mm on the sleeve and some small divisions on the thimble scale.

There are two types of thimble readings:

1. One with 50 divisions on the thimble scale and
2. The other with 100 divisions on the thimble scale.

However, the two types give the same reading when used to measure the same distance.

The diagram of the micrometer screw gauge is shown in the figure 2.5 below.

How to use the Micrometer screw gauge

• Place the object whose thickness is to be measured in between the jaws (the anvil and the spindle) of the micrometer.
• Rotate the ratchet clockwise until the jaws touch the object. As soon as the object is gripped tight enough, it starts to slip, making a characteristic sound.
• Take the reading in three steps as shown below.

  Step I:  Read and record the reading on the sleeve scale at the edge of the thimble in millimeters and half millimeters. E.g. 4.0 mm.

   

  Step II:  Read and record the reading on the thimble scale opposite to the centerline on the sleeve scale (i.e. where a division on the thimble scale coincides with the centerline on the sleeve scale) in hundredths of millimeters.

  E.g.  Let the 33^rd division coincide with the centerline.

In hundredths, 33 becomes    = 0.33 mm

Step III  Get the sum of the two readings i.e. sleeve scale reading and the thimble scale reading.

Sleeve scale reading = 4.00 mm

Thimble scale reading = + 0.33 mm

Total reading =
4.33 mm

Note:  If the answer is required in cm or m, you convert it as required.

Example 1

Find the reading on the micrometer screw gauge shown in the diagrams below.

Solution  (a)  Sleeve scale reading = 3.00 mm

Thimble scale reading,  = + 0.29 mm

Total reading =
3.29 mm

(b)  Sleeve scale reading = 7.50 mm

Thimble scale reading  ,  = + 0.23 mm

Total reading =
7.73 mm

SELF-CHECK 2.0

1.  (a)  Draw and label the diagram of the Vernier calipers.

(b)  Find the thickness of a text book measured using a vernier caliper if the main scale reading is 24 mm and the 8^th vernier mark coincides with one of the marks on the main scale.

(c)  Find the reading on the verniers shown in the diagrams below.

2.  (a)  Draw and label the diagram of the micrometer screw gauge.

(b)  Find the reading on the micrometer screw gauge shown in figures 2.10 below.

2.13  Measurement of Area (Two dimensions)

Regular Surfaces

The area of regular surfaces is found by measuring any two of the following dimensions and then applying the appropriate formula.

– Length (l), width (w), height (h), side (s), radius (r) and diameter

Units of area

The SI unit for area is square metre (m²).

Other units are:  – mm²,  cm²,  km² and hector.

1 m²  = 100 x 100  = 10,000 cm²

1 cm²  = 10 x 10  = 100 m m²

1 m²  = 1000 x 1000  = 1,000,000 mm²

Table 2.5 below shows the common regular surfaces and their respective formulae.

1. Measurement of Volume (Three dimensions)
   

(i)  Regular Solids

The volume of regular solids is determined by measuring the dimensions and then applying an appropriate formula.

Units of Volume

The SI unit of volume is metre cubed, (m³).

Other units are:  mm³, cm³.

1 m³ = 1 000 000 cm³

 1 cm³ = 0.000 001 m³

The table below shows the common regular solids and their respective formulae.

Table  2.6

(ii)  Irregular Solids

The volume of irregular solids is determined by using displacement method. In this method, the solid is fully or wholly immersed in a liquid and the volume of the liquid displaced is measured. This method operates on the principle that “A body fully or wholly immersed in a fluid (liquid) displaces its own volume“.

Apparatus/requirements used to measure the volume of irregular solid are:

 (i) Measuring cylinder, water and a piece of thin silk thread.

 (ii) Measuring cylinder, overflow (displacement) can, water and a piece of silk thread.

Measuring the volume of irregular solid

1.  Using a measuring cylinder

• Fill a measuring cylinder with water.
• Read and record the initial reading.
• Tie the irregular object with a piece of thin silk thread and lower it carefully into the water in the cylinder until it is fully immersed.
• Shake it gently to remove any air bubbles.
• Read and record the final reading.

Diagram showing measurement of volume of irregular object

Before immersing After immersing

The volume is calculated from the formula:

Volume of irregular object  = Final reading – Initial reading

2.  Using an Overflow can and measuring cylinder

• Fill an overflow can with water until water flows out through the spout and wait until the water ceases dripping and then place a dry measuring cylinder below the spout.
• Tie the irregular object with a piece of thin silk thread and lower it carefully into the water in the overflow can until it is fully immersed.
• Shake it gently to remove any air bubbles.
• Wait until the water ceases dripping into the measuring cylinder.
• Read and record the volume of the displaced water in the cylinder.

Diagram showing measurement of volume of irregular object

Volume of object  = Volume of displaced water

(iii)  Measurement of volume of Liquids

The volume of a given liquid is determined by using a measuring cylinder.

The liquid is carefully poured into the measuring cylinder and the volume is read off by placing the eye level in line with the bottom of the meniscus of the liquid surface in order to avoid error due to parallax.

Diagram showing correct reading of measuring cylinder

Specified volumes of liquids are accurately measured by using specific instruments such as:  – Burette, Pipette and Syringe.

Note:  The first two are commonly used in Volumetric analysis in Chemistry practical.

(iv)  Gases

The volume of gases is measured by using a syringe. The gas syringe is connected to the gas supply with a help of rubber tubing. Due to the gas pressure the piston moves backwards and the required volume is noted from the scale.

Unit for measuring Volume

The SI unit for volume is metre cubed (m³)

1 m³  = 1 000 000 cm³

1 dm³ = 1 000 cm³

1 litre  = 1 dm³

 \  1 litre = 1 000 cm³

2.2  Measurement of Mass

Mass is a measure of the amount of matter in an object. Or it is the quantity of matter in a body. It is constant everywhere. That is, it does not vary from place.

The mass of an object is commonly measured by comparing it with a standard (known) mass.

The instruments used to measure mass are:

• Beam balance,
• Triple balance,
• Lever arm,
• Electric balance,
• Chemical balance and
• Spring balance.

The unit for measuring mass

The SI unit for mass is the kilogram (kg).

Other units include:  milligrams (mg), grams (g) and tonne (ton).

1 kg = 1 000 g

1 g  = 1 000 mg

1 ton  = 1 000 kg

2.3  Measurement of Time

Time is measured using:  watches and stopwatches (both mechanical and digital).

Due to accuracy and easy in reading, the digital stopwatches are preferred. They measure to 0.01 second.

Units for measuring time

The SI unit for time is second (s).

Other metric units include:

• Minute (min), hour (hr), day, week, month, year, decade, century, millennium.

  1 hr  = 60 min

  1 min  = 60 s

  1 hr  = 60 x 60 = 3 600 s

\ 1 hr  = 3 600 s

Worked Examples

Conversion from one unit to another

1.  Convert the following as required.

(a)  (i)  5 000 m to km.  (ii)  60 m to cm

(b)  (i)  2 kg to gram. (ii)  500 g to kg.

(c)  (i)  24 hr to seconds.  (ii)  30minutes to seconds.

 (iii)  1800 s to hours.  (iv)  900 s to minutes.

Solution

(a)  (i)  1 km  = 1000 m (ii)   1 m  = 100 cm

y  = 5000 m  60 m  = r cm

1000 m x y  = 1 km x 5000 m 1 m x r  = 60 m x 100 cm  

y  = r  =

\ y  = 5 km \ r  = 6000 cm

(b)  (i) 1 kg  = 1000 g (ii)   1 kg  = 1000 g

2 kg  = t    z  = 500 g

1 kg x t  = 2 kg x 1000 g    1000 g x z  = 1 kg x 500 g

t  =       z  =

\
t  = 2000 g \z  = 0.5 Kg

(c)  (i)   1 hr  = (60 x 60) s (ii)  1 min  = 60 s

24 hr  = t 30 min  = v

  1 hr x t  = 24 hr x (60 x 60) s  1 min x v  = 30 min x 60 s

t = v  =

\t  = 86 400 s   \v  = 1 800 s

(iii) 1 hr  = 3600 s (iv)  1 min  = 60 s

  h  = 1800 s q  = 900 s

  3600 s x h  = 1 hr x 1800 s   60 s x q  = 1 min x 900 s

  h  =
q  =

\ h  = ½ hr \ q  = 15 min

Self-Check 2.1

1.  Convert the following as required.

(a)  (i)  10, 000 m to km.  (ii)  20 m to cm  

(b)  (i)  25 kg to gram. (ii)  2000 g to kg.

(c)  (i)  12 hr to seconds.  (ii)  ¼ hr to seconds.

(d)  (i)  20 000 cm³ to m³  (ii)  50 m³ to cm³

2.4  Significant Figures

Measurements are always given correct to a certain number of significant figures.

To determine the number of significant figures, in a measurement the following rules may be useful.

1. All non-zero digits (1, 2, 3, 4, 5, 6, 7, 8, 9) are significant.

   E.g.  5.23cm  has 3 significant figures (3 sgf)

2. All zeros between non-zero digits are significant.

   E.g.  4.002 has  4s.g.f.

3. All zeros to the right of a decimal point and following a non-zero digit are significant.

   E.g.  62.00  has  4 s.g.f.

4. A zero before a decimal point and zero(s) after a decimal point but before a non-zero digit are not significant.

   E.g.  0.2,  0.005,   0.000004  all have 1 s.g.f.

NB:  When handling calculations, the final answer should always be given with only as many significant figures as that number involved in the calculation which has the least number of significant figures.

Example:

A rectangular block of wood measures 5.24 cm by 3.64 cm by 0.63 cm respectively. Calculate the volume of the block of wood.

Solution:  l
= 5.24 cm, w = 3.64 cm, h = 0.63 cm.

Applying Volume  = lwh

= 5.24 x 3.64 x 0.63

= 12.016368

= 12 cm³

Explanation:  Since the measurent with the least number of sgf is 0.63, which is 2 sgf, the final answer must have 2 sgf.

Self-Check 2.2

1.  How many significant figures are there in each of the following numbers:

 (a) 4.02  (b) 0.008  (c) 8 600  (d) 1 049  (e) 0.0002

2.  The following values were taken as part of a set of experimental data:  

25.57 cm and 8.48 mm. Find the sum of the two figures.

3.  A water tank measures by 4.5 m by 3.25 m by 5.5 m. Calculate giving the answers with the correct number of sgf.

 (a)  Base area (b)  Volume of the tank.

2.5  SCIENTIFIC NOTATION  (Exponential or Standard Notation)  

Scientific Notation is a short way of writing or expressing very large or very small numbers using powers of 10.

The number is expressed in the form: M x 10ⁿ

Where: M  – is a positive number (1 ≤ M < 10.

 n  – is ± integer (number)

 (a)  Determination of the arithmetic sign of n

The arithmetic sign of n is determined by using any one of the following simple rule.  (i)  The direction of the movement of the decimal point.

(ii)  The value of the number given.

Rule:  

1. When the decimal point is moved from right to left, the power of 10 is positive and when moved from left to right, the power of 10 is negative.
   
2. When the value of the number given is greater than 10, the power of 10 is positive and negative when the number is less than 1 (i.e. when the number begins with
   0.…).
   

   (iii)  When the value of the number given is 1 or greater than 1 but less than 10, the power of 10 is zero.
   

    

Examples

Express the following numbers in exponential notation.

(i)  25000 (ii)  0.000024  (iii)  250 x 4  (iv)   ¼ x 1200

Solution

1. 25 000  
   

In (i), the decimal point is not shown but its position is understood. For such numbers, the position of the decimal number is always after the last digit. For this case the decimal point is after the last zero. We move the decimal point from that position and put it in between the first two digits in order to make it less than 10 but greater than 1.

To get the power of 10, we count the number of digits between the original position of the decimal point and the new position.  Then the number becomes  = 2.5 x 10⁴

(ii)  0.000024  

In (ii), the position of the decimal point is clearly seen. We move it from left to right and put it between the first two non-zero digits in order to make it less than 10 but greater than 1.

To get the power of 10, we again count the number of digits between the original position of the decimal point and the new position.  Then the number becomes  = 2.4 x 10^-5  

For the case of (iii) and (iv), we first work out the answers and then use the same procedure to express the answers.

(iii)  250 x 4 = 1 000 (iv)   ¼ x 1200 = 3 000

= 1 x 10³   = 3 x 10³

Try (iii) and (iv)? Answers  (i)  4200 (iii)   0.085

(b)  Working out numbers with powers of 10

In working out numbers expressed using powers of ten, we use the following rules.  (i)  When multiplying powers of ten, add the exponents together.

(ii)  When dividing powers of ten subtract the exponents.

Examples

Work out the following numbers

(a)  10⁴ x 10⁸  (b) 10⁶ x 10^-3  (c)     (d)  

Solution:  

(a)  10⁴ x 10⁸  = 10^{(4 + 8)} (d)   =

= 10¹²     = 0.75 x 10^{(5 + 2)}

(b) 10⁶ x 10^-3  = 10^{(6 + -3)} = 7.5 x 10^-1 x 10⁷

= 10³   = 7.5 x 10 ^{(-1 + 7)}

(c)    = = 7.5 x 10⁶

= 3 x10³

Self-Check 2.3

Write the following numbers in scientific notation.

(a)  (i)  0.000222  (ii)  0.0025 (b)  (i)  5620  (ii)  75000

(c)  (i)  120 million  (ii)  20 (d)  (i)  5  (ii)  55 x 20

(e)  (i)  25 km in m  (ii)  10 g in kg.

Self-Check 2.4

1.  Which of the following is a fundamental quantity?

A. Time B. Density C. Volume D. Area

2.  What does a beam balance measure?

A. Area B. Mass C. Length D. Density

3.  Which one of the following is not a method of science?

A. Measurement  B. Observation  

C. Experimentation  D. Presentation

4.  Which one of the following are not matter?

I. Steam II.   Pencil III.  Light IV.  Space

A. I and II B. II and III C. III and IV D. I and IV.

5.  How many cubic centimeters are there in a litre?

A. 500   B. 100 C. 2000 D. 1000

6.  The sides of a black board are 2m and 5m, What is the surface area in m² ?

A. 0.1 B. 10 C. 1 D. 7

7.  How many mm² are in 0.032 dm².

A. 3.2 B. 320 C. 32 D. 0.0032

8.  Find the correct expression.

A. Litre is a unit of length.

B. 1 A is equal to 1 ten-thousandth of a micron.

C. A day is equal to one complete rotation of the earth.

D. A graduated cylinder is used to measure volume.

9.  What is equivalent to 5 minutes?

A. 30s B. 60s C. 120s D. 300s

10.  How many minutes are there between 05: 30 and 21:15?

A. 715 B. 945 C. 1595 D. 900

11.  The width of a meter rule is accurately measured by a

A. micrometer screw gauge B. vernier caliper

C. tape measure   D. meter rule

12.  A set of apparatus that is suitable for measurement of the volume of an irregular object includes;

A. Over flow can, measuring cylinder, irregular object and a string.

B. Measuring cylinder, irregular object, over flow cans, flask

C. Overflow can,. Irregular objects, string, retort sand and burette

D. Burette, overflows can, irregular object, a string, measuring cylinder, and retort stand.

13.  Convert 25cm³ into m³

A. 02.5 x 10⁵ B. 2.5 x 10² C. 2.5 x 10^-1 D. 2.5 x 10^-5

14.  The figure shows vernier calipers. The diameter of the object is

A. 1.05 cm B. 1.06 cm C. 1.56 cm D. 1.60 cm

15.  Three of the fundamental physical quantities are:

A.  Density, mass and time B.  Length, time and mass

C.  Length, time and weight D.  Volume, density and mass

That was a good revision test. Do not be unduly worried if you made a slip or two in your working. Try to avoid doing so, of course, but you are doing fine.

Now on to the next Chapter.