Intermediate Physics "measurements" By Abubakkar Marwat

F.Sc Part-1 (PHY)

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MEASUREMENTS

Chapter No: 1

Science is the study of nature. It is divided into two branches: Biological sciences: deals with the study of living things Physical sciences: deals with the study of non-living things There are three main frontiers of fundamental science: • the world of extremely large (universe itself) • the world of extremely small (electrons, protons, neutrons, mesons etc) • the world of complex matter All these are studied in fundamental physics which is the heart of science. 1.1 INTRODUCTION TO PHYSICS: Physics is the branch of science which deals with the study of matter and energy and relationship between them. Physics is very vast so it is divided into branches, few of them are: • Nuclear physics: deals with the study of atomic nuclei • Particle physics: it deals with the study of particles of which, matter is composed (electron, proton, neutron etc.) • Relativistic Mechanics: deals with the velocities that approach velocity of light • Solid State physics: deals with the study of structure and properties of solids Physics plays an important role in the development of technology and engineering. The “chips” (used in computers) are made of silicon. Silicon can be obtained from sand. Now it is up to us whether we make a sandcastle or a computer out of it. 1.2 PHYSICAL QUANTITIES: The foundation of physics is based upon physical quantities and laws of physics are expressed in terms of these physical quantities. Mass, length, time, velocity, force, density, temperature, electric current etc all are physical quantities. Physical quantities are divided into two categories: • •

Base quantities: these are not defined in terms of other physical quantities e.g.; mass, length, time Derived quantities: these are derived from other base quantities. Examples are velocity (derived from length & time), acceleration, force etc.

1.3 INTERNATIONAL SYSTEM OF UNITS (S.I): In 1960, an international committee agreed on a set of definitions and standard to describe the physical quantities and a system was established known as “System International”. It is consists of three kinds of units: base units, derived units and supplementary units. 1.3.1 Base units: there are 7 base units for various physical quantities, described below: Electric Intensity of Temperature Current Light kilogram second ampere kelvin candela kg s A K cd

Physical Qt. Length Mass S.I unit Symbol

meter m

Time

Amount of Substance mole mol

1.3.2 Supplementary units: there are two units of purely geometrical quantities which are neither base units nor derived, known as supplementary units. Abu Bakkar Marwat (B.Sc. Yarn Manufacturing) NTU, Faisalabad. [email protected]

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Radian: it is plane angle between two radii of a circle which cut an arc on the circumference, equal to the radius Steradian: it is solid angle (3-dimensional) subtended at the centre of a sphere by an area of its surface equal to the square of radius of the surface. Physical Qt. SI unit Plane angle radian Solid angle steradian

Symbol rad sr

1.3.3 Derived units: these units are used for all other physical quantities and are derived from the base and supplementary units. Some of them are as follows: Physical Qt.

Force

Work

Power

Unit Symbol In terms of base units

newton joule watt N J W kg m s-2 N m = kg m2 s-2 J s-1 = kg m2 s-3

Electric charge coulomb C As

Pressure pascal Pa N m-2 = kg m-1 s-2

1.3.4 Scientific Notation: Numbers are expressed in standard form (in power of 10) called “scientific notation”. There should be only one non-zero digit at left of decimal. e.g.; 134.7 is written as 1.347x102 and 0.0023 is written as 2.3x10-3. 1.3.5 Conventions for Indicating Units: While using or writing units for physical quantities, following points should be kept in mind: 1. Full name of the unit does not begin with a capital letter even if named after a scientist e.g., newton (Newton is wrong). 2. If unit is named after a scientist the symbol of that unit is written in capital latter, N (for newton), Pa (for pascal). 3. There should be no space between unit and its prefix, e.g., 1x10 -3 m is equal to “1 mm” (1 m m is wrong), similarly “5 cm” (“5 c m” is wrong) 4. While writing a combination of two base units, a space is given between them, e.g., newton meter is written as N m (Nm is wrong), similarly, joule per second = J s-1 5. Compound prefix are not allowed. e.g., 1x10-12 F is written as 1 pF (1 µµF is wrong) 6. A number such as 5.0x104 cm may be expressed in scientific notation as 5.0x10 2 m. 7. When a multiple of base unit (e.g., kilo, mega, milli) is raised to a power, the power applies to the whole multiple and not the base unit alone. 1 km2 = 1 (km) 2 = 1 (10 3)2 = 1x106 m2 8. Measurement in practical work should be recorded immediately in the most convenient unit, e.g., micrometer screw gauge measurement in mm, and the mass of calorimeter in grams (g). But before calculation for the result, all measurements must be converted to the appropriate SI base units. 1.4 ERRORS AND UNCERTAINITIES: All physical measurements are uncertain or imprecise to some extent (results are doubtful). It is very difficult to eliminate all possible errors or uncertainties in a measurement. The error may occur due to: • Negligence or inexperience of the person (operator) • There is some fault in the apparatus • The method or technique used is inappropriate The uncertainty may occur due to limitation of an instrument. It is usually described as an error.

Abu Bakkar Marwat (B.Sc. Yarn Manufacturing) NTU, Faisalabad. [email protected]

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There are two major types of errors: i) Random error: Random error occurs when repeated measurements of the quantity, give different values under the same conditions. Cause: it is due to unknown causes. Reducing random error: its effect can be reduced by repeating the measurements several times and taking an average. ii) Systematic error: It affects all measurements equally. It produces a consistent difference in reading. Cause: zero error of instrument, poor calibration of instrument Reducing systematic error: this error can be reduced by comparing the instrument with another which is known to be more accurate. 1.5 SIGNIFICANT FIGURES: In any measurement, the accurately known digits and the first doubtful digit are called significant figures. Explanation: We are measuring the length of a straight line with meter rod with a least count of 0.1 cm (where 1m = 100 cm and each cm = 1mm). The length measured was found to be between 12.6 and 12.7 cm. It was also found that the line was longer than mid point of 12.6 cm (i.e. more than 12.65) so normally we take this reading as 12.7 cm not 12.6 cm. So in this reading, last digit “7” is not sure or in other words it is doubtful. These two sure digits “1, 2” and one doubtful digit “7” are known as significant figures. So uncertainty or accuracy in the value of a measured quantity can be indicated by using significant figures. If the above measurement is taken by a vernier caliper with a least count of 0.01 cm, it will be recorded as 12.70. So we can say: “as we improve the quality of our measuring instrument and techniques, we extend the measured result to more and more significant figures which improve the accuracy of the result”. Rules for deciding No of significant figures: 1) All digits 1,2,3,4,5,6,7,8,9 are significant. Zero may or may not be significant. In case of zeros, following rules may be followed: • A zero between two significant figures is significant. 1002, 20105 (all zeros are significant) • Zeros to the left of significant figures are not significant. e.g., 0.00467, 02.59 all zeros are non-significant. • Zeros to the right of a significant figure may or may not be significant. i. In case of decimal fraction (3.570, 74.00), zeros to the right of significant figures are significant. e.g., 7.2300, 23.00 all zeros are significant. ii. In case of integers (2000), the number of significant zeros depend upon the accuracy (or least count) of the measuring instrument. e.g., 8000 kg, if L.C is 1kg, then (8000/1 = 8000) there are 3-significant zeros, and if L.C is 10kg then (8000/10 = 800) there are 2-significant zeros. • In scientific notation, the figures other than the powers of ten are significant figures. 223x105 contains 3, and 2.3025x108 contains 5 significant figures. Rounding off data: In a measurement, insignificant figures should be deleted. While deleting the figures, the last significant figure that is retained is rounded off, for which following rules are followed. • If the first digit dropped is less than 5, the last digit remains unchanged. 43.23 is rounded off as 43.2, similarly 5.232 is rounded off as 5.23. Abu Bakkar Marwat (B.Sc. Yarn Manufacturing) NTU, Faisalabad. [email protected]

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• If the first digit dropped is more than 5, last digit is increased by 1. e.g., 43.26 is rounded off as 43.3, similarly 5.237 is rounded off as 5.24. • If the digit to be dropped is 5, then: i. the last digit is increased by 1 if it is odd. e.g., 43.35 is rounded off as 43.4, similarly 5.235 is rounded off as 5.24. ii. the last digit remains unchanged if it is even e.g., 43.45 is rounded off as 43.4 and 5.225 is rounded off as 5.22 2) In multiplying or dividing, the answer is rounded off according to the factor containing the least number of significant figures. e.g., (5.348x3.64)/1.336 = 14.5708982036 as the factor “3.64” contains the least number of significant figures i.e., 3, the answer is rounded off to 3 significant figures i.e., 14.6. 3) In adding or subtracting numbers, the number of decimal places retained in the answer should equal the smallest number of decimal places in any of the quantities given. e.g., adding 72.1+3.42+0.003 = 75.523 so it is rounded off as “75.5” because the least number of decimal place in the given data is “1” in 72.1, so decimal point is placed as one. Similarly 2.7543+4.10+1.273 = 8.1273 is rounded off as “8.13”. 1.6 PRECISION AND ACCURACY: Precision: it means that results obtained during an experiment are very close to each other. The precision of a measurement is determined by the instrument being used. Accuracy: it means that results obtained during an experiment are close to the accepted value. Accuracy of a measurement depends upon the fractional or percentage uncertainty. Precision or absolute uncertainty: it is equal to the least count of measuring instrument. For vernier caliper having L.C = 0.01 cm, absolute uncertainty is ±0.01 cm Fractional uncertainty: it the ratio of absolute uncertainty to the measured value. e.g., if a measurement of 25.5 cm is taken by a meter rod with a L.C or absolute uncertainty of 0.1 cm, the fractional uncertainty will be => 0.1/25.5 = 0.004 Percentage uncertainty: fractional uncertainty expressed as percentage is known as percentage uncertainty (i.e., fractional uncertainty x 100) e.g., for 25.5 cm reading with 0.1 cm L.C, percentage uncertainty will be =>

0.1

x100 =

0.4 %

25.5

So: a precise measurement is the one which has less precision or absolute uncertainty and an accurate measurement is one which has less fractional or %age uncertainty (or error) 1.7 ASSESMENT OF TOTAL UNCERTAINTY IN THE FINAL RESULT: The maximum possible uncertainty or error in the final result can be found as follows. 1.7.1 for addition or subtraction: Absolute uncertainties are added: e.g., x1 = 10.5 ±0.1 cm, x2 = 26.8 ±0.1 cm. then x1+ x2 =10.5 + 26.8 = 37.3 and total uncertainty = 0.1 + 0.1 = 0.2 cm So answer is “37.3 ± 0.2 cm” 1.7.2 for multiplication and division: Percentage uncertainties are added: e.g., for the formula R =

V

, V = 5.2 ± 0.1 V, and I = 0.84 ± 0.05A

I

%age uncertainty for V =

0.1 5.2

x100 = 2% ,

%age uncertainty for I =

0.05

x100 = 6%

0.84

Abu Bakkar Marwat (B.Sc. Yarn Manufacturing) NTU, Faisalabad. [email protected]

F.Sc Part-1 (PHY)

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Hence R = V/ I = 5.2/0.84 = 6.19 Ω and total uncertainty = 2% + 6% = 8%. Ans:[6.2 ± 0.5 Ω] {result is rounded off to 2-significant figures and 8%of 6.2 is 0.05}

1.7.3 for power factor: Multiply the percentage uncertainty by that power. e.g., in calculation the area of circle: A = π r2 so percentage uncertainty in A = 2 x % age uncertainty in radius “r”. if r = 2.25 cm with 0.01 cm L.C of vernier calliper, then: % age uncertainty in r =

0.01

x100 = 0.4 %

& total % uncertainty in A = 2x 0.4 = 0.8%

2.25

Area A= π r2 = 3.14x (2.25)2 = 15.9 ± 0.127 cm2

[0.8% of 15.9 is 0.127]

1.7.4 for uncertainty in the average value of many measurements: First of all, average of all the values is calculated. For reading 1.20,1.22,1.23,1.19,1.22,1.21 average will be =

1.20 + 1.22+1.23+1.19+1.22+1.21

= 1.21mm

6

Then deviation of these reading from the average is calculated, which are 0.01, 0.01, 0.02, 0.02, 0.01, 0. so mean of deviation =

0.01+0.01+0.02+0.02+0.01+0

= 0.01mm

which is

6

actually the uncertainty in the mean diameter. So answer 1.21 ± 0.01 mm 1.7.5 for the uncertainty in a timing experiment: The uncertainty in the time period of a vibrating body is found by dividing the least count of timing device by the number of vibrations. e.g., the time of 30 vibrations of a simple pendulum recorded by a stopwatch with L.C 0.1 s, is 54.6 s, then time period: T = 54.6/30 = 1.82s with uncertainty 0.1/30 = 0.003 s Thus time period is written as T = 1.82 ± 0.003 s Example 1.1: the length, breadth and thickness of a sheet are 3.233 m, 2.105 m, and 1.05 cm respectively. Calculate the volume up to the correct appropriate significant digits. Solution: Given that: length l = 3.233 m, breadth b = 2.105 m, and thickness t = 1.05 cm = 1.05 x 10-2 m So volume V = l x b x t = 3.233 x 2.105 x 1.05 x 10-2 = 7.14573825 x 10-2 m3 As the factor “1.05 cm” has minimum number of significant figures equal to 3, the answer is recorded up to 3 significant figures: V = 7.15 x 10 -2 m3 Example 1.2: the mass of a metal box measured by a lever balance is 2.2 kg. Two silver coins of masses 10.01 g and 10.02 g measured by a beam balance are added to it. What is now the total mass of the box correct up to the appropriate precision? Solution: Given that: box =2.2 kg, 1st coin = 10.01 g = 0.01001 kg, 2nd coin = 10.02 g = 0.01002 kg Total mass = 2.2 + 0.01001 + 0.01002 = 2.22003 kg since the factor 2.2 kg has the decimal point placed at 1, which is least of the all so answer is rounded of to place the decimal point at 1. Ans = 2.2 kg Example 1.3: the diameter and length of a metal cylinder measured with the help of vernier calliper of least cont 0.01 cm are 1.22 cm and 5.35 cm. Calculate the volume V of the cylinder and uncertainty in it. Solution: Given that: diameter d = 1.22 cm with L.C 0.01 cm, length l = 5.35 with L.C 0.01 cm.

Abu Bakkar Marwat (B.Sc. Yarn Manufacturing) NTU, Faisalabad. [email protected]

F.Sc Part-1 (PHY)

-MEASUREMENTS-

Absolute uncertainty in length =

0.01 5.53

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x100 = 0.2%

Absolute uncertainty in diameter =

0.01 1.22

x100 = 0.8%

Now total uncertainty in V = 2 x (%age uncertainty in d) + ( %age uncertainty in l) = (2 x 0.8) + 0.2 = 1.8 % 2

And volume V =

3.14 x (1.22) x 5.35 4

= 6.2509079cm with 1.8 % uncertainty

Thus V = 6.2 ± 0.1 cm3

3

[1.8 % of 6.2 = 0.1]

Define the terms: distance, displacement, speed, velocity, acceleration, mass & weight Distance: the separation between two points is known as distance, denoted by “l”. The S.I unit of distance is meter denoted by “m”. It is a scalar quantity. Displacement: the smallest distance between to points is known as displacement, denoted by l The S.I unit of displacement is meter denoted by “m”. It is a vector quantity. Speed: distance traveled per unit time is known as speed, denoted by “v”. s

Mathematically, v = , the units of speed: m/s. It is a scalar quantity. t

Velocity: rate of change of displacement is known as velocity, denoted by “v”. s

Mathematically, v = , the units of velocity: m/s. It is a vector quantity. t

Acceleration: rate of change of velocity is known as acceleration, denoted by “a”. v

Mathematically, a = , the units of acceleration is m/s2. It is a vector quantity. t

Mass: the quantity of matter in a body. m = W/g. S.I unit-kg (kilogram), a scalar quantity. Weight: it is the force with which earth attracts any object towards its centre. W = mg S.I unit of weight is N (newton) it is a vector quantity. 1.8 DIMENSIONS OF PHYSICAL QUANTITIES: Each base quantity Length , Mass and Time is considered as a dimension denoted by L, M, or T written within square brackets respectively. e.g., length, breadth, height, and diameter etc all are measured in meter so a common dimension [L] for them. Find the dimensions of speed, acceleration, and force. Speed: s

As we have, v = = length/time = [L]/ [T] = [LT-1] t

Acceleration: As we have, a =

v t

= velocity/time =

[ L] 1 [L] displacement 1 x = x = −2 = [LT-2] t t [T ] [T ] [T ]

Force: As we have, F = ma =

mass ×

velocity time

=

mass ×

1 time

×

displacement time

=

[M ] [T ]

×

[ L] [T ]

=

[ M ][ L] 2

[T ]

-2

= [MLT ]

Dimensions are used to: a) Check the homogeneity of physical equation: In order to check the correctness of an equation, we have to prove that the quantities on both sides have the same dimensions. Abu Bakkar Marwat (B.Sc. Yarn Manufacturing) NTU, Faisalabad. [email protected]

F.Sc Part-1 (PHY)

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Example 1.4: Check the correctness of the relation v =

7/7 F ×l m

Solution: Dimension of L.H.S of the equation v = [LT-1] Dimension of R.H.S of the equation = (

F ×l m

1

-1 1/2

) 2 = (F x l x m )

= ( [MLT-2] x [L] x [M-1] )1/2

= ( [L2T-2] )1/2 = [LT-1], hence proved L.H.S = R.H.S and equation is correct. b) Deriving a possible formula: Example 1.5: Derive a relation for the time period of a simple pendulum (Fig) using dimensional analysis. The various possible factors on which the time period T may depends are: i) Length of the pendulum (l) ii) Mass of the bob (m) iii) Angle θ which the thread makes with the vertical iv) Acceleration due to gravity (g) Solution: The relation for the time period T will be of the form T ∝ ma x lb x θc x gd T = constant (ma x lb x θc x gd) _________ (A) Now we have to find the values of powers a, b, c, and d, for which writing the dimensions of both sides we get: [T] = constant x [M]a [L]b [LL-1 ]c [LT-2]d Comparing the dimensions on both sides we have: [T] = [T]-2d [M]0 = [M]a [L]0 = [L]b+c-c+d Equating powers on both the sides we get: -2d = 1 or d = -1/2, a = 0, b+d =0 or b + (-1/2) = 0 or b = ½ and θ = [LL-1]c = [L0]c = 1 Now putting values of a, b, θ and d in Eq. (A) T = constant x m0 x l1/2 x 1 x g -1/2 T = constant (1 x l1/2 x g -1/2) l

T = constant

answer.

g

Example 1.6: Find the dimensions and S.I units of coefficient of viscosity “η” in the relation of Stokes law which is F = 6 π η r v Solution: 6π is a number having no dimensions so it is not accounted in dimensional analysis. F=ηrv η=

F r×v

Substituting the dimensions of F, r, and v in R.H.S η=

[ MLT

−2

]

−1

[ L][ LT ]

−1

=

[ MT ] [ L]

-1 -1

η = [ML T ] and S.I unit of coefficient of viscosity is kg m-1 s-1 -:-----------------------x----------------------:-

Abu Bakkar Marwat (B.Sc. Yarn Manufacturing) NTU, Faisalabad. [email protected]