UNITS AND MEASUREMENTS CHAPTER 2
Chapter Outline
2.1 Physical Quantities and Their Units
2.2 The International System of Units
2.3 Accuracy and Precision of Instruments and Errors in Measurement
2.4 Significant Figures
2.5 Measuring Instruments
2.6 Dimensions of Physical Quantities
2.7 Dimensional Formulae and Dimensional Equations
2.8 Dimensional Analysis and Its Applications
2.1 PHYSICAL QUANTITIES AND THEIR UNITS
Physics is an experimental science based on the measurement of physical quantities.
Understanding of different physical pheno-mena requires the need to measure relevant physical quantities. Physical quantities are those that can be defined and measured. Without measurement, there can be no advancement in Physics. Experimental measurements are essential to verify theoretical laws. We use a number of physical quantities (in daily life), like length, time, area, volume, etc., to describe an event or a phenomenon.
For measuring a physical quantity, a standard reference is necessary. This reference is known as a unit
2.1.1
Physical Quantities
“All quantities in terms of which laws of physics can be expressed and which can be measured directly or indirectly are called physical quantities.”
Physical quantities can be classified into two types:
1. Fundamental physical quantities
2. Derived physical quantities
Fundamental Physical Quantities
The physical quantities that are independent of any other quantities are called fundamental physical quantities.
Examples: Mass, length, time, electric current, thermodynamic temperature, amount of substance, and luminous intensity
Derived Physical Quantities
The physical quantities that can be derived from fundamental quantities are called derived quantities.
Examples: Area, volume, speed, velocity, etc.
2.1.2 Unit
Unit of measurement of a physical quantity is the standard reference of the same physical quantity, which is used for comparison of the given physical quantity.
The meaning of the measurement of a physical quantity is to find out the number of t imes its unit is contained in that
physical quantity. Therefore, the process of measurement of a physical quantity involves: i) the selection of a unit and ii) to find out the number of times that unit is contained in the physical quantity.
For example, if we are asked to measure the length of a table, the unit to be selected must be that of length.
Suppose, we use metre as the unit. We place a metre rod successively along the length of the table and find out the number of times the metre rod is contained in the length of the table.
Suppose, the length of the table is covered by the metre rod in three successive placings.
Then, 3 is called the numerical value of length of the table and metre is the unit of length.
We may write:
Length of the table = 3 × 1 m = 3 m
It may be pointed out that it is not sufficient to say that the length of the table is 3. The unit of the physical quantity has to be stated along with the result of the measurement.
Measurement of a physical quantity = (Numerical value of the physical quantity) × (size of its unit)
If ‘ u ’ is the size of the unit and ‘ n ’ is the numerical value of the physical quantity X (the quantity to be measured) for the selected unit, then the measure of the physical quantity is X = nu.
It follows that, if the size of the chosen unit is small, then the numerical value of the quantity will be large and vice-versa. It is obvious that the measure of the physical quantity is always the same, i.e., nu = constant
If n1 is the numerical value of the physical quantity for a unit u 1, and n 2 for a unit u 2, then n1u1 = n2u2
2.2 THE INTERNATIONAL SYSTEM OF UNITS
To measure the fundamental base quantities, length, mass, and time, there are three systems of units, namely
1. FPS system (British system),
2. CGS system (metric system), and
3. MKS system
The basic units in these systems are given in the following table.
FPS foot pound second
CGS centimetre gram second
MKS metre kilogram second
The system of units which is, at present, internationally accepted for measurement is the Systeme Internationale d’Unites, abbreviated as SI.
An international organisation, the Conference General des Poids at Measures, or CGPM, is internationally recognised as the authority on the definition of units. In English, this body is known as the ‘General Conference on Weights and Measures.’ The System International de Units, or SI units, was set up in 1960, by the CGPM.
Requirements of a Standard Unit
A standard unit must have the following features:
i) It should be of suitable size, i.e., neither too large nor too small, when compared to the quantity to be measured.
ii) It should be very well defined.
iii) It should be independent of time and place.
iv) It should be easily available so that all laboratories can duplicate and use it as per requirement.
v) It should be independent of physical conditions, like temperature, pressure, humidity, etc.
In SI, there are seven fundamental (base) physical quantities, namely length, mass, time, electric current, thermodynamic temperature, luminous intensity, and quantity of substance. In addition to these, two more physical quantities, plane angle and solid angle, were introduced as supplementary fundamental quantities. Fundamental quantities and supplementary quantities, along with their units and symbols, are given in the following tables.
SI fundamental (base) quantities and units
Base quantity
SI units
Name Symbol
Length metre m
Mass kilogram kg
Time second s
Electric current ampere A
Thermodynamic temperature kelvin K
Luminous intensity candela cd
Amount of substance mole mol
SI supplementary quantites and units
Basic quantity
SI units
Name Symbol
Plane angle Radian rad
Solid angle Steradian sr
2.2.1 Fundamental Units
The units of the fundamental physical quantities are called fundamental units.
Examples: metre, kilogram, second, etc.
2.2.2 Derived Units
The units of derived physical quantities are called derived units.
Examples: newton, erg, joule, etc.
Physical Quantity Derived unit Symbol of Unit
Force newton N = kg m/s2
Energy, work, heat joule J = Nm = kg m2/ s2
Power watt W = J/s = Nm/s = kgm2/s3
Pressure, stress pascal Pa = N/m2 = kg/ ms2
Frequency hertz Hz = s-1
Electric charge coulomb C = As
Electric potential Volt V = J/C
Capacitance farad F = coulomb/volt = A2s2/Nm
Resistance ohm Ω = V/A = J/A2s
Self inductance henry H = J/A2 = Nm/ A2
Magnetic flux weber Wb
Magnetic flux density tesla T = Wb/m2
2.2.3 Multiples and Submultiples of SI Units
To measure very low or very high values of physical quantities, prefixes are used to represent them. For example, 10 – 3 m is written as 1 millimetre, 10–9 seconds is written as nanosecond, and 106 watts is written as 1 megawatt.
2.2.4 Rules for Using Symbols for SI Units/Prefixes
1. Full names of units, even if they are named after scientists, should not be written with initial capital letter.
Example: newton but not Newton
2. Symbols for a unit named after a scientist should have a capital letter.
Examples: ‘N’ for newton; ‘W’ for watt; ‘A’ for ampere
3. Punctuation marks should not be used after symbol of unit.
Examples: 100 kg but not 100 kg. 100 mm but not 100 mm. (or) 100 m.m.
4. Symbols for units should not take plural form.
Examples: 100 newtons is 100 N but not 100 Ns; 50 metres is 50 m but not 50 ms or 50 mts.
TEST YOURSELF
1. Candela is the unit of (1) electric intensity (2) luminous intensity (3) sound intensity (4) magnetic field intensity
2. The unitless quantity is (1) velocity gradient (v/l) (2) pressure gradient (P/l) (3) distance gradient (s/l) (4) force gradient (F/l)
3. Which of the following is not a derived unit? (1) joule (2) watt (3) kilogram (4) newton
4. Hertz is the unit for (1) Frequency (2) Force (3) Electric charge (4) Power
5. In the following which is not the unit of time?
(1) Leap year
(2) Lunar month
(3) Light year
(4) sidereal year
Answer Key (1) 2 (2) 3 (3) 3 (4) 1 (5) 3
2.3 ACCURACY AND PRECISION OF INSTRUMENTS AND ERRORS IN MEASUREMENT
We can completely understand a physical quantity only if we are able to measure it. An accurate measurement of physical quantities plays a very important role in the world of science.
2.3.1 Accuracy
Accuracy is a measure of the closeness of the measured value to the true value.
Smaller the uncertainty in the measured value, greater is its accuracy.
2.3.2 Precision
The precision of an instrument gives the minimum value that can be measured by it. This is called the resolution or limit of measurement of the instrument. Accuracy depends on the resolution of the instrument. Measurement with more precise instrument need not be accurate. For example, let the diameter of a rod be nearly 1.78 cm. It is measured as 1.6 cm by an instrument. Again, it is measured by another precise instrument as 1.48 cm. The first measurement (1.6 cm) is closer to 1.78 cm than that (1.48 cm) measured by the second precise instrument. The second measurement is more precise but not accurate. The first measurement is accurate but less precise.
However, in spite of great technological advancements and sophistication in the development of instrumentation, no measurement can be free from errors. Errors may be because of several reasons, both controllable and uncontrollable.
Hence, a knowledge of errors is essential to interpret the results and evaluate the measurements on a scientific basis.
1. The length of a straight line is measured a number of times by a number of observers. The following are the results of these measurements. Decide precision and accuracy.
Actual length = 3.785 cm ± 0.001 cm
1st set of measurements: 3.8 cm, 3.9 cm, 3.7 cm
2nd set of measurements: 3.478 cm, 3.479 cm, 3.478 cm, 3.478 cm, 3.479 cm
3rd set of measurements: 3.55 cm, 3.65 cm, 3.45 cm, 3.35 cm
4th set of measurements: 3.784 cm,3.785 cm, 3.784 cm, 3.785 cm, 3.784 cm
Sol: First set of measurements is accurate because it is closest to the actual value, but it is not precise.
Second set of measurements is not accurate but it is the most precise because the readings are reproducible.
Third set of measurements is neither accurate nor precise.
Fourth set of measurements is accurate as well as precise.
Try yourself:
1. Two clocks are being tested against a standard clock located in a national laboratory. At 12:00:00 noon, by the standard clock, the readings of the two clocks are as follows:
Clock 1 Clock 2
Monday 12:00:06 10:15:08
Tuesday 12:01:25 10:14:59
Wednesday 11:59:06 10:15:28
Thursday 12:01:50 10:15:07
Friday 11:59:25 10:14:42
Saturday 12:01:40 10:15:14
Sunday 12:01:29 10:15:21
If you are doing an experiment that requires precision time interval measurements, which of the two clocks will you prefer?
Ans: The range of variation over the seven days of observations is 164 s for clock 1, and 32 s for clock 2. The average reading of clock 1 is much closer to the standard time than the average reading of clock 2. The important point is that a clock with zero error is not as significant for precision work as its variation, because a ‘zero-error’ can always be easily corrected. Hence, clock 2 is to be preferred to clock 1.
Key Insights:
■ The accuracy of a measurement is a measure of how close the measured value is to the
true value of the quantity. Precision tells us to what resolution or limit the quantity is measured.
2.3.3 Errors in Measurement
Error is the amount of uncertainty that is present in the measurement made with a measuring instrument. It is the difference between the measured value and the true value.
Errors creep in the observed or calculated values due to various reasons. They are categorised under two heads:
(a) Systematic errors
(b) Random errors
Systematic Errors
Er ror that is always unidirectional is called systematic error.
If some error is occurring in the positive direction only (or in the negative direction only), all the time, it is a systematic error.
Example: If measured value is always 2 units more than true value, it is a systematic error (positive).
If measured value is always 3 units less than true value, it is a systematic error (negative).
i) Instrumental errors: Instrumental errors arise from the errors due to imperfect design or calibration of the measuring instrument.
Example: Zero-error of a measuring instrument
If a scale is worn off at the starting end up to 0.3 cm reading and we do not notice it and treat it as zero, then every length we measure will always be 0.3 cm more than the true value.
The instrumental error can be detected by interchanging two similar instruments or by using different methods for measuring the same physical quantity.
ii) C onstant errors: Whe n the result of a series of measurements are in er ror by
the same amount, such an error is called constant error. Zero error is also a constant error.
iii) Environmental errors: These arise due to external conditions, that is, due to changes in the environment. During the experimental measurement, there may be changes in external conditions, i.e., changes in temperature, pressure, humidity, wind velocity, etc. For example, if a metal scale is calibrated in winter and used in summer, due to expansion in summer, the measured values will be less than true values.
iv) Personal errors: These are entirely due to the personal peculiarities of the experimenter. Individual bias, lack of proper setting of the apparatus, carelessness in taking observations (without taking the required necessary precautions), etc., are the causes for this type of error. A person may be habituated to hold his eyes (head) always a bit too far to the right (or left) while taking the reading with a scale. This will give rise to parallax error.
If a person keeps his eye-level below the level of mercury in a barometer, all the time, his readings will have systematic error. These errors can be minimised by obtaining several readings carefully and then taking their arithmetical mean.
v) Imperfection in experimental technique or procedure: Sometimes, even when we know the nature of the error, it cannot be eliminated due to imperfection in experimental arrangement. Examples include, in calorimetry, loss of heat due to radiation, the effect on weighing due to buoyancy of air, etc. These errors will always exist.
vi) Minimisation of systematic errors: We can minimise systematic errors,
■ by selecting better instruments with higher resolution,
■ by taking care to avoid personal bias
■ by improving the experimental technique
■ by estimating the error and applying the necessary correction.
Random Errors
The errors that are irregular, whose cause is not known, and that are random in nature in their sign and size are called random errors. Unlike systematic errors, these errors are at random with respect to the sign and magnitude. They occur irregularly; sometimes increasing, sometimes decreasing, with changing magnitudes.
These errors arise due to the random fluctuations in experimental conditions that cannot be predicted. For example, the random fluctuations in temperature, voltage supply, and mechanical vibrations of experimental setup, etc., are the causes of random errors.
The diameter of a wire may be varying from position to position along its length due to non-uniform cross section. Hence, we get random errors in the measurement of diameter. At some places, the value will be more, and at other places, value will be less with different magnitudes.
The uneven tightening of the screw in a screw gauge at different places of a wire of uniform cross section also results in random errors.
Thus, even when the same person repeats the same observation, he may get different reading each time.
Example: Suppose, true length of a body is 2.5 cm. Same person measures it three times as 2.3 cm, 2.7 cm, and 2.9 cm. Then, the errors are –0.2 cm, +0.2 cm, and +0.4 cm, respectively.
Random errors are also called indeterminate errors.
Random errors cannot be eliminated but can be minimised by increasing the number of observations.
The total random errors are determined by statistical methods.
Random errors cannot be minimised by taking precautions.
Key Insights:
■ Almost all measurements contain both systematic and random errors.
Gross Errors
Gross errors are the result of sheer carelessness on the part of the observer.
Causes of gross errors are (a) neglect of the sources of error and (b) reading the instrument incorrectly.
Examples:
1) In a tangent galvanometer experiment, the coil is to be placed exactly in the magnetic meridian and care should be taken so that no other magnetic material is present in the vicinity.
2) In the measurement of length of a pendulum, the observer may read it as 52.4 cm but may record it as 25.4 cm. This is gross error due to careless recording.
No corrections can be applied to these gross errors. Taking all precautions that are required, we can minimise gross errors.
Least Count Error
The smallest value that can be measured by the measuring instrument is called its least c ount. All the readings or measured values are good only up to this value.
The least count error is the error associated with the resolution of the instrument. For example, a vernier callipers has the least count as 0.01 cm; a spherometer may have a least count of 0.001 cm. Least count error belongs to the category of random errors but within a limited size; it occurs with both systematic and random errors. If we use a metre scale for measurement of length, it may have
graduations at 1 mm division scale spacing or interval.
By using instruments of higher precision and improving experimental techniques, the least count error can be reduced. When a large number of observations are taken, the arithmetic mean would be very close to the true value.
True Value
The arithmetic mean of a large number of repeated measurements is taken as the true value or actual value.
If x 1 , x 2 ............ x N are the results of an experiment repeated N times, the arithmetic mean X is given by X xx x N x N N i N 12 1 ....
Key Insights:
Formulae for least counts:
1) Vernier Callipers
LC Main Scale Division No. of Vernier Scale Divisions = 1
or, LC = 1 MSD – 1 VSD
2) Screw Gauge
Pitch of Screw Distance Moved No. of Rotations
LC = Pitch of S = c crew No. of Head Scale Divisions
2.3.4 Absolute Error, Relative Error, and Percentage of Error
The same error can be expressed in different ways as an absolute error, a relative error, or a percentage error.
a) Error (Dx): Difference between measured value and actual value (or true value) is called error Dx = measured value – actual value.
So, it can be +ve or –ve.
Example:
Suppose, true length of a body is 2.5 cm. If it is measured as 2.7 cm, then the error is + 0.2 cm.
If it is measured as 2.2 cm, then the error is – 0.3 cm.
b) Absolute error (|D x|): The magnitude of the difference between the true value of the measured physical quantity and the value of individual measurement is called the absolute error of that measurement.
If x mean = Mean value, and xi = ith observed value, then the absolute error (| D x|) in the ith observed value is defined as: |D xi| = |xi – xmean|
Absolute error is always positive and has same units as that of the quantity being measured.
c) Mean absolute error ( D x mean ): The arithmetic mean of all the absolute errors is considered the mean absolute error or final absolute error of the value of the physical quantity concerned.
Suppose x1, x2, x3, ...xn are measured values of a quantity and their mean value is xmean.
Each individual observed value has an error attached with it, as follow s.
D x1 = |x1 – xmean|
D x2 = |x2 – xmean|
D x 3 = |x3 – xmean|
D x n = |xn – xmean|
D x mean = x n i i n 1
So, we write actual value, x = xmean ± Dx mean
d) Relative error: The relative (or proportional) error of a measured physical quantity is the ratio of the mean absolute error to the mean value of the quantity measured.
Relative error =
Mean absolute error
Mean value x x mean mean
Relative error is a pure number having no units.
Percentage error = Relative error × 100 x x mean mean 100
2. In an experiment, the values of refractive index of glass were found to be 1.54, 1.53, 1.44, 1.54, 1.56 and 1.45 in successive measurements.
Calculate
(i) Mean value of refractive index of glass
(ii) Absolute error in each measurement
(iii) Mean absolute error
(iv) Relative error and (v) Percentage error.
Sol (i) Mean value of refractive index, mean 1541 53 1441 54 1561 45 6 =1.51
(ii) Taking µ mean as the true value, the errors in the six measurements are, 1.54 – 1.51 = +0.03, 1.53 – 1.51 = +0.02, 1.44 – 1.51 = –0.07, 1.54 – 1.51 = + 0.03, 1.56 – 1.51 = + 0.05 and 1.45 – 1.51 = – 0.06. The absolute errors are 0.03, 0.02, 0.07, 0.03, 0.05, and 0.06.
(iii) Mean absolute error in the value of µ is mean 0030 02 0070 03 0050 06 6 == 026 6 004
(iv) Relative error in the value of µ, mean mean 004 151 0 02649 ..003
(v) Percentage error in the value of µ = 0.03 × 100 = 3%
Try yourself:
2. Readings of length of a pole are 2.63 m, 2.56 m, 2.42 m, 2.71 m, and 2.80 m. Calculate the mean absolute errors and relative errors and percentage errors. What do you think of the actual value of the length and its limits?
Ans: 0.11 m, 0.04, 4% and (2.51 m to 2.73 m)
2.3.5 Combination or Propagation of Errors
Suppose, we want to calculate acceleration due to gravity using the formula,
g = 4 2 2 p L T
We can measure length L using scale and we can measure time period T using clock.
There may be errors in all measuring instruments and measurements.
We should know the errors in individual measurements of length ( L ) and time ( T ), then combine them to find the error in the calculation of ‘g’.
So, we must know how errors combine (or propagate) in addition, subtraction, multiplication and division.
Case (i)
Combination of errors in case of addition:
Suppose, a physical quantity Z is the sum of the physical quantities X and Y.
Z = X + Y
Let D X and D Y be the respective absolute errors in X and Y. Then, the values of X and Y will be X ± D X and Y ± D Y.
Let the error in Z be D Z.
Since Z = X + Y,
Z ± D Z = (X ± D X) + (Y ± D Y)
Z ± D Z = (X + Y) ± D X ± D Y
Maximum possible error in Z is
D Z = D X + D Y
Percentage error in Z XY X + Y 100
Case (ii)
Combination of errors in case of substraction:
Suppose, a physical quantity Z is the difference of the physical quantities X and Y.
Z = X – Y
Let D X and D Y be the respective absolute errors in X and Y. Then, the values of X and Y will be X ± D X and Y ± D Y.
Let the error in Z be D Z.
Since Z = X – Y,
Z ± D Z = (X ± D X) – (Y ± D Y)
Z ± D Z = (X – Y) ± D X ± D Y
Maximum possible error in Z is D Z = D X + D Y.
Percentage error in Z
Key Insights:
100 X – Y
■ Wh en two quantities are added or subtracted, the maximum possible error in the final result is the sum of absolute errors in the individual quantities.
3. If L = 2.06 cm ± 0.02 cm and B = 1.11 cm ± 0.03 cm, what are the values of (L + B) and (L – B)?
Sol. L + B = 3.17 cm ± 0.05 cm
L – B = 0.95 cm ± 0.05 cm
Please note that actual values, i.e., 2.06 cm and 1.11 cm, are added in case of (L + B) and subtracted in case of (L – B), but absolute errors are added in both cases.
Try yourself:
3. Two objects A and B are of lengths 5 cm and 7cm, determined with errors 0.1 cm and 0.2 cm, respectively. What is the error in determining (a) the total length and (b) the difference in their lengths?
Ans: (a) (12 ± 0.3); (b) (2 ± 0.3) cm
Key Insights:
■ If a ‘constant’ is multiplied with an observed/measured value, the result will have its absolute error equal to ‘constant’ times the absolute error in the observed/ measured value. This means, the percentage error in the result will remain unchanged.
4. If L = 2.01 m ± 0.01 m, what is 3L?
Sol. Here, we should equate 3L with (L + L + L) to find out the absolute error in 3L.
Thus, 3L = (3 × 2.01) m ± 3 × 0.01 m = 6.03 m ± 0.03 m.
Try yourself:
4. If L1= 2.02 m ± 0.01 m, L2 = 1.02 m ± 0.01 m, determine L1 + 2L2
Ans: 4.06 m ± 0.03 m
Case (iii)
Combination of errors in case of multiplication:
Suppose, a physical quantity Z is the product of the physical quantities X and Y.
Z = XY
Let D X and D Y be the respective absolute errors in X and Y. Then, the values of X and Y will be X ± D X and Y ± D Y.
Y will be X ± D X and Y ± D Y.
Let the error in Z be D Z
Z = XY
logZ = logX + logY
Differentiating both sides, dZ Z = dX X + dY Y
Maximum relative error: DDD Z Z = X X + Y Y
Case (iv)
Combination of errors in case of division:
Suppose X, Y, Z are physical quantities such that Z= X Y
Let D X and D Y be the respective absolute errors in X and Y. Then, the values of X and Y will be X ± D X and Y ± D Y
Let the error in Z be D Z.
Z= X Y
logZ = logX – logY
Differentiating both sides, dZ Z = dX XdY Y
Maximum relative error DDD Z Z = X X + Y Y
Key Insights:
■ Both in multiplication and division, maximum relative error in the result is the sum of the relative errors of the quantities being multiplied or divided. Similarly, both in multiplication and division, Maximum percentage error in the result is the sum of the percentage errors of the quantities being multiplied or divided.
5. The length and breadth of a rectangular object are 25.2 cm and 16.8 cm, respectively, and have been measured to an accuracy of 0.1 cm. Find the relative error and percentage error in the area of the object.
Sol Area A = l × b; l = 25.2 cm and D l = 0.1 cm b = 16.8 cm and D b = 0.1 cm
\ Relative error in area
b b = 0.1 25.2 + 0.1 16.8
= 0.004 + 0.006 = 0.01
Percentage error = 1%.
6. In an experiment, to determine the value of acceleration due to gravity g using a simple pendulum, the measured value of length of the pendulum is 31.4 cm known to 1 mm accuracy and the time period for 100 oscillations of pendulum is 112.0 s, known to 1 resolution. Find the accuracy in determining the value of g.
Sol (Accuracy is to be taken as the error involved)
l = 31.4 cm and D l = 1mm = 0.1 cm
T= 112.0
100 =1.12 s and T=0.01 s D Formula for g is
g = 4 T 2 2
g g =+2 T T 01 31 4 2 001 112 . . . .
= 0.003 + 0.02 = 0.023
Relative error in determining g is 0.023 and percentage error = 0.023 × 100 = 2.3%
Try yourself:
5. L = 20.04 m ± 0.01 m B = 2.52 m ± 0.02 m, what are the values of (L× B) and (L/B)?
Ans: 50.50 2m ± 0.42 ,2m 27.95m ± 0.07 2m
Key Insights:
■ The result should be rounded to which place of decimal is explained later in this chapter.
Similarly,
L B = 20.04 m 2.52 m 0050 79
7950 845 795 084 100 795 .. %
7950 067
Case (v)
Combination of errors in case of a quantity raised to different powers:
Suppose the physical quantity Z depends on the quantity X as Z = kX n, where k is a constant and n is a real number.
Let D X and D Z be the respective absolute errors in X and Z.
Let Z = kXn log Z = log k + n log X
Differentiating both sides, DD Z Z =n X X
Maximum relative error, DD Z Z =n X X
Percentages error in Z = n times percentage error in X
Similarly, Let Z = A.B C pq r
Where A, B, C and Z are variables, and p, q and r are constants.
log Z = p log A + q log B – r log C
Differentiating both sides, dZ Z = p. dA A +q. dB B -r. dC C Z Z =p A A +q B B -r C C DDDD
Maximum fractional error in Z is DDDD Z Z =p A A +q B B +r C C
Key Insights:
■ All quantities are positive, Maximum % error in Z = p (%error in A) + q(% error in B)+ r(% error in C)
7. A rectangular metal slab of mass 33.333 g has its length 8.0 cm, breadth 5.0 cm and thickness 1 mm. The mass is measured with accuracy up to 1 mg with a sensitive balance. The length and breadth are measured with a vernier caliper having a least count of 0.01 cm. The thickness is measured with a screw gauge of least count 0.01 mm. Calculate the percentage accuracy in density calculated from above measurements.
Sol The percentage error itself gives us the percentage accuracy.
d= mass(M) length (l) breadth (b)height (h) = M lbh ××
The relative error is given by d d M M b b h h
Given, ℓ = 8.0 cm and D ℓ = 0.01 cm; b = 5.0 cm and D b = 0.01 cm
h = 1 mm and D h = 0.01 mm and M = 33.333 g and M = 1 mg = 0.001 g
Hence, DM M = 0.001 33.333 001 80 001 5 . . , . b b and h h 001 1
The percentage error is given by M m b b h h 100 0 001 33 333 001 80 001 5 001 1 . . . . .. 100
= 0.003+0.125+0.2+1.0 = 1.328. 1.328 ≅ 1.3
The percentage error is 1.3%
8. Two resistors of resistances R1 = 100 ± 3 and R2 = 200 ± 4 ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination. Use for (a) the relation R = R1 + R2 and for (b) 1 R = 1 R + 1
Sol (a) The equivalent resistance of series combination R = R1 +R2 = (100 ±3) ohm + (200 ± 4) ohm = 300 ± 7 ohm (b) The equivalent resistance of parallel combination
′ R= RR R+R = 200 3 =66.7ohm 12 12
Then, from 1 R = 1 R + 1 R 12 ′ , we get,
Then, R= 66.7 1.8 ohm (Here, DR is expressed as 1.8 instead of 2 to keep in conformity with the rules of significant figures.)
Try yourself:
6. The period of oscillation of a simple pendulum is T 2 Lg/. Measured value of L is 20.0 cm known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90 s using a wrist watch of 1 s resolution. What is the accuracy in the determination of g?
Ans: 2.7%
TEST YOURSELF
1. If absolute error is 0.05 m for a measured length of 5 m. What is the percentage error? (1) 1% (2) 2% (3) 3% (4) 4%
2. The measured mass and volume of a body are 53.63 g and 5.8 cm 3 respectively, with possible errors of 0.01 g and 0.1 cm 3. The maximum percentage error in density is about (1) 0.2% (2) 2% (3) 5% (4) 10%
3. In the measurement of volume of solid sphere using the formula Vr 4 3 3 if the error committed in the measurement of the radius r is 1.5%, the percentage error in the volume measurement is (1) 3% (2) 1.5% (3) 4.5% (4) 0.75%
4. A physicist performs an experiment and takes 200 readings. He repeats the same experiment and now takes 800 readings. By doing so the probable error (1) remains the same (2) is four times (3) is halved (4) is reduced by a factor of 1/4
5. The percentage error in the measurement of mass and speed are 2% and 3% respectively. Maximum estimate of percentage error of K.E is (1) 11% (2) 8% (3) 5% (4) 1%
6. Zero errors of measuring instruments are called (1) Indeterminate errors (2) Random errors (3) Disproportional errors (4) Instrumental errors
7. If the length of a cylinder is measured to be 6.42 cm with an error of 0.02 cm. The percentage error in the measured length is nearly (1) 0.4% (2) 0.3% (3) 0.2% (4) 0.1%
8. In an experiment, four quantities , b, c and d are measured with percentage error 1%, 2%, 3% and 4% respectively. Quantity P is calculated as follows P ab cd = 32 .% error in P is (1) 7% (2) 4% (3) 14% (4) 10%
2.4 SIGNIFICANT FIGURES
The digits of a number, representing a measurement that is definitely known, plus one more digit added at the end which is estimated, are called significant digits or significant figures.
Example: Suppose least count of a scale used to measure length of a rod is 0.1 cm.
Suppose one end of the rod is coinciding with zero of the scale and the other end is between two marks reading 14.5 cm and 14.6 cm.
So, length of the rod is more than 14.5 cm but less than 14.6 cm.
One observer may think that the correct measurement is perhaps 14.52 cm.
Here 1, 4 and 5 are reliable and last digit 2 is a guess and, thus, not reliable.
So, in 14.52 cm, there are 4 significant figures.
The last significant figure is called the ‘first doubtful figure’ or the ‘least significant figure’. It is also called ‘uncertain digit’. It is 2 in 14.52 cm.
2.4.1 Rules to Count Significant Figures
1) All non–zero digits in a given number are significant without any regard to the location of the decimal point, if any.
6,482 has four significant digits.
648.2 or 64.82 or 6.482 all have the same number (four) of significant digits.
2) All zeros occurring between two non–zero digits are significant without any regard to the location of decimal point, if any.
206008 has six significant figures.
206.008 or 2.06008 has also got six significant digits.
3) If the number is less than one, all the zeros to the right of the decimal point but to the left of the first non–zero digit are not significant.
In the number 0.000608, the three zeros to the right of decimal point and up to 6 are not significant. The zero that is conventionally placed before decimal point is obviously not significant. The given number has three significant figures. The zero between 6 and 8 is significant from rule 2.
4) (a) All zeros to the right of a decimal point are significant if the number is more than one. 60.00 contains four significant digits.
4) (b) All zeros to the right of the last non–zero digit after the decimal point are significant.
0.007800 has four significant figures. The zero between decimal point and 7 is not significant, from rule-3. But the zeroes after eight are significant. Otherwise it would have been reported as 0.0078, only with two significant digits.
5) (a) All zeros to the right of the last non–zero digit in a number having no decimal point are not significant.
4060 contains three significant figures as the zero to the right of 6 is not significant.
40,6000 has also got only three significant figures.
5) (b) However, there is an exception to rule 5(a). All zeros to the right of the last non–zero digit in a number having no decimal point will be significant if they come from an actual measurement.
Suppose a distance measured in metres is reported as 4.500 m, having four significant figures. Expressed in cms, it will be 450 cm. It has only two significant figures. To have four significant figures, we should express it as 450.0 cm. Similarly, if we want to express the value in km, we have to write the result as 0.004500 km and not as 0.0045 km. When expressed in mm, the value becomes 4500 mm. It has only two significant figures from rule 5(a). Now arises a problem. We cannot write 4500 mm in any way, as we have done with cm and km, to have four significant figures.
To avoid such ambiguity, we write the number in scientific notation as powers of 10. The exponent or power of 10 is called order of magnitude.
Now, 4500 mm is written as 4.500×103 mm and by rule 4(a), the number has 4 significant figures and order of magnitude 3. Similarly, the result is expressed as 4.500 ×10 2 cm with
order of magnitude 2 and 4.500 ×10–3 km with order of magnitude –3.
Key Insights:
■ Very large and very small quantities of physical measurement are usually expressed in scientific notation to denote accuracy of measurements.
■ A choice of change of different units doesn’t change the number of significant digits or figures in a measurement.
9. Write down the number of significant figures in the following.
(i) 0.007 (ii) 2. 64 × 1024
(iii) 0.2370 (iv) 6.320
(v) 6.032 (vi) 0.0006032
Sol (i) 0.007 has one significant figure.
(ii) 2. 64 × 1024 has three significant figures.
(iii) 0.2370 has four significant figures.
(iv) 6.320 has four significant figures.
(v) 6.032 has four significant figures.
(vi) 0.0006032 has four significant figures.
Try yourself:
7. Write down the number of significant figures in the following:
(a) 6428 (b) 62.00 m (c) 0.00628 cm (d) 1200 N
Ans: (a) 4 (b) 4 (c) 3 (d) 4
2.4.2 Rules for Arithmetic Operations with Significant Figures
The final result of arithmetic operations should never have more significant figures than the least number of significant figures in the original components.
This results in the following rules.
Rule 1: In addition or subtraction, the final result should retain only that many decimal places as are there in the number with the least decimal places.
Key Insights:
■ In addition and subtraction, the rule is in terms of decimal places (and not in terms of significant figures).
Rule 2 : In multiplication or division, the final result should retain only that many significant figures as are there in the original number with the least number of significant figures.
2.4.3 Rounding off the Uncertain Digits
The process of omitting the non–significant digits and retaining only the desired number of significant digits, incorporating the required modifications to the last significant digit, is called ‘rounding off the number’.
Rules for Rounding off Numbers
i) Th e preceding digit is raised by 1 if the immediate insignificant digit to be dropped is more than 5.
Example: 5728 is to be rounded off to three significant figures. Here, 8 is the non–significant digit to be dropped, and it is more than 5. Hence, 5728 = 5730.
ii) The preceding digit is to be left unchanged if the immediate insignificant digit to be dropped is less than 5.
Example: 5728 is to be rounded off to two significant figures. Here, the immediate or first non–significant digit to be dropped is 2 (along with 8 also), and it is less than 5. Hence, 5728 = 5700. In the above two examples, 1 and 2, zeros are to be placed in the places of digits dropped. Otherwise, the values will change drastically.
iii) If the immediate insignificant digit to be dropped is 5, then there will be two different cases.
(a) If the preceding digit is even, it is to be unchanged, and 5 is dropped.
Example: 6.7258 is to be rounded off to two decimal places (or three significant
figures). The digit to be dropped here is 5 (along with 8) and the preceding digit 2 is even. Hence, it is to be retained as two only.
6.7258 = 6.72
(b) If the preceding digit is odd, it is to be raised by 1. 6.7158 is to be rounded off to two decimal places (or three significant figures). As the preceding digit 1 is odd, it is to be raised by 1 as 2.
6.7158 = 6.72
In the above two examples, iii(a) and iii(b), zeros should not be placed in places of decimal digits dropped. Otherwise, the number of significant digits will drastically change.
10. Round off to 3 significant figures:
(i) 20.96 (ii) 0.0003125
Sol (i) 20.96 has four significant figures. The fourth significant figure is more than 5 and, hence, on rounding off to three significant figures, the given measurement will become 20.9 + 0.1 i.e., 21.0.
(ii) 0.0003125 has four significant figures. The fourth significant figure is 5 and hence on rounding off to three significant figures, the given measurement will become 0.000312 or 3.12×10–4. This is because 2 before 5 is an even number
Try yourself:
8. Round off to four significant figures: (a) 45.689 (b) 2.0082
Ans: (a) 45.69 (b) 2.008
2.4.4 Rules for Determining the Uncertainty in the Result of Arithmetic Calculations
(i) Addition and subtraction
In addition or subtraction of numbers, before performing the operation we must identify the
number with the least number of digits after decimal point (say n ) and round off other numbers to one extra digit in the decimal part (say n + 1). The final result is again rounded off, such that it contains same number of decimal places as that of the number with least number of decimal places (say n) among all other numbers.
11. A stick has a length of 12.132 cm and another stick has a length of 12.4 cm.
(a) If the two sticks are placed end to end, what is their total length?
(b) If the two sticks are placed side by side, what is the difference in their lengths?
Sol
(a) Let lengths of the sticks are named as
l1 = 12.132 cm, l2 = 12.4 cm
Here, l2 has one decimal place and l1 has to be rounded off to have only two decimal places.
l1 + l2 = 12.13 + 12.4 = 24.53
This is to be rounded off to have one decimal place only.
\ The total length is 24.5 cm
(b) l1 = 12.132, l2 = 12.4.
l1 – l2 = 12.4 – 12.132
Here 12.4 has only one decimal place and hence 12.132 should have only two decimal places.
\ l1 - l2 = 12.4 - 12.13 = 0.27
This should be rounded off to have only one decimal place.
\ l1 - l2 = 0.3
Hence difference of their lengths is 0.3 cm.
12. Find the value of 2.2 + 4.08 + 3.125 + 6.3755.
Sol Out of all the four numbers, 2.2 has got the least number of decimal places i,e. one.
Hence we should retain only two decimal places in the remaining numbers. Hence, 4.08 remains as it is. 3.125 is to be rounded off as 3.12 (as 2 before 5 is even), 6.3755 is to be rounded off as 6.38 (as 7 before 5 is odd). Now, adding 2.2 + 4.08 + 3.12 + 6.38 = 15.78.
Finally, we should have only one decimal place and, hence, 15.78 is to be rounded off to 15.8
Try yourself:
9. Solve with due regard to significant figures. (i) 46.7 – 10.4 = ...........
Ans: (i) 36.7 (ii) 4.5 × 10-6
(ii) Multiplication and division
In multiplication or division of numbers, before performing the operation, we must identify the number with the least number of significant figures (say n) and round off other numbers to one extra digit (say n + 1). The final result is again rounded off, such that it contains same number of significant figures (say n) as that of the number with least number of significant figures among all other numbers.
Key Insights:
■ The final res ult of arithmetic operations should never have more significant figures than the least number of significant figures in the original components.
(iii) In square root value of the given number, there can be same number of significant figures that are present in the given number.
Example: 58 97 7 ..679 =
(iv) In case of transcendental functions:
Transcen dental functions have the same number of significant figures as their arguments.
Non-algebraic functions like sine, sin –1 , exponential and logarithmic functions are called transcendental functions.
Arguments are quantities on which some operation is performed, e.g., in case of sin 30°, 30° is argument; in case of log x, x is argument, etc.
Example:
sin (60°) = 0.86
sin (60.00°) = 0.8660
sin (60.0°) = 0.866
sin (60.000°) = 0.86603
13. Find the product of 1.2, 2.54 and 3.257 with due regard to significant figures.
Sol Out of the three numbers, 1.2 has got the least number of significant figures-two. We should round off the other numbers to 2 + 1 = 3 significant figures and carry the multiplication. 2.54 has three significant figures and, hence, needs no rounding off. 3.257 is to be rounded off to 3.26.
1.2 × 2.54 × 3.26 = 9.93648. But the result should be limited to the least number of significant digits, i.e., two digits only. The final answer should be written as 9.9 after rounding off.
14. Find the value of p /53.2 with due regard to significant figures.
Sol. Out of the two numbers, 53.2 has three significant digits. So, p should be written with 3 + 1 = 4 significant figures.
p = 3.1415 = 3.142
(as 1 is odd it is raised by one)
Now, 3 142 53 2 0 0590601 . . . =
This is to be rounded off to three significant figures.
0.059060 = 0.0591
Try yourself:
10. Find out the results of the following operations.
(i) 117.3 × 0.0024 (ii) 9.27 ÷ 41 (iii) 42 × 0.041 (iv) 124.2 + 52.487 (v) 124.2 – 52.487 (vi) 58 97 (vii) (17.5)2 Ans: (i)
TEST YOURSELF
1. Write down the number of significant figure of 6.032 (1) 4 (2) 3 (3) 2 (4) 6
2. The significant figures in 300.500 are (1) 6 (2) 5 (3) 4 (4) 3
3. The number of the significant figures in 11.118 × 10–6 is (1) 3 (2) 4 (3) 5 (4) 6
4. Product of 2.78 and 0.996 is divided by 1.527, the result has significant figures (1) one (2) two (3) three (4) four
5. When 0.0003125 is reduced to 3 significant figures, its value is (1) 0.00312 (2) 3.125 (3) 0.000312 (4) 0.000313
6. Round off to 3 significant figures a) 20.96 and b) 0.0003125 (1) 21.0; 312 × 10–4 (2) 21.0; 3.12 × 10–4 (3) 2.10; 312 × 10–4 (4) 210; 3.12 × 10–4
7. If 3.8 × 10–6 is added to 4.2 × 10–5 giving due regard to significant figures, then the result will be:
(1) 4.58 × 10–5 (2) 4.6 × 10–5
(3) 4.5 × 10–5 (4) none of these
8. The sum of the numbers 436.12, 227.2 and 0.301 in appropriate significant figure is (1) 663.621 (2) 664
(3) 663.6
(4) 663.62
Answer Key
(1) 1 (2) 1 (3) 3 (4) 3
(5) 3 (6) 2 (7) 2 (8) 3
2.5 MEASURING INSTRUMENTS
To measure very small length accurately two instruments are mainly used in the laboratory. They are Screw gauge and Vernier Calipers.
2.5.1
Screw Gauge (Micrometer)
Description
of Screw Gauge
The object to be measured is put between the jaws. The sleeve is hollow part, fixed with the frame, and main scale is printed on it.
The spindle and thimble are welded, and move together by means of a screw. The circular scale is printed on the thimble, as shown. It generally consists of 100 divisions (sometime, 50 divisions also).
The main scale has mm marks (Sometimes, it also has 1/2 mm marks below mm marks).
(Usually, if pitch of the screw gauge is 1 mm, then before there are 1 mm marks on main scale and if pitch is 1/2 mm, then there are 1/2 mm marks also.)
This instrument can read upto 0.01 mm (10 µ m) accuracy, that is why it is called micrometer.
Screw gauge is used to measure closely upto
How can it divide 1 mm in 100 parts?
To divide 1 mm in 100 parts, a screw is used.
In one rotation, the screw (spindle) moves forward by 1 mm. (Called pitch of the screw)
The rotation of the screw (spindle) is divided in 100 parts (called circular scale).
Hence, 1 mm is divided in 100 parts.
20 10 90 80 Spindle 0
1 rotation = 1 mm, 100 circular parts = 1 mm
So 1 circular part = 1 100 mm
= Least count of screw gauge
So, let’s generalise it.
Least count of a screw gauge = 1 100 mm
Pitch of screw
Number of divisions on circular scale
How to Find Thickness by a Screw Gauge?
Object objective is inserted between the jaws
47 circular part = 1 mm 1 circular part =
Objective thickness = 7.------= 7 mm + 47
)
Main scale Reading Circular scale Reading Least Count 1 mm 100 = 7 mm + 47 circular divisions
Thickness of object
Reading of Screwgauge = = + Pitch Numbers of divisions on Circular Scale Least Count Circular scale Reading Main scale Reading
15. A wire of resistance R = 100.0 Ω and length l = 50.0 cm is put between the jaws of screw gauge. Its reading is shown in figure. Pitch of the screw
gauge is 0.5 mm and there are 50 divisions on circular scale. Find its resistvity in correct significant figures and maximum permissible error in ρ (resistivity).
40
054 () .( .%%)
Try yourself:
11. To measure diameter of a wire, a screw gauge is used. The main scale division is 1 mm. In a complete rotation, the screw advances by 1 mm and the circular scale 100 divisions. The reading of screwgauge is as shown in figure. (mm) object
If there is no error in mass measurement, but error in length, measurement is 1%, then find max. possible error in density.
Ans: 1.65%
Zero Error
If there is no object between the jaws (i.e., jaws are in contact), the screwgauge should give zero reading. But due to extra material
on jaws, even if there is no object, it gives some excess reading. This excess reading is called zero error. 0 0 10
Excess reading = 0.07 mm (zero error)
If we put a wire between jaws object 90 0
It gives 3.07 mm, in which there is 0.07 mm excess reading, which has to be subtracted. So, actual thickness,
3070 ..07
Observed reading Excess reading Zero error
300 . mm
Actual reading = observedexcess reading readding zero error
16. Find the thickness of the wire. The main scale division is of 1 mm. In a complete rotaion, the screw advances by 1 mm and the circular scale has 100 divisions. 0 0 10
If we put a wire between jaws
Sol: Excess reading (Zero error) = 0.07 mm
It is given 7.67 mm in which there is 0.07 mm excess reading, which has to be removed (substracted)
So, actual reading = 7.67 – 0.07 = 7.60 mm.
17. Find the thickness of the wire. The main scale divisionis of 1 min. In a complete rotation, the screw advance by 1 mm and the circular scale has 100 divisions. If no object is placed between the jaws, the zero of main scale is barely visible and 93 rd circular division matches with the main scale line . object 90 0
If we put a wire between jaws (mm) object 0 90 0
Sol: Excess reading (Zero error) = (–1mm )+ (93) (0.01) = – 0.07 mm
It is giving 7.95 mm in which there is –0.07 mm excess reading, which has to be removed (subtracted) so actual reading = 7.95 – (–0.07) = 8.02 mm
Try yourself:
12. Find the thickness of the wire. The main scale division is of 1 mm. In a complete rotaion, the screw advances by 1 mm and the circular scale has 100 devisions. If no object is placed between the jaws, the zero of main scale is barely visible and 97rd circular matches with the main scale line.
If we put a wire between jaws
object
On the upper plate, main scale is printed, which is simply an mm scale.
On the lower plate, vernier scale is printed, which is a bit compressed scale. Its one part is of 0.9 mm.
(10 vernier scale divisions = 9 mm ⇒1 vernier scale division = 0.9 mm)
The object which is to be measured is fitted between the jaws, as shown.
How to Read Vernier Callipers
Ans: 9.00 mm
Zero Correction
Zero correction is invert of zero error.
Zero Correction = – (zero error)
Actual reading = Observed reading – zero error = Observed reading + zero correction
2.5.2 Vernier Callipers
It is us used to measure accuratly upto 0.1 mm.
Thickness of the object = 24. ------= 24 mm + 7 (0.1 mm) + –Main Scale Reading Vernier Scale Reading least count
Formula says Thickness of the object = 1 main Scale devision 1 vernier Scale devision
Mark on vernier scale which exactly co-insides with some mark of main = 1 mm – 0.9 mm = 0.1 mm
Now lets see how the slight difference between 1 MSD and 1 VSD reflects as least count
18. Read the vernier. 10 division of vernier scale are matching with 9 divisions of main scale.
Required length = 13 mm + x = ?
At point ‘A’, main scale and vernier scale are matching.
So, length OA along main scale = length OA along vernier scale
13mm + 3 (main scale division) = (13 mm + x) + 3(vernier scale division)
Get 13 mm + x = 13 mm + 3(Main sclae division - vernier scale division)
= 13 mm + 3(1 mm - 0.9 mm)
= 13 mm + 3 (0.1 mm) = 13.3 mm
(1 main scale division – 1 vernier scale division) + Main Scale Reading Vernier Scale Reading least count
Hence, the slight difference between 1 MSD (1 mm) and 1 VSD ( 0.9 mm) reflects as least count (0.1 mm)
Thickness of object = Reading of vernier callipers + Main Scale Reading Vernier Scale Reading least count =
Sol: 10 vernier scale divisions = 9 mm 1 vernier scale division = 0.9 mm
⇒ Least count = (Main scale division - vernier scale division)
= 1 mm - 0.9 mm = (from figure)
= 0.1 mm
Thickness of the object = (main scale reading ) + (vernier scale reading )(least count)
So, thickness of the object = 15 mm + (6) (0.1 mm) = 15.6 mm
Try yourself:
13. Read the special type of vernier. 20 divisions of vernier scale are matching with 19 divisions of main scale.
Ans: 13.7 mm
Zero Error
Even if there is no object between the jaws, it gives some excess reading. This excess reading is called zero error.
19. In the vernier callipers, 9 main scale divisions match with 10 vernier scale divisions. What will be the thickness of the object using the defected vernier callipers
Sol: From first figure, excess reading (zero error) = 0.6 mm. If an object is placed, vernier gives 14.6, mm in which there is 0.6 mm excess reading, which has to be subtracted. So, actual thickness = 14.6 - 0.6 = 14.0 mm.
we can also do it using the formula,
Actual reading = observed reading - excess reading (zero error)= 14.6 - 0.6 = 14.0 mm
Try yourself:
14. The least count of main scale is 1 mm. In the vernier calliper, 9 main scale divisions matches with 10 vernier scale divisions. When no object is placed between the jaws, the zero of vernier scale is slightly behind the zero of main scale. When a sphere is placed between the jaws, the reading of the vernier is shown in the figure. The thickness of the object using the defected vernier calliper will be
Zero Correction:
Zero correction is invert of zero error. Zero correction = –(zero error)
Actual reading = observed reading–excess reading (zero error) = Observed reading + zero correction
Example: If zero error was 0.6 mm, zero correction will be – 0.6 mm
Example: If zero error was – 0.4 mm, zero correction will be + 0.4 mm
20. The main scale of a vernier callipers reads 10 mm in 10 divisions. 10 divisions of vernier scale coincide with 9 divisions of the main scale. When the two jaws of the callipers touch each other, the fifth division of the vernier coincides with some main scale divisions and the zero of the vernier is to the right of the zero of the main scale. When a cylinder is tightly placed between the two jaws, the zero of vernier scale lies slightly behind 3.2 cm and the fourth vernier division coincides with a main scale division. What is the diameter of the cylinder?
Sol: Zero error = 0.5 mm = 0.05 mm
Observed reading of cylinder diameter = 3.1 mm + (4) (0.01 cm) = 3.14 cm
Actual thickness of cylinder = (3.14) – (0.05) = 3.09 cm
Try yourself:
15. In the previous question, if the length of the cylinder is measured as 25 mm, and mass of the cylinder is measured as 50.0 gm, find the density of the cylinder (g/cm3) in proper significant figures.
Ans: 2.7 g/cm3 (in two S.F.)
TEST YOURSELF
1. The vernier constant of a travelling microscope is 0.001 cm. If 49 main scale divisions coincide with 50 vernier scale divisions. Then the value of 1 main scale division is (1) 0.1 mm (2) 0.5 mm (3) 0.4 mm (4) 1 mm
2. A screw gauge gives the following reading when used to measure the diameter of a wire. Main scale reading : 0 mm Circular scale reading : 52 divisions Given that 1 mm on main scale corresponds to 100 divisions of the circular scale. The diameter of wire from the above data is:
(1) 0.052 cm
(2) 0.026 cm
(3) 0.005 cm
(4) 0.52 cm
3. When the circular scale of a screw gauge completes 2 rotations, it covers 1 mm over the pitch scale. The total number of the circular scale divisions is 50. The least count of the screw gauge in metre is (1) 10–5 (2) 10–2 (3) 10–3 (4) 10–2
4. A vernier callipers has 1 mm marks on the main scale. It has 20 equal divisions on the vernier scale, whcih match with 16 main scale divisions. For this vernier callipers, the least count is :
(1) 0.02 mm (2) 0.05 mm (3) 0.1 mm (4) 0.2 mm
2.6 DIMENSIONS OF PHYSICAL QUANTITIES
Definit ion: The dimensions of a physical quantity are the powers (exponents) to which the fundamental (base) quantities are raised to represent that physical quantity.
Note that using the square brackets [ ] around the quantity means that we are dealing with the dimensions of the quantity. In mechanics, all the physical quantities can be written in terms of the dimensions [L], [M] and [T]
For example, the dimensional formula of force is [M1 L1 T–2]. The dimensions of mass is 1 and that of length is 1 and that of time is –2.
2.7 DIMENSIONAL FORMULAE AND DIMENSIONAL EQUATIONS
Definition: The expression showing the powers to which the fundamental quantities must be raised to represent a physical quantity is called dimensional formula.
The fundamental physical quantities, namely mass, length, time, temperature and electric current are indicated by the symbols [M], [L], [T], [K] or [θ], [I] or [A] etc., respectively.
Example:
1. Force = Mass × acceleration
So, dimensional formula of force is [ MLT–2 ] = [M1 L1 T–2]
In the dimensional formula of force, dimension of mass = 1, dimension of length = 1, dimension of time = –2.
2. Work = F.S = m a S;
So, dimensional formula of work = [M LT–2] L=[M1L2T–2]
In the dimensional formula of work, dimension of mass = 1, dimension of length = 2, dimension of time = –2.
3. Energy gradient = Energy Length
So, dimensional formula = ML T L =M LT 2-2 11 2
In the dimensional formula of energy gradient, dimension of mass = 1
dimension of length = 1 dimension of time = –2
2.7.1 Dimensional Equation
Definition: It is an equation equating the physical quantity with its dimensional formula.
The equation in which physical quantity is related to fundamental quantities raised to appropriate powers is called a dimensional formula.
Examples:
1) Force [F] = [MLT –2 ] is dimensional equation of force.
2) Work [W] = [ML 2 T –2 ] is dimensional equation of work.
2.7.2 Dimensional Constants and Dimensionless Quantities
Dimensional constants: Constants having dimensional formula are called dimensional constants.
Examples: Planck’s constant, universal gravitational constant, universal gas constant, Boltzmann’s constant, Stefan’s constant, Wien’s constant, speed of light
Dimensionless quantities: Physical quantities having no dimensional formula are called dimensionless quantities.
Ex amples: Angle, strain, relative density, coefficient of friction, coefficient of restitution, and Poisson’s ratio have zero dimensions and are dimensionless quantities. Numbers and ratio of same quantities have no dimensional formula because they do not depend on fundamental quantities
TEST YOURSELF
1. Which one of the following is the dimensionless quantity?
(1) momentum (2) force
(3) work (4) strain
2. Which of the following has the dimensions of pressure?
3. Plane angle and solid angle have:
(1) Dimensions but no units
(2) No units and no dimensions
(3) Both units and dimensions
(4) Units but no dimensions
4. Two physical quantities of which one is a vector and the other is a scalar having the same dimensional formula are
(1) work and energy
(2) torque and work
(3) impulse and momentum
(4) power and pressure
5. Choose the false statement
(1) Relative permittivity is a dimensionless constant
(2) Angular displacement has neither units nor dimensions
(3) Refractive index is dimensionless variable
(4) Permeability of vacuum is a dimensional constant
6. The dimensions of power are:
(1) [ML2/T3] (2) [ML2/T2] (3) [ML2/T] (4) [ML/T2]
7. The dimensions of a b in the equation P at bx 2 where P is pressure, x is distance and t is time, are (1) [M2LT–2] (2) [ML0T–2] (3) [ML3T–1] (4) [M0LT–3]
8. Given that ya t p qx sin , where t represents time in second and x represents distance in meter. Which of the following statement is true?
(1) Dimension of t is same as that of q
(2) Dimension of x is same as that of p
(3) Dimension of x is same as that of q
(4) Dimension of t is same as that of p
9. Obtain the dimensions of K in the equation WKx = 1 2 2 . Here, W is work, while x is change in length.
(1) [M1L0T–2] (2) [M0L1T–1] (3) [M1L1T–2] (4) [M1L0T–1]
10. The force is given in terms of time t and displacement x by the equation F = A cos Bx + C sin Dt. The dimensional formula of AD B is
(1) [M0LT–1] (2) [ML2T–3] (3) [M1L1T–2] (4) [M2L2T–3]
11. The amplitude of damped oscillation of mass m varies with time t as AA e bt m 0 2 . Then dimensional formula of ‘ b’ is
(1) M–1L1T–2 (2) ML–1T–2 (3) ML0T–1 (4) ML0T–2
12. If the centripetal force is of the form mavbrc, find the values of a, b and c
(1) a = 2, b = –1, c = 1
(2) a = 2, b = 1, c = 1
(3) a = 1, b = 2, c = –1
(4) a = 1, b = –1, c = 2
Answer Key
(1) 4 (2) 4 (3) 4 (4) 2 (5) 2 (6) 1 (7) 2 (8) 4 (9) 1 (10) 2 (11) 3 (12) 3
2.8 DIMENSIONAL ANALYSIS AND ITS APPLICATIONS
1) D ime nsional formulae can be used to convert one system of units into another system.
2) Dimensional formulae can be used to check the correctness of an equation.
3) Dimensional formulae can be used to derive relationship among different physical quantities.
4) Dimensional formulae can be used to find unit of a given physical quantity.
5) Di mensional formulae can be used to design our own new system of units.
To convert one system of units into another system, this is based on the fact that, for a given physical quantity, numerical value × unit = constant, i.e., N 1U1= N2U2.
So, when units change, numerical value also may change. To convert a physical quantity from one system to another, we write its units in terms of mass, length and time, and then convert into another system. Let the dimensional formula of a physical quantity be [MaLbTc]. Let n1 be its numerical value in one system in which units of fundamental quantities are M 1 , L 1 , and T 1 . Then, the numerical value n2 in another system in which units of fundamental quantities are M2, L2 and T2, is given by
21. Convert newton into dyne. Sol Dimensional formula of force is M 1L1T-2 1 N = n dynes ⇒ n = 1 N/dynes
Alternate method : [F] = [MLT–2] ; 1 newton=1 (kg)(m) (s)–2 = 1(103 g) (102 cm)(s)–2 = 105(g) (cm)(s)–2 1 newton = 105 dynes
Try yourself:
16. Convert the unit of work done from MKS system to CGS system.
Ans: 1 joule = 107 ergs
2.8.1 To Check the Dimensional Consistency of a Given Equation
This is based on the principle of homogeneity.
Principle of homogeneity of dimensions:
Statement: Only physical quantities having same dimensions can be added, subtracted, or equated.
If x = y + z is dimensionally correct and if x represents the physical quantity, the force, then y and z also must represent the same physical quantity, i.e., force. It means that the terms on both sides of a dimensional equation should have same dimensions.
If the dimensions of each term on both sides are same, the equation is dimensionally correct, otherwise, it is not correct.
22. Verify the correctness of the equation v = u
+ at
Sol In the dimensional form,
[L1T-1] = [L1T-1] + [L1T-2] [T1] = [L1T-1]
The dimension for L on both sides is 1
The dimension for T on both sides is –1
Hence, from the principle of homogeneity of dimensions, the given equation is correct.
Try yourself:
are same, the given equation is dimensionally correct.
Since, the dimensions of each term on both sides
17. Check the correctness of the formula S = ut + at2 where S is the distance, u is velocity, a is acceleration, and t is time. Ans: [L]
Key Insights:
■ We cannot say anything about the physical correctness of the formula. We know that the correct formula is
S = ut + ½ at2
Hence, we conclude that dimensional correctness is no guarantee for physical correctness of the formula. However, dimensional incorrectness guarantees the physical incorrectness of the formula. In spite of the above limitation, the method is still helpful to a great extent.
23. Consider the equation T g 2 and check whether it is correct or not.
Sol We know, dimensional formulae of T, ℓ, and g are [T], [L], [LT–2], respectively.
\ Dimensional formula of g = L LT =T =T -2 2
Dimensional formula of left side is also [T].
As 2 p is a dimensionless quantity, the dimensional formula of left hand side of the equation is equal to dimensional formula of right hand side. Hence, we conclude that the given equation is a correct one.
Try yourself:
18. Check the correctness of the relation s = ut + ½ at 2 , where u is initial velocity, a the acceleration, t the time and s the displacement.
Ans: LHS = RHS, formula is dimensionally correct
2.8.2
Deducing Relation among the Physical Quantities
Th is is also based on the principle of homogeneity. If one knows the quantities on which a particular physical quantity depends and if one guesses that this dependence is of product type, the method of dimensional analysis may be helpful in the derivation of the relation.
24. Derive an expression for the time period (T) of a simple pendulum which may depend upon the mass ( m ) of the bob, length ( l ) of the pendulum, and acceleration due t o gravity (g).
Sol. Let T = kma lb gc where k is a dimensionless constant.
Writing the equation in dimensional form, we have
[M0L0T1] = [M]a [L]b [LT-2]c = [Ma Lb+c T-2c]
Equating exponents of M, L, and T on both sides, we get a = 0, b + c = 0, –2c = 1.
Solving the eq., we get a = 0, b = 1/2, c = -1/2
Hence, T = k g , where k is constant.
Try yourself:
19. A gas bubble from an explosion under water oscillates with a period proportional to PaρbEc, where P is the static pressure, ρ is the density of water and E is the total energy of the explosion. Find the values of a, b, and c.
Ans:5 6 1 2 1 3 ,,
2.8.3
Finding Unit of a Given Physical Quantity
By using the formula of a physical quantity in terms of basic quantities, we can find its dimensional formula. So, in the dimensional formula, by replacing M, L, and T by the fundamental units, we get the unit of physical quantity.
Example: Force = mass × acceleration
[F] = [M] [LT–2] = [MLT –2]
Hence, units of force will be kg ms–2, which is also called newton (N).
25. The dimensional formula of the product of two physical quantities P and Q is ML2T–2. The dimensional formula of P/Q is ML 0T–2. Then, what are the units of physical quantities P and Q?
Sol Given, P × Q has dimensions ML 2T–2
P/Q has dimensions ML0T–2
Hence, (PQ) P Q =MLT ML T 2-20 -2
P2 = M2L2T–4
⇒ P = ML1T–2 = Force ......(1)
Given,
P Q = M LT ; ML T Q = MLT
10 -2 1-2 0-2
Q = ML T ML T = L = Displa
c cement 2
\ P and Q are, respectively, the force and the displacement.
Hence, their units are newton and metre.
26. If the equation of state of a gas is expressed as P + a V Vb = RT, 2 - where P is the pressure, V is the volume, and T is the absolute temperature, and a, b, R are constants, then find the dimensions of ‘a’ and ‘b’.
Sol. By principle of homogeneity of dimensions, P can be added to P only. It means a V 2 also gives pressure.
Dimension formulae for pressure (P) = M1L–1T–2 and volume (V) = M0L3T0.
Since a V 2 = pressure
\ a = M1L5T–2
Similarly, b will have same dimensions as volume [V – b] = [Volume] \ b = M0L3T0
Try yourself:
20. If E, M, L, and G denote energy, mass, angular momentum, and universal gravitational constant, respectively, prove that EL MG 2 52 is a dimensionless quantity.
= A dimensionless quantity.
Ans: ML
2.8.4 To Design Our Own New System of Dimensions
In place of M, L, and T, we can take any other three physical quantities as the fundamental physical quantities with the condition that they
should be independent of each other and not derivable mutually. Suppose, in a new system, force (F), length (L), and time (T) are taken as fundamental physical quantities. Then, dimensional formulae of other quantities, like mass, velocity, acceleration, etc., can be derived in terms of F, L, and T.
27. If pressure P, velocity V, and time T are taken as the fundamental (base) quantities, find the dimensional formula of force.
Sol. Let force = F = PaVbTc
(F) = M1L1T–2, (P) = M1L–1T–2, (V) = M0L1T–1, (T) = M0L0T1
\ MLT–2 = (ML–1T–2)a (M0L1T–1)b(M0L0T1)c
MLT–2 = MaL–a+bT–2a-b+c
\ Comparing powers of M, L, T on both sides,
a = 1, –a + b = 1 \ b = 2 and
– 2a – b + c = – 2; – 2 – 2 + c = – 2; c = 2
\ Force (F) = PV2T2
Try yourself:
21. If unit of mass is taken as 1 kg, of time as 1 minute, and acceleration due to gravity is taken as 9.81 ms–2, what is the unit of energy?
Ans: 3.464 × 105 joules
2.8.5 Limitations on
the Uses of Dimensional Analysis
There is no limitation as far as the first use (conversion of units) is concerned. As for the second use (checking the dimensional accuracy of the given relationship) is concerned, if both the LHS and RHS are dimensionless, we cannot know whether the quantities in the numerator and denominator are in correct or reverse order. The rest of the uses have the following limitations.
(i) Proportionality constants cannot be determined by dimensional analysis.
(ii) Formulae containing non-algebraic functions, i.e., sin, cos, log, exponential, etc., cannot be derived.
(iii) Formulae which contain two or more terms in the RHS cannot be derived.
Example: S = ut + ½ at2 .
(iv)Formulae containing dimensional constants cannot be derived.
Example: F= Gm m d 12 2
(v) Dimensional analysis does not differentiate between a scalar and a vector quantity.
(vi) If we do not know the physical quantities on which a physical quantity depends, we cannot even proceed to derive a formula.
(vii) It is difficult to apply dimensional methods if the physical quantity depends upon more than three fundamental quantities, the length, ma ss, and time.
TEST YOURSELF
1. If A B Bx C log( ) is dimensionally true, then (here η is the coefficient of viscosity and x is the distance) (1) C is dimensionless constant (2) B has dimensions of –1 in length (3) the dimensional formula of A is ML–2T–1 (4) all the above are true
2. Velocity of sound in a medium is given by A1/2d–1/2 where d is density of medium. Then A represents (1) Elastic constant (2) Energy (3) Volume (4) Compressibility
3. In a particular system, the unit of length, mass and time are chosen to be 10 cm, 10 g and 0.1 s respectively. The unit of force in this system will be equivalent to: (1) 0.1 N (2) 1 N (3) 10 N (4) 100N
4. If the unit of length is doubled, unit of time is halved and unit of force is quadrupled, the unit of power would change by the factor (1) 1/4 (2) 16 (3) 1/16 (4) 8
5. The density of mercury in cgs system is 13.6 g cm–3. Its value in SI is (1) 136 kg/m3 (2) 1360 kg/m3 (3) 13600 kg/m3 (4) 1.36 kg/m3
Answer Key
(1) 4 (2) 1 (3) 1 (4) 2 (5) 3
Table
3.1: Dimensional Formulae of Physical Quantities
S.No.
1. Volume length × breadth × height
2. Density mass / volume
3. Linear density mass / length
4. Relative density densityofsubstance densityofwater
5. a) Energy density Energy / volume
b) Pressure, stress force/Area
c) Young’s, Rigidity, Bulk moduli of elasticity Stress / strain
6. a) Velocity displacement / time
b) Speed distance / time
MºL3Tº m3 S
ML–3Tº kg m–3 S
ML–1Tº kg m–1 S
MºLºTº S
ML–1T–2 N m–2 S
ML–1T–2 N m–2 S,Tensor
ML–1T–2 N m–2 S
MºLT–1 m s–1 V
MºLT–1 m s–1 S
c) Acceleration force / mass MºLT–2 m s–2 V
7. Hubbles’s Constant (H) Velocity / Distance T-1 s–1 –
8. a) Areal velocity Area / time MºL2T–1 V
b) Coefficient of kinematic viscosity Coefficientofviscosity density
9. a) Momentum (linear) mass × velocity
b) Impulse Force × time
10.
a) Moment of force or moment of couple or torque Force × displacement
b) Work Force × displacement
c) All energies Force × displacement
11. Power Force × velocity
12. Moment of inertia I = MK2
13. Angular displacement q = d/r
14. a) Angular velocity w = q /t
b) Velocity gradient dv/dx
c) Frequency u = c/ l
d) Decay constant number of fissions/ time
15. Angular acceleration d w /dt
MºL2T–1 m2 s–1 S
ML1T–1 kg m s–1 V
ML1T–1 kg m s–1 V
ML2T–2 N m V
ML2T–2 joule S
ML2T–2 joule
ML2T–3 watt S
ML2Tº kg m2 Tensor
MºLºTº rad V
MºLºT–1 rad/s V
MºLºT–1 s–1 V
MºLºT–1 Hz S
MºLºT–1 s–1 S
MºLºT–2 rad s–2 V
S.No.
16. a) Angular momentum I w
b) Plank’s constant h = Energy frequency
c) Angular impulse t × t
17. Gravitational constant Fd m 2 2
18. Gravitational field intensity F
19. a) Gravitational potential W m
b) Latent heat Q m Q = Heat energy
c) Calorific value Q m Q = Heat energy
20. a) Force constant b) Spring constant c) Surface tension
21. Compressibility
22. Coefficient of viscosity
23. a) Coefficient of linear expansion b) All coefficients of expansions of solids liquids and gases
24. a) Thermal capacity
J s S
J s V
J kg–1 S
J kg–1 S
J
S b) Entropy
S.No. Physical Quantity
c) Boltzmann Constant PV NT ML2T–2K–1 J K-1 S
25. a) Specific heat Q m L2T–2K–1 J kg –1K-1 S
b) Specific gas constant r P dT = L2T–2K–1 J kg –1K-1 S
26. a) Universal gas constant R = PV nT
ML 2 T –2 K –1mol–1 J mol–1 K–1 S
b) Molar specific heat CP–CV = R ML 2 T –2 K –1mol–1 J mol –1 K–1 S
27. Stefan’s constant E At q 4 MLºT–3K–4 W m –2 K–4 S
28. Coefficient of thermal conductivity Qd At 21
29. a) Thermal resistance L
W m–1K–1 S
W–1K S b) Temperature gradient
Thermal conductance
Permeability
Pole strength
Magnetic moment
Magnetic induction
H m-1 S
S.No.
35. a) Intensity of magnetic field b) Intensity of magnetisation H = B µ I = M V IL–1 A m–1 V
36. Magnetic flux φ=BA= W I
ML2T–2I–1 weber S
37. a) Intensity of sound b) Intensity of heat c) Emissive power Power Area M T–3 watt/ metre2 S
38. Spectral emissive power Power L3 ML–1T–3 watt/ metre3 S
39. Pressure gradient Pressure Length =P/L ML–2T–2 Pa m–1 V
40. Electric current i MºLºTºI1 ampere S
41. Electric charge Q = it IT coulomb S
42. Intensity of electric field E = F/Q MLT–3I–1 N C-1 V
43. a) Electric potential b) Electromotive force (emf) W Q
ML2T–3I–1 volt S
44. Electric capacity C= Q V = Q W 2 M–1L–2T4I2 farad S
45. Electric resistance R= V I = Power I2
46. Electric conductance K= 1 R
47. Resistivity ρ= RA L
48. Conductivity 1
ML2T–3I–2 ohm S
M–1L–2T3I2 ohm or siemen S
ML3T–3I–2 ohm metre S
M–1L–3T3I2 Sm-1 S
49. Permittivity 0 2 2 Q Fd
50. a) Self inductance b) Mutual inductance Energy Current = emf di/dt 2
farad/ metre S
henry S 51. Linear charge density q L
S 52. Surface charge density q A
Volume charge density q V
55. Electrochemical equivalent m it
kgA –1s–1 S 56. Activity dN dt
57. Mobility v E d
CHAPTER REVIEW
■ Quantities which can be measured directly or indirectly are called physical quantities.
■ There are seven fundamental physical quantities, e.g., mass, length, time, thermodynamic temperature, electric current, amount of substance, and luminious intensity.
■ If ‘ P ’ is a physical quantity, ‘ n ’ is its numerical value and ‘ u’ is its unit, then P = nu = n1u1 = n2u2 = n3u3
■ Accuracy is a measure of closeness of the measured value to the true value.
■ The precision of an instrument gives the minimum value that can be measured by it.
■ If an experiment is repeated n times and the measured values are x1, x2, x3, . . . xn, then mean value x mean absolute error, x mn m n xx xx 1 12 3 ... x xx xx xx xx n x x mm mn m m m 12 3 relative error = percentaage error = x x m m 100%
relative error =
percentaage error = x x m m 100%
■ Probable value of the measured quantity, = (x m ± D xm)
■ If X, Y and Z are three physical quantities such that Z = X ± Y, then maximum possible error D Z = ( D X + D Y).
Percentage error in Z = XY XY 100%
■ If a physical quantity ‘ P ’ depends upon other physical quantities A, B and C, such that P ∝ ( A ) m ( B ) n ( C ) ℓ , then maximum possible relative error is given by P P m A A n B B C C
■ Only physical quantities having same dimensions can be added, subtracted or can be equated.
■ The terms on both sides of a dimensional equation should have same dimensions.
■ When a physical quantity depends upon other physical quantities and if their relationship is of product type, then only method of dimensional analysis can be applied to find their relationship.
■ An equation can be physically correct, but it may be dimensionally incorrect.
■ An equation may be dimensionally correct but it may be physically incorrect.