Vernier Callipers
 The Vernier callipers has two scales.The main scale is contained on the steel frame and is graduated in centimeters but also has millimeters divisions. The Vernier scale is contained on the sliding jaw and has 10 equal divisions.
 The length of Vernier scale is 0.9cm implying that each division of the vernier scale is 0.09cm.
 The difference between the main scale division and the Vernier scale division is called the least count. This is the accuracy of the Vernier callipers i.e. (0.90.09)cm=0.01cm.
 Vernier callipers has inside jaws used to measure internal diameters and outside jaws used to measure external diameters.
Using Vernier Calipers
 Place the object whose diameter(length) is to be measured between the outside jaws.
 Close the jaws till they just grip the object.
 Record the reading of the main scale, opposite and to the left of the zero mark of the vernier scale.
 Read the vernier scale mark that coincides exactly with a main scale mark and multiply it with the least count(accuracy) of the Vernier callipers. This is the Vernier scale reading.
 The sum of the vernier scale reading and the main scale reading gives the diameter(length) of the object.
Vernier callipers reading = vernier scale reading + main scale reading
Example
Main scale reading: 10.0 cm
Vernier scale reading: 0.02 cm (Alignment of scale lines)
Vernier Callipers reading = Main scale reading + Vernier scale reading: = 10.0 cm + 0.02 cm = 10.02 cm
Exercise
 Describe how you would measure the internal diameter of 100cm^{3} beaker using vernier callipers.
 Write down the vernier calipers reading in diagram (a)(b) and (c) showed below.
Zero Error of the Vernier Callipers
 Vernier callipers is said to have a zero error if the zero marks of the main scale and vernier scale do not coincide when the jaws of the calipers are closed without an object.
 There are two types of errors:
i) Positive Error
 Occurs when the zero mark of the main scale is to the left of the zero mark of the vernier scale.
Example
This vernier callipers has a zero error of +0.13cm
Correction of the Positive Error
 The positive error is corrected by subtracting the zero error from the reading obtained.
(ii) Negative Error
 Occurs when the zero mark of the main scale is to the right of the zero mark of the vernier scale.
Example
This vernier callipers has a negative error of 0.02 cm
Correction of the Negative Error
 The negative error is corrected by adding zero error to the reading obtained.
Exercise
The figure below shows a vernier callipers
State the correct reading of the scale if the instrument has a zero error of –0.02cm.
Micrometer Screw Gauge
 It is used to measure very small lengths such as the diameter of a thin wire.
 The micrometer screw gauge consist of a thimble which carries a circular rotating scale known as thimble scale and a spindle which moves forward and backwards when the thimble is rotated.
 The sleeve has a linear scale in millimeters and half millimeter called sleeve scale and the thimble has a circular scale of 50 or 100 equal divisions.
 The ratchet at the end of the thimble prevents the user from exerting more pressure on an object when the micrometer screw gauge is in use.
 The distance moved by the spindle in one complete rotation of the thimble is called the pitch of the micrometer. A spindle moves forward or backwards by 0.5mm per a complete rotation of the thimble with 50 divisions.
 Therefore each division of thimble scale represents a spindle travel of
^{0.5mm}/_{50} = 0.01mm  This means that if the thimble rotates through one division, the spindle moves backwards or forward by 0.01mm. This is the least count(accuracy) of the micrometer screw gauge.
 Least count of the screw gauge is defined as the distance moved bythe spindle when the thimble rotates through one division.
Using a micrometer screw gauge
 Place the object whose diameter/length is to be measured between the anvil and the spindle.
 Close the micrometer using ratchet until the object is held gently between the anvil and the spindle. Note that the ratchet should slip only once when the grip is firm enough to give accurate reading.
 Read the sleeve scale and record it as:
Sleeve scale reading = mm  Read the thimble scale and multiply it by the least count of the screwgauge(0.01mm) and record it as:
Thimble scale reading =……x0.01=………….mm  Micrometer reading=sleeve scale reading +thimble scale reading
Example
Sleeve scale reading: 2.5 mm
Thimble scale reading: 0.38 mm
Micrometer reading: 2.88 mm
The zero error of the micrometer screwgauge
It occurs if the zero mark of the thimble scale does not coincide with the horizontal(centre) line of the sleeve scale when the micrometer is closed without an object.
Positive error of micrometer screw gauge
 Occurs when the zero mark of the thimble scale is below the horizontal line.
 The positive error is corrected by subtraction of the erro rfrom the reading given by the micrometer screw gauge.
Negative error
 It occurs when the zero mark of the thimble scale is above the horizontal line of the sleeve scale.
 The negative error is corrected by adding the error to the reading obtained by the screw gauge.
Significant Figures
 Significant figures refer to the number of digits used to specify the accuracy of a value.
Note:
 The digits 19 are all significant when they appear in a number.
 The first digit from the left of a number is the first significant figures.
 The number of significant figures is determined by counting the number of digits from the first significant figure on the left.
 Zero may be significant or not depending on the position of the digit.
 If zero occurs between non zero digits it is significant e.g.1004(4sf),15607(5sf),180.45(5sf)
 When zero occurs at the left end of a number it is not significant e.g.0.00546(3sf),0.0002(1sf)
 If the zero occurs at the right hand end of an integer it may or may not be significant. E.g.60000. It can be correct to 1 significant figure therefore the zeros are not significant. If all the zeros are counted(ended) then it will be correct to 6 significant figures.
 If the zero occurs at the right hand end after the decimal point,it is always significant e.g.2.000(4sf), 3.0(2sf)
Exercise
Write down the number of significant figures in each of the following
 40000
 609
 0.000675
 5237.8
 0.0000600
 0.002304
Standard Form
 This is a way of writing a number especially a very large or very small number in which only one integer appears before the decimal point.
 A positive number is said to be in standard form when written as A x 10^{n}, where A is such that 1≤A<10 and the index n is an integer e.g.3567=3.567x10^{3}
 If the number lies between zero and 1 then the index n becomes a negative e.g.0.0003567=3.567x10^{4}
Exercise
Express the following in cm giving the answers in standard form
 0.1mm
 125mm
 3.8m
 0.015m
 7.8km
Decimal places
 Refer to number of digits to the right of the decimal point and this determines the accuracy of the number e.g.6.0345(4d.p)
Exercise
Find the volume of a cube whose side is 2.22cm. Express your answer correct to 3d.p
Standard Prefixes Used With SI Units
 The table below shows multiples and submultiples used with SI units, their prefixes and symbol for the prefixes.
Submultiple/ multiple prefix Symbol for prefix 10^{1} deci d 10^{2} centi c 10^{3} milli m 10^{6} micro μ 10^{9} nano n 10^{12} pico p 10^{15} femto f 10^{18} atto a 10^{1} deca Da 10^{2} hecto H 10^{3} kilo K 10^{6} mega M 10^{9} giga G 10^{12} tera T 10^{15} peta P 10^{18} exa E
The Oil Drop Experiment
 This is an experimentused in the estimation of diameter/size/thickness of a molecule.
 In this experiment, a tray is filled with water to the brim, and lycopodium powder is lightly sprinkled on the water surface.
 An oil drop is carefully placed at the centre of the tray and allowed to spread on the surface of water until it is one molecule thick. This forms a patch whose diameter is measured.
 Thickness of oil molecule is estimated as
volume of oil drop = volume of oil patch
^{4}/_{3}πr^{3} = π(^{d}/_{2})^{2} x thickness, t, of oil patch(or molecule)
Functions of lycopodium powder
 It breaks surface tension
 it clearly shows the extent of spread of the oil drop
Function of beams:
 Used to estimate diameter of the spread oil patch
Assumptions made in oil drop experiment
 The oil drop is perfectly spherical
 The oil patch is perfectly cylindrical
 The oil patch is one molecule thick.
Possible Sourcesof Error in the Experiment
 Error in measuring the diameter(or volume) of oil drop
 Error in measuring diameter of oil patch
Exercise
 In an experiment to estimate the size of an oil molecule,the diameter of the patch was measured to be 200mm for an oil drop of radius 0.25mm. Determine the diameter of the molecule of the oil.
 In an experiment to estimate the diameter of oil molecule 100 drops of oil are released from burette and level of oil in burette changes from 0.5cm^{3 }to 20.5cm^{3}. One of the drops is placed on water and spreads over a circular patch of diameter 20cm.
 Determine:
 The volume of the oil drop
 The area of the patch covered by the oil
 The diameter of the oil molecule
 State:
 Assumptions made in this experiment
 Two possible sources of errors in this experiment
 Determine:
Revision Questions
 What are the zero errors of the micrometer screw gauges shown in the figures below?(the micrometers are closed). If the micrometers were used to measure the diameter of a wire whose diameter is 1.00mm, what would be the reading on each?
 Compare and contrast the scales of two micrometer screw gauges with a pitch of 0.5mm and 1.0mm.
 What are the two limitations of the micrometer screw gauge?
 List down the advantages and disadvantages of the micrometer screw gauge over the vernier callipers.
 Sketch a micrometer screw gauge scale reading:
 0.23mm
 5.05mm

 What are the zero errors of the vernier callipers in figures (a) and (b) below?
 If the correct diameter of an object is 4.01cm, what would be the readings of both calipers for this diameter?
 The callipers in figure (a) was used to measure the diameter of a cylindrical object and recorded 4.55cm while the one in figure (b) was used to measure the diameter of a sphere and recorded 5.05cm. Calculate correct volumes of these objects in m^{3}.(Takeπ=3.142)
 What are the zero errors of the vernier callipers in figures (a) and (b) below?