Chapter 2: Units, Dimensions, and Measurement PDF
Document Details
Uploaded by NourishingRoseQuartz
Tags
Summary
This chapter provides an overview of units, dimensions, and measurements in physics. It covers physical quantities, their magnitude and units, types of physical quantities (scalar and vector), fundamental and derived quantities and units, and the concept of dimensions. It also describes different systems of units like CGS, MKS, FPS, and SI units, along with their prefixes and practical units.
Full Transcript
E3 60 50 Units, Dimensions and Measurement 1.1 Physical Quantity. A quantity which can be measured and by which various physical happenings can be ID explained and expressed in form of laws is called a physical quantity. For example length, mass, time, force etc. On the other hand various happenings...
E3 60 50 Units, Dimensions and Measurement 1.1 Physical Quantity. A quantity which can be measured and by which various physical happenings can be ID explained and expressed in form of laws is called a physical quantity. For example length, mass, time, force etc. On the other hand various happenings in life e.g., happiness, sorrow etc. are not physical quantities because these can not be measured. U Measurement is necessary to determine magnitude of a physical quantity, to compare two similar physical quantities and to prove physical laws or equations. D YG A physical quantity is represented completely by its magnitude and unit. For example, 10 metre means a length which is ten times the unit of length 1 kg. Here 10 represents the numerical value of the given quantity and metre represents the unit of quantity under consideration. Thus in expressing a physical quantity we choose a unit and then find that how many times that unit is contained in the given physical quantity, i.e. Physical quantity (Q) = Magnitude × Unit = n × u Where, n represents the numerical value and u represents the unit. Thus while expressing U definite amount of physical quantity, it is clear that as the unit(u) changes, the magnitude(n) will also change but product ‘nu’ will remain same. n u = constant, ST i.e. or n1 u 1 n 2 u 2 constant ; n 1 u i.e. magnitude of a physical quantity and units are inversely proportional to each other.Larger the unit, smaller will be the magnitude. 1.2 Types of Physical Quantity. (1) Ratio (numerical value only) : When a physical quantity is a ratio of two similar quantities, it has no unit. e.g. Relative density = Density of object/Density of water at 4 oC Refractive index = Velocity of light in air/Velocity of light in medium Units, Dimensions and Measurement 51 Strain = Change in dimension/Original dimension Note : Angle is exceptional physical quantity, which though is a ratio of two similar physical quantities (angle = arc / radius) but still requires a unit (degrees or radians) to specify it along with its numerical value. 60 (2) Scalar (Magnitude only) : These quantities do not have any direction e.g. Length, time, work, energy etc. Magnitude of a physical quantity can be negative. In that case negative sign indicates that E3 the numerical value of the quantity under consideration is negative. It does not specify the direction. ID Scalar quantities can be added or subtracted with the help of following ordinary laws of addition or subtraction. (3) Vector (magnitude and direction) : e.g. displacement, velocity, acceleration, force etc. U Vector physical quantities can be added or subtracted according to vector laws of addition. These laws are different from laws of ordinary addition. Note : There are certain physical quantities which behave neither as scalar nor as vector. D YG For example, moment of inertia is not a vector as by changing the sense of rotation its value is not changed. It is also not a scalar as it has different values in different directions (i.e. about different axes). Such physical quantities are called Tensors. 1.3 Fundamental and Derived Quantities. (1) Fundamental quantities : Out of large number of physical quantities which exist in nature, there are only few quantities which are independent of all other quantities and do not require the help of any other physical quantity for their definition, therefore these are called U absolute quantities. These quantities are also called fundamental or base quantities, as all other quantities are based upon and can be expressed in terms of these quantities. ST (2) Derived quantities : All other physical quantities can be derived by suitable multiplication or division of different powers of fundamental quantities. These are therefore called derived quantities. If length is defined as a fundamental quantity then area and volume are derived from length and are expressed in term of length with power 2 and 3 over the term of length. Note : In mechanics Length, Mass and time are arbitrarily chosen as fundamental quantities. However this set of fundamental quantities is not a unique choice. In fact any three quantities in mechanics can be termed as fundamental as all other quantities in mechanics can be expressed in terms of these. e.g. if speed and time are taken as fundamental quantities, length will become a derived quantity because then length will be expressed as Speed Time. 52 Units, Dimensions and Measurement and if force and acceleration are taken as fundamental quantities, then mass will be defined as Force / acceleration and will be termed as a derived quantity. 1.4 Fundamental and Derived Units. Normally each physical quantity requires a unit or standard for its specification so it 60 appears that there must be as many units as there are physical quantities. However, it is not so. E3 It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. Arbitrarily the physical quantities mass, length and time are choosen for this purpose. So any unit of mass, length and time in mechanics is called a fundamental, absolute or base unit. Other units which can be expressed in terms of fundamental units, are called derived units. For example light year or km is a fundamental units as it is a unit of length while s–1, m2 or kg/m are derived units as these are derived from units of time, mass and length respectively. ID System of units : A complete set of units, both fundamental and derived for all kinds of physical quantities is called system of units. The common systems are given below – (1) CGS system : The system is also called Gaussian system of units. In it length, mass and U time have been chosen as the fundamental quantities and corresponding fundamental units are centimetre (cm), gram (g) and second (s) respectively. D YG (2) MKS system : The system is also called Giorgi system. In this system also length, mass and time have been taken as fundamental quantities, and the corresponding fundamental units are metre, kilogram and second. (3) FPS system : In this system foot, pound and second are used respectively for measurements of length, mass and time. In this system force is a derived quantity with unit poundal. ST U (4) S. I. system : It is known as International system of units, and is infact extended system of units applied to whole physics. There are seven fundamental quantities in this system. These quantities and their units are given in the following table Quantity Name of Unit Symbol Length metre m Mass kilogram kg Time second s Electric Current ampere A Temperature Kelvin K Amount of Substance mole mol Luminous Intensity candela cd Units, Dimensions and Measurement 53 Besides the above seven fundamental units two supplementary units are also defined – Radian (rad) for plane angle and Steradian (sr) for solid angle. Note : Apart from fundamental and derived units we also use very frequently practical 60 units. These may be fundamental or derived units e.g., light year is a practical unit (fundamental) of distance while horse power is a practical unit (derived) of power. E3 Practical units may or may not belong to a system but can be expressed in any system of units e.g., 1 mile = 1.6 km = 1.6 × 103 m. 1.5 S.I. Prefixes. Power of 10 Prefix Symbol exa E D YG 1018 U ID In physics we have to deal from very small (micro) to very large (macro) magnitudes as one side we talk about the atom while on the other side of universe, e.g., the mass of an electron is 9.1 10–31 kg while that of the sun is 2 1030 kg. To express such large or small magnitudes simultaneously we use the following prefixes : 1015 peta P 1012 tera T 109 giga G 10 mega M 103 kilo k 102 hecto h 101 deca da 10–1 deci d 10–1 centi c 10–3 milli m 10–6 micro 10–9 nano n 10–12 pico p 10–15 femto f 10–18 atto a ST U 6 1.6 Standards of Length, Mass and Time. 54 Units, Dimensions and Measurement (1) Length : Standard metre is defined in terms of wavelength of light and is called atomic standard of length. The metre is the distance containing 1650763.73 wavelength in vacuum of the radiation corresponding to orange red light emitted by an atom of krypton-86. 60 Now a days metre is defined as length of the path travelled by light in vacuum in 1/299,7792, 458 part of a second. (2) Mass : The mass of a cylinder made of platinum-iridium alloy kept at International Bureau of Weights and Measures is defined as 1 kg. E3 On atomic scale, 1 kilogram is equivalent to the mass of 5.0188 1025 atoms of 6C12 (an isotope of carbon). ID (3) Time : 1 second is defined as the time interval of 9192631770 vibrations of radiation in Cs-133 atom. This radiation corresponds to the transition between two hyperfine level of the ground state of Cs-133. 1.7 Practical Units. (i) 1 fermi = 1 fm = 10–15 m D YG (ii) 1 X-ray unit = 1XU = 10–13 m U (1) Length : (iii) 1 angstrom = 1Å = 10–10 m = 10–8 cm = 10–7 mm = 0.1 mm (iv) 1 micron = m = 10–6 m (v) 1 astronomical unit = 1 A.U. = 1. 49 1011 m 1.5 1011 m 108 km (vi) 1 Light year = 1 ly = 9.46 1015 m (vii) 1 Parsec = 1pc = 3.26 light year U (2) Mass : (i) Chandra Shekhar unit : 1 CSU = 1.4 times the mass of sun = 2.8 1030 kg ST (ii) Metric tonne : 1 Metric tonne = 1000 kg (iii) Quintal : 1 Quintal = 100 kg (iv) Atomic mass unit (amu) : amu = 1.67 10–27 kg mass of proton or neutron is of the order of 1 amu (3) Time : (i) Year : It is the time taken by earth to complete 1 revolution around the sun in its orbit. (ii) Lunar month : It is the time taken by moon to complete 1 revolution around the earth in its orbit. 1 L.M. = 27.3 days Units, Dimensions and Measurement 55 (iii) Solar day : It is the time taken by earth to complete one rotation about its axis with respect to sun. Since this time varies from day to day, average solar day is calculated by taking average of the duration of all the days in a year and this is called Average Solar day. average solar day or 1 the part of solar year 365.25 60 1 Solar year = 365.25 average solar day (iv) Sedrial day : It is the time taken by earth to complete one rotation about its axis with respect to a distant star. E3 1 Solar year = 366.25 Sedrial day = 365.25 average solar day Thus 1 Sedrial day is less than 1 solar day. 1 Shake = 10– 8 sec ID (v) Shake : It is an obsolete and practical unit of time. 1.8 Dimensions of a Physical Quantity. U When a derived quantity is expressed in terms of fundamental quantities, it is written as a product of different powers of the fundamental quantities. The powers to which fundamental D YG quantities must be raised in order to express the given physical quantity are called its dimensions. To make it more clear, consider the physical quantity force Force = mass × acceleration .... (i) mass length/tim e mass velocity = mass × length × (time)–2 time time Thus, the dimensions of force are 1 in mass, 1 in length and – 2 in time. U Here the physical quantity that is expressed in terms of the base quantities is enclosed in square brackets to indicate that the equation is among the dimensions and not among the magnitudes. ST Thus equation (i) can be written as [force] = [MLT–2]. Such an expression for a physical quantity in terms of the fundamental quantities is called the dimensional equation. If we consider only the R.H.S. of the equation, the expression is termed as dimensional formula. Thus, dimensional formula for force is, [MLT –2 ]. 1.9 Important Dimensions of Complete Physics. Mechanics S. N. Quantity (1) Velocity or speed (v) m/s [M0L1T (2) Acceleration (a) m/s2 [M0LT –2] Unit Dimension –1 ] 56 Units, Dimensions and Measurement S. N. Quantity (3) Momentum (P) kg-m/s [M1L1T –1 (4) Impulse (I) Newton-sec or kg-m/s [M1L1T –1 (5) Force (F) Newton [M1L1T –2 (6) Pressure (P) Pascal [M1L–1T –2 (7) Kinetic energy (EK) Joule [M1L2T –2 (8) Power (P) Watt or Joule/s [M1L2T –3 (9) Density (d) kg/m3 (10) Angular displacement () Radian (rad.) [M0L0T 0] (11) Angular velocity () Radian/sec [M0L0T –1 (12) Angular acceleration () Radian/sec2 [M0L0T –2 (13) Moment of inertia (I) kg-m2 (14) Torque () Newton-meter [M1L2T –2 (15) Angular momentum (L) Joule-sec [M1L2T –1 (16) Force constant or spring constant Newton/m (k) [M1L0T –2 (17) Gravitational constant (G) (18) Dimension ] ] ] ] 60 Unit ] ] U ID E3 [M1L– 3T 0] ] ] [M1L2T0] ] ] ] [M–1L3T – 2] Intensity of gravitational field (Eg) N/kg [M0L1T –2 (19) Gravitational potential (Vg) Joule/kg [M0L2T –2 (20) Surface tension (T) N/m or Joule/m2 [M1L0T –2 (21) Velocity gradient (Vg) Second–1 [M0L0T –1 (22) Coefficient of viscosity () kg/m-s [M1L– 1T –1 (23) Stress N/m2 [M1L– 1T –2 (24) Strain No unit (25) Modulus of elasticity (E) N/m2 (26) Poisson Ratio () No unit [M0L0T 0] (27) Time period (T) Second [M0L0T1] (28) Frequency (n) Hz ST U D YG N-m2/kg2 ] ] ] ] ] ] [M0L0T 0] [M1L– 1T [M0L0T Heat S. N. Quantity (1) Temperature (T) Kelvin [M0L0T0 1] (2) Heat (Q) Joule [ML2T– 2] (3) Specific Heat (c) Joule/kg-K [M0L2T– 2 –1] Unit Dimension –2 –1 ] ] Units, Dimensions and Measurement 57 S. N. Quantity (4) Thermal capacity Joule/K [M1L2T –2 (5) Latent heat (L) Joule/kg [M0L2T –2 (6) Gas constant (R) Joule/mol-K [M1L2T– 2 – 1] (7) Boltzmann constant (k) Joule/K [M1L2T– 2 – 1] (8) Coefficient of thermal conductivity (K) Joule/m-s-K (9) Stefan's constant () Watt/m2-K4 (10) Wien's constant (b) Meter-K (11) Planck's constant (h) Joule-s (12) Coefficient of Linear Expansion () Kelvin–1 [M0L0T0 –1] (13) Mechanical eq. of Heat (J) Joule/Calorie [M0L0T0] (14) Vander wall’s constant (a) (15) Vander wall’s constant (b) –1] 60 ] [M1L1T– 3 –1 [M1L0T– 3 –4 ] ] E3 [M0L1To1] ID [M1L2T–1] Newton-m4 [ML5T– 2] m3 [M0L3T0] U Electricity Dimension Unit Quantity (1) Electric charge (q) Coulomb [M0L0T1A1] (2) Electric current (I) Ampere [M0L0T0A1] (3) Capacitance (C) Coulomb/volt or Farad [M–1L– 2T4A2] (4) Electric potential (V) Joule/coulomb M1L2T–3A–1 (5) Permittivity of free space (0) (6) Dielectric constant (K) Resistance (R) Dimension Coulomb 2 Newton - meter 2 [M–1L–3T4A2] Unitless [M0L0T0] Volt/Ampere or ohm [M1L2T– 3A– 2] Resistivity or Specific resistance Ohm-meter () ST (8) U (7) Unit D YG S. N. [M1L3T– 3A– 2] (9) Coefficient of Self-induction (L) volt second or henery or ohm-second ampere [M1L2T– 2A– 2] (10) Magnetic flux () Volt-second or weber [M1L2T–2A–1] (11) Magnetic induction (B) (12) Magnetic Intensity (H) (13) Magnetic Dipole Moment (M) Joule newton ampere meter ampere meter volt second or Tesla meter 2 Ampere/meter Ampere-meter2 2 [M1L0T– 2A– 1] [M0L– 1T0A1] [M0L2T0A1] 58 Units, Dimensions and Measurement S. N. Quantity Unit Newton ampere 2 or Dimension Joule ampere 2 meter or (14) Permeability of Free Space (0) (15) Surface charge density () Coulomb metre 2 (16) Electric dipole moment (p) Coulomb meter (17) Conductance (G) (1/R) ohm 1 (18) Conductivity () (1/) (19) (20) Current density (J) Intensity of electric field (E) (21) Rydberg constant (R) ohm 1meter 1 Ampere/m2 Volt/meter, Newton/coulomb m–1 [M0L–2T1A1] [M0L1T1A1] E3 [M–1L–2T3A2] [M–1L–3T3A2] M0L–2T0A1 M1L1T –3A–1 M0L–1T0 ID 1.10 Quantities Having Same Dimensions. [M1L1T–2A–2] 60 Ohm sec ond henery Volt second or or meter meter ampere meter Dimension Quantity (1) [M0L0T–1] Frequency, angular frequency, angular velocity, velocity gradient and decay constant (2) [M1L2T–2] Work, internal energy, potential energy, kinetic energy, torque, moment of force (3) [M1L–1T–2] Pressure, stress, Young’s modulus, bulk modulus, modulus of rigidity, energy density (4) [M1L1T–1] Momentum, impulse (5) 0 1 –2 Acceleration due to gravity, gravitational field intensity 1 1 –2 Thrust, force, weight, energy gradient (7) 1 2 –1 [M L T ] Angular momentum and Planck’s constant (8) 1 0 –2 [M L T ] Surface tension, Surface energy (energy per unit area) (9) [M0L0T0] Strain, refractive index, relative density, angle, solid angle, distance gradient, relative permittivity (dielectric constant), relative permeability etc. (10) [M0L2T–2] Latent heat and gravitational potential (11) [M L T 1 ] Thermal capacity, gas constant, Boltzmann constant and entropy (12) [M0L0T1] (13) 0 0 [M L T ] (14) 2 D YG U S. N. [M L T ] [M L T ] U (6) –2 – ST 0 2 1 –2 [ML T ] l g , m k , R g , where l = length g = acceleration due to gravity, m = mass, k = spring constant L/R, LC , RC where L = inductance, R = resistance, C = capacitance V2 q2 t, VIt, qV , LI 2 , , CV 2 where I = current, t = time, q = charge, R C L = inductance, C = capacitance, R = resistance I 2 Rt , 1.11 Application of Dimensional Analysis. (1) To find the unit of a physical quantity in a given system of units : Writing the definition or formula for the physical quantity we find its dimensions. Now in the dimensional Units, Dimensions and Measurement 59 formula replacing M, L and T by the fundamental units of the required system we get the unit of physical quantity. However, sometimes to this unit we further assign a specific name, e.g., Work = Force Displacement [W] = [MLT–2] [L] = [ML2T–2] So be kg m2/s2 which is called joule. Sample problems based on unit finding a The equation P 2 (V b) = constant. The units of a is V (a) Dyne cm 5 [MNR 1995; AFMC 1995] E3 Problem 1. 60 So its units in C.G.S. system will be g cm2/s2 which is called erg while in M.K.S. system will (b) Dyne cm 4 (d) Dyne / cm 2 (c) Dyne / cm 3 ID a Solution : (b) According to the principle of dimensional homogenity [ P ] 2 V [a] [P] [V 2 ] [ML1T 2 ] [L6 ] [ML5 T 2 ] or unit of a = gm × cm 5 sec–2= Dyne cm4 If x at bt 2 , where x is the distance travelled by the body in kilometre while t the time in U Problem 2. seconds, then the units of b are (c) km/s2 (b) km-s D YG (a) km/s (d) km-s2 x Solution : (c) From the principle of dimensional homogenity [ x ] [bt 2 ] [b ] 2 t Problem 3. The unit of absolute permittivity is (a) Farad - meter Unit of b = km/s2. [EAMCET (Med.) 1995; Pb. PMT 2001] (c) Farad/meter 2 (b) Farad / meter Solution : (b) From the formula C 4 0 R 0 (d) Farad C 4 R U By substituting the unit of capacitance and radius : unit of 0 Farad/ meter. Problem 4. Unit of Stefan's constant is ST (a) Js 1 Solution : (b) Stefan's formula Problem 5. (b) Jm 2 s 1 K 4 (c) Jm 2 (d) Js Joule Q Q Unit of 2 T 4 = Jm 2 s 1 K 4 4 4 At m sec K AtT The unit of surface tension in SI system is [MP PMT 1984; AFMC 1986; CPMT 1985, 87; CBSE 1993; Karnataka CET (Engg/Med.) 1999; DCE 2000, 01] (a) Dyne / cm 2 (b) Newton/m Solution : (b) From the formula of surface tension T (c) Dyne/cm (d) Newton/m2 F l By substituting the S.I. units of force and length, we will get the unit of surface tension = Newton/m 60 Units, Dimensions and Measurement Problem 6. A suitable unit for gravitational constant is (a) kg metre sec 1 Solution : (c) As F Gm 1m 2 (b) Newton metre 1 sec (c) Newton metre 2 kg 2 G r2 [MNR 1988] (d) kg metre sec 1 Fr 2 m1m 2 Problem 7. 60 Substituting the unit of above quantities unit of G = Newton metre 2 kg 2. The SI unit of universal gas constant (R) is [MP Board 1988; JIPMER 1993; AFMC 1996; MP PMT 1987, 94; CPMT 1984, 87; UPSEAT 1999] (b) Newton K 1 mol 1 (c) Joule K 1 mol 1 Solution : (c) Ideal gas equation PV nRT [R] (d) Erg K 1 mol 1 E3 (a) Watt K 1 mol 1 [P][V ] [ML1 T 2 ][L3 ] [ML 2 T 2 ] [mole ] [K ] [nT ] [mole ] [K] ID So the unit will be Joule K 1mol 1. U (2) To find dimensions of physical constant or coefficients : As dimensions of a physical quantity are unique, we write any formula or equation incorporating the given constant and then by substituting the dimensional formulae of all other quantities, we can find the dimensions of the required constant or coefficient. G Fr 2 m 1m 2 D YG (i) Gravitational constant : According to Newton’s law of gravitation F G Substituting the dimensions of all physical quantities [G] or [MLT 2 ][L2 ] [M 1 L3 T 2 ] [M ][M ] Substituting the dimensions of all physical quantities [h] ST r2 E U (ii) Plank constant : According to Planck E h or h m1m 2 [ML2 T 2 ] [ML2 T 1 ] [T 1 ] (iii) Coefficient of viscosity : According to Poiseuille’s formula Substituting the dimensions of all physical quantities [] dV pr 4 pr 4 or 8l dt 8 l(dV / dt) [ML1 T 2 ][L4 ] [ML1 T 1 ] 3 [L][L / T ] Sample problems based on dimension finding Problem 8. X 3YZ 2 find dimension of Y in (MKSA) system, if X and Z are the dimension of capacity and magnetic field respectively (a) M 3 L2 T 4 A 1 [MP PMT 2003] (b) ML2 (c) M 3 L2 T 4 A 4 (d) M 3 L2 T 8 A 4 Units, Dimensions and Measurement 61 1 Dimensions of (a) [ LT 1 0 0 , where symbols have their usual meaning, are (c) [L2 T 2 ] (b) [ L1 T ] ] Solution : (d) We know that velocity of light C 1 0 0 1 1 2 2 2 So [LT ] = [L T ]. 0 0 1 0 0 [AIEEE 2003] (d) [L2 T 2 ] 60 Problem 9. [X ] [M 1 L2 T 4 A 2 ] [M 3 L2 T 8 A 4 ]. 2 2 1 2 [Z ] [MT A ] C2 E3 Solution : (d) X 3YZ 2 [Y ] Problem 10. If L, C and R denote the inductance, capacitance and resistance respectively, the dimensional formula for C 2 LR is (b) [M 0 L0 T 3 I 0 ] R R Solution : (b) [C 2 LR ] = C 2 L2 = (LC )2 L L (c) [M 1 L2 T 6 I 2 ] ID (a) [ML 2 T 1 I 0 ] is equal to [T 2 ] U and we know that frequency of LC circuits is given by f 1 2 (d) [M 0 L0 T 2 I 0 ] 1 LC i.e., the dimension of LC D YG L L and gives the time constant of L R circuit so the dimension of is equal to [T]. R R R By substituting the above dimensions in the given formula (LC )2 [T 2 ]2 [T 1 ] [T 3 ]. L Problem 11. A force F is given by F at bt 2 , where t is time. What are the dimensions of a and b [BHU 1998; AFMC 2001] (a) MLT 3 and ML2 T 4 (b) MLT 3 and MLT 4 (c) U MLT 1 and MLT 0 ST 2 F MLT Solution : (b) From the principle of dimensional homogenity [F] [at] [a] [MLT t T 2 F MLT Similarly [F] [bt 2 ] [b] 2 [MLT 2 t T 4 3 (d) ] ]. Problem 12. The position of a particle at time t is given by the relation x (t) v0 t (1 c ), where v 0 is a constant and 0. The dimensions of v 0 and are respectively (a) M 0 L1 T 1 and T 1 (b) M 0 L1 T 0 and T 1 (c) M 0 L1 T 1 and LT 2 (d) M 0 L1 T 1 and T 1 Solution : (a) From the principle of dimensional homogeneity [ t] = dimensionless [ ] [T 1 ] t Similarly [x ] v0 ] [v 0 ] [x ][ ] [L][T 1 ] [LT 1 ]. [ ] 62 Units, Dimensions and Measurement Problem 13. The dimensions of physical quantity X in the equation Force (c) M 2 L2T 2 (b) M 2 L2 T 1 (a) M 1 L4 T 2 X is given by Density (d) M 1 L2 T 1 Solution : (c) [X] = [Force] × [Density] = [MLT 2 ] [ML3 ] = [M 2 L2 T 2 ]. n 2 n1 crossing a unit area perpendicular to X- axis x 2 x1 60 Problem 14. Number of particles is given by n D in unit time, where n 1 and n 2 are number of particles per unit volume for the value of x meant to x 2 and x 1. Find dimensions of D called as diffusion constant (b) M 0 L2 T 4 (c) M 0 LT 3 (d) M 0 L2 T 1 E3 (a) M 0 LT 2 Solution : (d) (n) = Number of particle passing from unit area in unit time = [L2 T 1 ] No. of particl e [M 0 L0 T 0 ] = At [L2 ] [T ] Now from the given formula [D] ID [n1 ] [n2 ] No. of particle in unit volume = [ L3 ] [n][ x 2 x1 ] [L2 T 1 ][L] [L2 T 1 ]. 3 [n2 n1 ] [L ] U Problem 15. E, m, l and G denote energy, mass, angular momentum and gravitational constant El 2 are m 5 G2 D YG respectively, then the dimension of (a) Angle (b) Length (c) Mass (d) Time Solution : (a) [E] = energy = [ML T ] , [m] = mass = [M], [l] = Angular momentum = [ML2 T 1 ] 2 2 [G] = Gravitational constant = [M 1 L3 T 2 ] Now substituting dimensions of above quantities in El 2 [ML2 T 2 ] [ML2T 1 ]2 = = [ M 0 L0 T 0 ] m 5G2 [M 5 ] [M 1 L3 T 2 ]2 i.e., the quantity should be angle. Problem 16. The equation of a wave is given by Y A sin k where is the angular velocity and U x v ST v is the linear velocity. The dimension of k is (a) LT (c) T 1 (b) T (d) T 2 L x Solution : (b) According to principle of dimensional homogeneity [k] = [T ]. v LT 1 Problem 17. The potential energy of a particle varies with distance x from a fixed origin as U where A and B are dimensional constants then dimensional formula for AB is (a) ML7/2T 2 (b) ML11 / 2 T 2 (c) M 2 L9 / 2T 2 Solution : (b) From the dimensional homogeneity [x 2 ] [B] [B] = [L2] As well as U [ A] [ x 1 / 2 ] [ x 2 ] [ B] [ML2 T 2 ] [ A][L1 / 2 ] [L2 ] [ A] [ML7 / 2 T 2 ] (d) ML13 / 2T 3 A x , x2 B Units, Dimensions and Measurement 63 Now [ AB] [ML7 / 2 T 2 ] [L2 ] [ML11 / 2 T 2 ] Problem 18. The dimensions of 1 0 E 2 ( 0 = permittivity of free space ; E = electric field ) is 2 (b) ML 2 T 2 (a) MLT 1 (d) ML2T 1 ML2 T 2 1 Energy 1 2 0 E2 [ML T ] 3 2 Volume L 60 Solution : (c) Energy density = (c) ML 1 T 2 Problem 19. You may not know integration. But using dimensional analysis you can check on some (a) 1 dx x an sin 1 1 the value of n is 2 1/2 a (2ax x ) (b) – 1 E3 results. In the integral (c) 0 Solution : (c) Let x = length [ X ] [L] and [dx ] [L] (d) 1 2 ID x By principle of dimensional homogeneity dimensionless [a] [x ] [L] a Problem 20. A physical quantity P D YG Solution : (b) F = BIL Dimension of [B] [Ln ] n 0 where B= magnetic induction, l= length and m = mass. The (b) ML2T 4 I–2 (a) MLT 3 [L L ] 2 1/ 2 2 2 B l m dimension of P is [L] 2 U By substituting dimension of each quantity in both sides: (c) M 2 L2 T 4 I (d) MLT 2 I 2 [F] [MLT 2 ] = [MT 2 I 1 ] [I][L] [I][L] B 2l 2 [MT 2 I 1 ] 2 [L2 ] [ML2 T 4 I 2 ] m [M ] Now dimension of [P] 2ct 2x cos , which of the following U Problem 21. The equation of the stationary wave is y= 2a sin statements is wrong (a) The unit of ct is same as that of (b) The unit of x is same as that of ST (c) The unit of 2c / is same as that of 2x /t (d) as that of x / Solution : (d) Here, 2ct as well as 2x The unit of c/ is same 2ct 2x 0 0 0 are dimensionless (angle) i.e. M L T 2c 2x So (i) unit of c t is same as that of (ii) unit of x is same as that of (iii) t and (iv) x is unit less. It is not the case with c . (3) To convert a physical quantity from one system to the other : The measure of a physical quantity is nu = constant 64 Units, Dimensions and Measurement If a physical quantity X has dimensional formula [MaLbTc] and if (derived) units of that physical quantity in two systems are [M 1a Lb1 T1c ] and [M 2a Lb2 T2c ] respectively and n1 and n2 be the numerical values in the two systems respectively, then n1 [u1 ] n 2 [u 2 ] a b M L T n 2 n1 1 1 1 M 2 L 2 T2 60 n1 [M 1a Lb1 T1c ] n 2 [M 2a Lb2 T2c ] c E3 where M1, L1 and T1 are fundamental units of mass, length and time in the first (known) system and M2, L2 and T2 are fundamental units of mass, length and time in the second (unknown) system. Thus knowing the values of fundamental units in two systems and numerical value in one system, the numerical value in other system may be evaluated. ID Example : (1) conversion of Newton into Dyne. The Newton is the S.I. unit of force and has dimensional formula [MLT–2]. So 1 N = 1 kg-m/ sec2 a b c 1 M L T kg m sec By using n 2 n1 1 1 1 1 gm cm sec M 2 L 2 T2 2 U 1 D YG 1 N = 105 Dyne 10 3 gm 10 2 cm sec 2 10 5 1 gm cm sec 1 1 (2) Conversion of gravitational constant (G) from C.G.S. to M.K.S. system The value of G in C.G.S. system is 6.67 10–8 C.G.S. units while its dimensional formula is [M–1L3T–2] So G = 6.67 10–8 cm3/g s2 a b 1 c ST U M L T gm cm sec By using n 2 n1 1 1 1 6.67 10 8 kg m sec M 2 L 2 T2 3 1 6.67 10 8 2 gm cm sec 3 2 10 gm 10 cm sec 3 2 6.67 10 11 G = 6.67 10–11 M.K.S. units Sample problems based on conversion Problem 22. A physical quantity is measured and its value is found to be nu where n numerical value and u unit. Then which of the following relations is true (a) n u 2 (b) n u [RPET 2003] (c) n u Solution : (d) We know P nu constant n1u1 n2u 2 or n 1. u (d) n 1 u Units, Dimensions and Measurement 65 Problem 23. In C.G.S. system the magnitude of the force is 100 dynes. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is [EAMCET 2001] (a) 0.036 (b) 0.36 (c) 3.6 (d) 36 60 Solution : (c) n1 100 , M1 g , L1 cm , T1 sec and M 2 kg , L2 meter , T2 m inute , x 1 , y 1 , z 2 x y M L T By substituting these values in the following conversion formula n2 n1 1 1 1 M 2 L2 T2 1 gm n2 100 3 10 gm 1 cm 2 10 cm 1 2 sec 60 sec E3 1 gm cm sec n 2 100 kg meter minute 2 2 3.6 ID Problem 24. The temperature of a body on Kelvin scale is found to be X K. When it is measured by a Fahrenheit thermometer, it is found to be X F. Then X is (a) 301.25 (b) 574.25 (c) 313 K 273 F 32 5 9 U Solution : (c) Relation between centigrade and Fahrenheit X 273 X 32 X 313. 5 9 D YG According to problem (d) 40 Problem 25. Which relation is wrong (a) 1 Calorie = 4.18 Joules (b) 1Å =10–10 m (c) 1 MeV = 1.6 × 10–13 Joules (d) 1 Newton =10–5 Dynes Solution : (d) Because 1 Newton = 10 5 Dyne. F L. ; where L= length, A= A l area of cross- section of the wire, L Change in length of the wire when stretched with a force F. The conversion factor to change it from CGS to MKS system is U Problem 26. To determine the Young's modulus of a wire, the formula is Y (a) 1 (b) 10 (c) 0.1 (d) 0.01 1 2 ST Solution : (c) We know that the dimension of young's modulus is [ML T ] C.G.S. unit : gm cm 1 sec 2 and M.K.S. unit : kg. m–1 sec–2. 1 M 1 L1 M 2 L2 By using the conversion formula: n 2 n1 1 T1 T2 2 1 gm cm kg meter 1 sec sec 2 1 2 1 n 2 gm cm sec 0.1 3 2 n1 10 gm 10 cm sec 10 1 Conversion factor Problem 27. Conversion of 1 MW power on a new system having basic units of mass, length and time as 10kg, 1dm and 1 minute respectively is (a) 2.16 10 12 unit (b) 1.26 10 12 unit (c) 2.16 1010 unit (d) 66 Units, Dimensions and Measurement Solution : (a) [P] [ML2 T 3 ] x Using the y z M L T n 2 n1 1 1 1 1 10 6 M 2 L 2 T2 relation 1 2 1 kg 1m 1s 10 kg 1 dm 1 min 3 [As 1MW 10 6 W ] 2 3 60 1kg 10 dm 1sec 10 6 10 kg 1 dm 60 sec 2.16 10 12 unit Problem 28. In two systems of relations among velocity, acceleration and force are respectively F 2 v1 , a2 a1 and F2 1. If and are constants then relations among mass, length E3 v2 and time in two systems are 2 3 T1 M1 , L2 2 L1 , T2 (c) M 2 3 2 M , L L , T T1 1 2 3 2 1 2 2 2 [L 2 T21 ] [L1 T11 ] (d) M 2 2 2 M1 , L2 3 L , T T1 2 3 1 2 2 3 M , L L , T T1 1 2 1 2 2 2 3......(i) a 2 a1 [L 2 T22 ] [L1 T12 ]......(ii) D YG and F2 1 U Solution : (b) v 2 v1 (b) M 2 ID (a) M 2 F1 [M 2 L 2 T2 2 ] [M 1 L1 T1 2 ] 1 ......(iii) Dividing equation (iii) by equation (ii) we get M 2 M1 M 2 12 ( ) B Squaring equation (i) and dividing by equation (ii) we get L 2 L1 U Dividing equation (i) by equation (ii) we get T 2 T1 3 3 2 ST Problem 29. If the present units of length, time and mass (m, s, kg) are changed to 100m, 100s, and 1 kg then 10 (a) The new unit of velocity is increased 10 times 1 force is decreased times 1000 (b) The new unit of (c) The new unit of energy is increased 10 times pressure is increased 1000 times (d) The new unit of Solution : (b) Unit of velocity = m/sec ; in new system = Unit of force kg m sec 2 ; in new system 100 m m (same) 100 sec sec 1 kg m 1 100 m kg 10 100 sec 100 sec 1000 sec 2 Units, Dimensions and Measurement 67 Unit of energy kg m 2 Unit of pressure sec 2 ; in new system kg m sec 2 kg m 2 1 100 m 100 m kg 10 sec 2 10 100 sec 100 sec ; in new system 1 1 1 kg kg m 10 7 10 100 100 sec 100 sec m sec 2 60 Problem 30. Suppose we employ a system in which the unit of mass equals 100 kg, the unit of length equals 1 km and the unit of time 100 s and call the unit of energy eluoj (joule written in reverse order), then (a) 1 eluoj = 104 joule (b) 1 eluoj = 10-3 joule 1 eluoj = 10-4 joule (d) E3 Solution : (a) [E] [ML2 T 2 ] (c) 1 eluoj [100 kg] [1km] 2 [100 sec]2 100 kg 10 6 m 2 10 4 sec 2 10 4 kg m 2 sec 2 10 4 Joule Problem 31. If 1gm cms–1 = x Ns, then number x is equivalent to (a) 1 10 1 (c) 6 10 4 (d) 1 10 5 ID (b) 3 10 2 Solution : (d) gm - cm s 1 10 3 kg 10 2 m s 1 10 5 kg m s 1 = 10–5 Ns D YG If X A (BC)2 DEF , U (4) To check the dimensional correctness of a given physical relation : This is based on the ‘principle of homogeneity’. According to this principle the dimensions of each term on both sides of an equation must be the same. then according to principle of homogeneity [X] = [A] = [(BC)2] [ DEF ] If the dimensions of each term on both sides are same, the equation is dimensionally correct, otherwise not. A dimensionally correct equation may or may not be physically correct. Example : (1) F mv 2 / r 2 By substituting dimension of the physical quantities in the above relation – [MLT 2 ] [MT 2 ] ST i.e. U [MLT 2 ] [M ][LT 1 ]2 /[L]2 As in the above equation dimensions of both sides are not same; this formula is not correct dimensionally, so can never be physically. (2) s ut (1 / 2)at 2 By substituting dimension of the physical quantities in the above relation – [L] = [LT–1][T] – [LT–2][T2] i.e. [L] = [L] – [L] As in the above equation dimensions of each term on both sides are same, so this equation is dimensionally correct. However, from equations of motion we know that s ut (1 / 2)at 2 Sample problems based on formulae checking 68 Units, Dimensions and Measurement Problem 32. From the dimensional consideration, which of the following equation is correct Solution : (a) T 2 R3 GM R3 GM GM (b) T 2 R3 2 gR 2 2 R3 (c) T 2 GM (d) T 2 R2 R g R2 GM [As GM = gR2] 60 (a) T 2 Now by substituting the dimension of each quantity in both sides. 1/2 [T ] E3 L [T ] LT 2 L.H.S. = R.H.S. i.e., the above formula is Correct. Problem 33. A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of the same dimensions and of low modulus of rigidity such that the M L (b) 2 L M (c) 2 U (a) 2 ID lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn block A executes small oscillations. The time period of which is given by [IIT-JEE 1992] ML (d) 2 M L D YG Solution : (d) Given m = mass = [M], = coefficient of rigidity = [ML 1 T 2 ] , L = length = [L] By substituting the dimension of these quantity we can check the accuracy of the given formulae [M ] [T ] 2 [][L] 1/2 M = 1 2 ML T L 1/2 = [T]. L.H.S. = R.H.S. i.e., the above formula is Correct. Problem 34. A small steel ball of radius r is allowed to fall under gravity through a column of a viscous U liquid of coefficient of viscosity. After some time the velocity of the ball attains a constant value known as terminal velocity v T. The terminal velocity depends on (i) the mass of the ST ball. (ii) (iii) r and (iv) acceleration due to gravity g. which of the following relations is dimensionally correct (a) vT mg r [CPMT 1992; CBSE 1992; NCERT 1983; MP PMT 2001] Solution : (a) Given v T = terminal velocity = [ LT [ LT 2 vT (b) 1 r mg (c) vT rmg (d) ] , m = Mass = [M], g = Acceleration due to gravity = ] r = Radius = [L], = Coefficient of viscosity = [ ] By substituting the dimension of each quantity we can check the accuracy of given formula mg vT r Units, Dimensions and Measurement 69 [LT 1 ] [M ][LT 2 ] [ML1 T 1 ][L] = [ LT 1 ] L.H.S. = R.H.S. i.e., the above formula is Correct. Problem 35. A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity pr 4 8l (b) V Solution : (a) Given V = Rate of flow = l 8 pr 4 (c) V 8 pl r (d) V 4 p 8lr 4 Volume [L3 T 1 ] , P = Pressure = [ ML1 T 2 ] , r = Radius = [L] sec E3 (a) V 60 flowing per second through a tube of radius r and length l and having a pressure difference p across its end, is = Coefficient of viscosity = [ML1 T 1 ] , l = Length = [L] V P r 4 8l [ML1T 2 ] [L4 ] 1 1 [ML T ] [L] = [L3 T 1 ] U [L3 T 1 ] ID By substituting the dimension of each quantity we can check the accuracy of the formula L.H.S. = R.H.S. i.e., the above formula is Correct. 1 2 D YG Problem 36. With the usual notations, the following equation S t u a(2t 1) is (a) Only numerically correct correct (b) (c) Both numerically and dimensionally correct nor dimensionally correct Only dimensionally (d) Neither numerically Solution : (c) Given S t = distance travelled by the body in tth sec.= [LT 1 ] , a = Acceleration = [LT 2 ] , U v = velocity = [LT 1 ] , t = time = [T] By substituting the dimension of each quantity we can check the accuracy of the formula 1 a (2t 1) 2 ST St u [LT 1 ] [LT 1 ] [LT 2 ] [T ] [LT 1 ] [LT 1 ] [LT 1 ] Since the dimension of each terms are equal therefore this equation is dimensionally correct. And after deriving this equation from Kinematics we can also proof that this equation is correct numerically also. Problem 37. If velocity v, acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v, A and F would be (a) FA1 v (b) Fv3 A 2 (c) Fv 2 A 1 (d) F 2 v 2 A 1 Solution : (b) Given, v = velocity = [LT 1 ] , A = Acceleration = [LT 2 ] , F = force = [MLT 2 ] By substituting, the dimension of each quantity we can check the accuracy of the formula 70 Units, Dimensions and Measurement [Angular momentum] = Fv3 A 2 [ML 2 T 1 ] [MLT 2 ] [LT 1 ]3 [LT 2 ]2 = [ML2 T 1 ] L.H.S. = R.H.S. i.e., the above formula is Correct. 60 Problem 38. The largest mass (m) that can be moved by a flowing river depends on velocity (v), density ( ) of river water and acceleration due to gravity (g). The correct relation is 2v 4 g 2 (b) m v 6 g (c) m 2 v 4 g 3 (d) m v 6 g3 E3 (a) m Solution : (d) Given, m = mass = [M], v = velocity = [LT 1 ] , = density = [ML 3 ] , g = acceleration due to gravity = [LT–2] By substituting, the dimension of each quantity we can check the accuracy of the formula g3 [ML3 ][ LT 1 ]6 [LT 2 ] 3 = [M] ID [M ] v6 U mK D YG L.H.S. = R.H.S. i.e., the above formula is Correct. (5) As a research tool to derive new relations : If one knows the dependency of a physical quantity on other quantities and if the dependency is of the product type, then using the method of dimensional analysis, relation between the quantities can be derived. Example : (i) Time period of a simple pendulum. Let time period of a simple pendulum is a function of mass of the bob (m), effective length U (l), acceleration due to gravity (g) then assuming the function to be product of power function of m, l and g ST i.e., T Km x l y g z ; where K = dimensionless constant If the above relation is dimensionally correct then by substituting the dimensions of quantities – [T] = [M]x [L]y [LT–2]z or [M0L0T1] = [MxLy+zT–2z] Equating the exponents of similar quantities x = 0, y = 1/2 and z = – 1/2 So the required physical relation becomes T K l g The value of dimensionless constant is found (2 ) through experiments so T 2 l g Units, Dimensions and Measurement 71 (ii) Stoke’s law : When a small sphere moves at low speed through a fluid, the viscous force F, opposing the motion, is found experimentally to depend on the radius r, the velocity of the sphere v and the viscosity of the fluid. So F = f (, r, v) If the function is product of power functions of , r and v, where K is 60 F K x r y v z ; dimensionless constant. or [MLT 2 ] [M x L x y z T x z ] E3 If the above relation is dimensionally correct [MLT 2 ] [ML1 T 1 ] x [L]y [LT 1 ]z Equating the exponents of similar quantities x = 1; – x + y + z = 1 and – x – z = – 2 Solving these for x, y and z, we get x = y = z = 1 F = Krv ID So eqn (i) becomes On experimental grounds, K = 6; so F = 6rv This is the famous Stoke’s law. U Sample problem based on formulae derivation Problem 39. If the velocity of light (c), gravitational constant (G) and Planck's constant (h) are chosen D YG as fundamental units, then the dimensions of mass in new system is (a) c 1 / 2 G 1 / 2 h 1 / 2 (b) c 1 / 2 G 1 / 2 h 1 / 2 (d) c 1 / 2 G 1 / 2 h 1 / 2 (c) c 1 / 2 G 1 / 2 h 1 / 2 Solution : (c) Let m c x G y h z or m K c x G y h z By substituting the dimension of each quantity in both sides [M 1 L0 T 0 ] K[LT 1 ] x [M 1 L3 T 2 ]y [ML2 T 1 ]z [M y z Lx 3 y 2 z T x 2 y z ] By equating the power of M, L and T in both sides : y z 1 , x 3 y 2 z 0 , x 2 y z 0 By solving above three equations x 1 / 2 , y 1 / 2 and z 1 / 2. U m c 1 / 2 G 1 / 2 h 1 / 2 Problem 40. If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius ST (r) of the drop and density ( ) of the liquid, then the expression of T is (a) T K r 3 / S (b) T K 1 / 2 r 3 / S (c) T K r 3 / S 1 / 2 (d) None of these Solution : (a) Let T S x r y z or T = K S x r y z By substituting the dimension of each quantity in both sides [M 0 L0 T 1 ] K [MT 2 ] x [L]y [ML3 ]z [M x z Ly 3 z T 2 x ] By equating the power of M, L and T in both sides x z 0 , y 3 z 0 , 2 x 1 By solving above three equations x 1 / 2 , y 3 / 2 , z 1 / 2 So the time period can be given as, T K S 1 / 2 r 3 / 2 1 / 2 K r 3 S. 72 Units, Dimensions and Measurement Problem 41. If P represents radiation pressure, C represents speed of light and Q represents radiation energy striking a unit area per second, then non-zero integers x, y and z such that P x Q y C z is dimensionless, are [AFMC 1991; CBSE 1992; CPMT 1981, 92; MP PMT 1992] (b) x 1, y 1, z 1 (c) x 1, y 1, z 1 Solution : (b) [P x Q y C z ] M 0 L0 T 0 (d) x 1, y 1, z 1 60 (a) x 1, y 1, z 1 By substituting the dimension of each quantity in the given expression [ML 1 T 2 ] x [MT 3 ] y [LT 1 ] z [M x y L x z T 2 x 3 y z ] M 0 L0 T 0 E3 by equating the power of M, L and T in both sides: x y 0 , x z 0 and 2 x 3y z 0 by solving we get x 1, y 1, z 1. Problem 42. The volume V of water passing through a point of a uniform tube during t seconds is related (a) (b) ID to the cross-sectional area A of the tube and velocity u of water by the relation V A u t , which one of the following will be true (c) (d) Solution : (b) Writing dimensions of both sides [L3 ] [L2 ] [LT 1 ] [ T ] [L3 T 0 ] [L2 T ] 1 (3 ) i.e. . 2 D YG Which give and U By comparing powers of both sides 2 3 and 0 Problem 43. If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional formula for mass will be (a) V 2 F0 E (b) V 0 FE2 (c) VF2 E0 (d) V 2 F0 E Solution : (d) Let M V a Fb Ec Putting dimensions of each quantities in both side [M ] [LT 1 ]a [MLT 2 ]b [ML2 T 2 ]c Equating powers of dimensions. We have b c 1, a b 2c 0 and a 2b 2c 0 U Solving these equations, a 2, b = 0 and c = 1 So M [V 2 F 0 E] ST Problem 44. Given that the amplitude A of scattered light is : (i) Directly proportional to the amplitude (A0) of incident light. (ii) Directly proportional to the volume (V) of the scattering particle (iii) Inversely proportional to the distance (r) from the scattered particle (iv) Depend upon the wavelength ( ) of the scattered light. then: (a) A Solution : (b) Let A 1 (b) A 1 2 (c) A 1 3 KA0 Vx r By substituting the dimension of each quantity in both sides (d) A 1 4 Units, Dimensions and Measurement 73 [L].[ L3 ][L x ] [ L] [ L] [L] [L3 x ] ; 3 x 1 or x 2 A 2 60 1.12 Limitations of Dimensional Analysis. Although dimensional analysis is very useful it cannot lead us too far as, is [ML2 T 2 ]it may be work or energy or torque. E3 (1) If dimensions are given, physical quantity may not be unique as many physical quantities have same dimensions. For example if the dimensional formula of a physical quantity (2) Numerical constant having no dimensions [K] such as (1/2), 1 or 2 etc. cannot be deduced by the methods of dimensions. or y a sin t U s u t (1 / 2) a t 2 ID (3) The method of dimensions can not be used to derive relations other than product of power functions. For example, cannot be derived by using this theory (try if you can). However, the dimensional correctness of these can be checked. D YG (4) The method of dimensions cannot be applied to derive formula if in mechanics a physical quantity depends on more than 3 physical quantities as then there will be less number (= 3) of equations than the unknowns (>3). However still we can check correctness of the given equation dimensionally. For example T 2 1 mgl can not be derived by theory of dimensions but its dimensional correctness can be checked. (5) Even if a physical quantity depends on 3 physical quantities, out of which two have same dimensions, the formula cannot be derived by theory of dimensions, e.g., formula for the U frequency of a tuning fork f (d / L2 ) v cannot be derived by theory of dimensions but can be checked. ST 1.13 Significant Figures. Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true. The following rules are observed in counting the number of significant figures in a given measured quantity. (1) All non-zero digits are significant. Example : 42.3 has three significant figures. 243.4 has four significant figures. 74 Units, Dimensions and Measurement 24.123 has five significant figures. (2) A zero becomes significant figure if it appears between to non-zero digits. Example : 5.03 has three significant figures. 5.604 has four significant figures. 60 4.004 has four significant figures. (3) Leading zeros or the zeros placed to the left of the number are never significant. 0.543 has three significant figures. 0.045 has two significant figures. 0.006 has one significant figures. E3 Example : (4) Trailing zeros or the zeros placed to the right of the number are significant. 4.330 has four significant figures. ID Example : 433.00 has five significant figures. 343.000 has six significant figures. Example : U (5) In exponential notation, the numerical portion gives the number of significant figures. 1.32 10–2 has three significant figures. D YG 1.32 104 has three significant figures. 1.14 Rounding Off. While rounding off measurements, we use the following rules by convention: (1) If the digit to be dropped is less than 5, then the preceding digit is left unchanged. Example : x 7.82 is rounded off to 7.8, again x 3. 94 is rounded off to 3.9. (2) If the digit to be dropped is more than 5, then the preceding digit is raised by one. U Example : x = 6.87 is rounded off to 6.9, again x = 12.78 is rounded off to 12.8. (3) If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is ST raised by one. Example : x = 16.351 is rounded off to 16.4, again x = 6.758 is rounded off to 6.8. (4) If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even. Example : x = 3.250 becomes 3.2 on rounding off, again x = 12.650 becomes 12.6 on rounding off. (5) If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd. Example : x = 3.750 is rounded off to 3.8, again x = 16.150 is rounded off to 16.2. 1.15 Significant Figures in Calculation. Units, Dimensions and Measurement 75 In most of the experiments, the observations of various measurements are to be combined mathematically, i.e., added, subtracted, multiplied or divided as to achieve the final result. Since, all the observations in measurements do not have the same precision, it is natural that the final result cannot be more precise than the least precise measurement. The following two rules should be followed to obtain the proper number of significant figures in any calculation. 33.3 (has only one decimal place) E3 (i) 3.11 + 0.313 (ii) 3.1421 0.241 + 0.09 3.4731 (iii) (has 2 decimal places) (answer should be reported to 2 decimal places) D YG Answer = 3.47 ID Answer = 36.7 (answer should be reported to one decimal place) U 36.723 60 (1) The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as are present in the number having the least number of decimal places. The rule is illustrated by the following examples : 62.831 (has 3 decimal places) – 24.5492 38.2818 rounding off) (answer should be reported to 3 decimal places after Answer = 38.282 ST U (2) The answer to a multiplication or division is rounded off to the same number of significant figures as is possessed by the least precise term used in the calculation. The rule is illustrated by the following examples : (i) 142.06 0.23 32.6738 (two significant figures) (answer should have two significant figures) Answer = 33 (ii) 51.028 1.31 66.84668 Answer = 66.8 (three significant figures) 76 Units, Dimensions and Measurement 0.90 0.2112676 4.26 (iii) Answer = 0.21 1.16 Order of Magnitude. 60 In scientific notation the numbers are expressed as, Number M 10 x. Where M is a number lies between 1 and 10 and x is integer. Order of magnitude of quantity is the power of 10 required to represent the quantity. For determining this power, the value of the quantity has E3 to be rounded off. While rounding off, we ignore the last digit which is less than 5. If the last digit is 5 or more than five, the preceding digit is increased by one. For example, (1) Speed of light in vacuum 3 10 8 ms 1 10 8 m / s (ignoring 3 < 5) (2) Mass of electron 9.1 10 31 kg 10 30 kg (as 9.1 > 5). ID Sample problems based on significant figures Problem 45. Each side a cube is measured to be 7.203 m. The volume of the cube up to appropriate significant figures is (b) 373.71 (c) 373.7 U (a) 373.714 (d) 373 Solution : (c) Volume a 3 (7.023)3 373.715 m 3 D YG In significant figures volume of cube will be 373.7 m 3 because its side has four significant figures. Problem 46. The number of significant figures in 0.007 m 2 is (a) 1 Solution : (a) (b) 2 (c) 3 (d) 4 Problem 47. The length, breadth and thickness of a block are measured as 125.5 cm, 5.0 cm and 0.32 cm U respectively. Which one of the following measurements is most accurate (a) Length (b) Breadth (c) Thickness (d) Height ST Solution : (a) Relative error in measurement of length is minimum, so this measurement is most accurate. Problem 48. The mass of a box is 2.3 kg. Two marbles of masses 2.15 g and 12.39 g are added to it. The total mass of the box to the correct number of significant figures is (a) 2.340 kg (b) 2.3145 kg. (c) 2.3 kg (d) 2.31 kg Solution : (c) Total mass 2.3 0.00215 0.01239 2.31 kg Total mass in appropriate significant figures be 2.3 kg. Problem 49. The length of a rectangular sheet is 1.5 cm and breadth is 1.203 cm. The area of the face of rectangular sheet to the correct no. of significant figures is : (a) 1.8045 cm 2 (b) 1.804 cm 2 (c) 1.805 cm 2 (d) 1.8 cm 2 Solution : (d) Area 1.5 1.203 1.8045 cm 2 1.8 cm 2 (Upto correct number of significant figure). Units, Dimensions and Measurement 77 Problem 50. Each side of a cube is measured to be 5.402 cm. The total surface area and the volume of the cube in appropriate significant figures are : (a) 175.1 cm 2 , 157 cm 2 (b) 175.1 cm 2 , 157.6 cm 3 (c) 175 cm 2 , 157 cm 2 (d) 175.08 cm 2 , 157.639 cm 3 60 Solution : (b) Total surface area = 6 (5.402)2 175.09 cm 2 175.1 cm 2 (Upto correct number of significant figure) Total volume (5.402)3 175.64 cm 3 175.6 cm 3 (Upto correct number of significant figure). (a) 10.00 m E3 Problem 51. Taking into account the significant figures, what is the value of 9.99 m + 0.0099 m (b) 10 m (c) 9.9999 m (d) 10.0 m Solution : (a) 9.99 m 0.0099 m 9.999 m 10.00 m (In proper significant figures). Problem 52. The value of the multiplication 3.124 4.576 correct to three significant figures is (b) 14.3 (c) 14.295424 ID (a) 14.295 (d) 14.305 Solution : (b) 3.124 4.576 14.295 =14.3 (Correct to three significant figures). Problem 53. The number of the significant figures in 11.118 10 6 V is (b) 4 (c) 5 U (a) 3 (d) 6 Solution : (c) The number of significant figure is 5 as 10 6 does not affect this number. D YG Problem 54. If the value of resistance is 10.845 ohms and the value of current is 3.23 amperes, the potential difference is 35.02935 volts. Its value in significant number would be (a) 35 V (b) 35.0 V (c) 35.03 V (d) 35.025 V Solution : (b) Value of current (3.23 A) has minimum significant figure (3) so the value of potential difference V( IR) have only 3 significant figure. Hence its value be 35.0 V. 1.17 Errors of Measurement. U The measuring process is essentially a process of comparison. Inspite of our best efforts, the measured value of a quantity is always somewhat different from its actual value, or true value. This difference in the true value of a quantity is called error of measurement. ST (1) Absolute error : Absolute error in the measurement of a physical quantity is the magnitude of the difference between the true value and the measured value of the quantity. Let a physical quantity be measured n times. Let the measured value be a1, a2, a3, ….. an. The a a 2 ....an arithmetic mean of these value is am 1 n Usually, am is taken as the true value of the quantity, if the same is unknown otherwise. By definition, absolute errors in the measured values of the quantity are a1 a m a1 a 2 am a 2 …………. 78 Units, Dimensions and Measurement an am an The absolute errors may be positive in certain cases and negative in certain other cases. (2) Mean absolute error : It is the arithmetic mean of the magnitudes of absolute errors in all the measurements of the quantity. It is represented by a. Thus | a1 | | a 2 | ..... | an | n 60 a Hence the final result of measurement may be written as a am a E3 This implies that any measurement of the quantity is likely to lie between (am a) and (am a). ID (3) Relative error or Fractional error : The relative error or fractional error of measurement is defined as the ratio of mean absolute error to the mean value of the quantity measured. Thus Relative error or Fractional error mean absolute error a mean value am a 100 % am D YG Percentage error U (4) Percentage error : When the relative/fractional error is expressed in percentage, we call it percentage error. Thus 1.18 Propagation of Errors. (1) Error in sum of the quantities : Suppose x = a + b Let a = absolute error in measurement of a b = absolute error in measurement of b U x = absolute error in calculation of x i.e. sum of a and b. The maximum absolute error in x is x (a b) ST Percentage error in the value of x (a b) 100 % ab (2) Error in difference of the quantities : Suppose x = a – b Let a = absolute error in measurement of a, b = absolute error in measurement of b x = absolute error in calculation of x i.e. difference of a and b. The maximum absolute error in x is x (a b) Percentage error in the value of x (a b) 100 % ab (3) Error in product of quantities : Suppose x = a b Let a = absolute error in measurement of a, Units, Dimensions and Measurement 79 b = absolute error in measurement of b x = absolute error in calculation of x i.e. product of a and b. The maximum fractional error in x is x a b x b a Let a = absolute error in measurement of a, b = absolute error in measurement of b a b E3 (4) Error in division of quantities : Suppose x 60 Percentage error in the value of x = (Percentage error in value of a) + (Percentage error in value of b) x = absolute error in calculation of x i.e. division of a and b. x a b x b a ID The maximum fractional error in x is U Percentage error in the value of x = (Percentage error in value of a) + (Percentage error in value of b) (5) Error in quantity raised to some power : Suppose x an bm D YG Let a = absolute error in measurement of a, b = absolute error in measurement of b x = absolute error in calculation of x The maximum fractional error in x is x b a n m x b a U Percentage error in the value of x = n (Percentage error in value of a) + m (Percentage error in value of b) Note : The quantity which have maximum power must be measured carefully because it's ST contribution to error is maximum. Sample problems based on errors of measurement Problem 55. A physical parameter a can be determined by measuring the parameters b, c, d and e using the relation a = b c / d e . If the maximum errors in the measurement of b, c, d and e are b1 %, c 1 %, d1 % and e1 %, then the maximum error in the value of a determined by the experiment is [CPMT 1981] (a) ( b1 c1 d1 e1 )% (b) ( b1 c1 d1 e1 )% (c) ( b1 c1 d1 e1 )% (d) ( b1 c1 d1 e1 )% Solution : (d) a b c /d e So maximum error in a is given by 80 Units, Dimensions and Measurement b c d e a 100 . 100 . 100 . 100 . 100 b c d e a max b1 c1 d1 e1 % Problem 56. The pressure on a square plate is measured by measuring the force on the plate and the Solution : (d) P (b) 2% (c) 6% F F , so maximum error in pressure (P ) A l2 (d) 8% E3 (a) 1% 60 length of the sides of the plate. If the maximum error in the measurement of force and length are respectively 4% and 2%, The maximum error in the measurement of pressure is F l P 100 100 2 100 = 4% + 2 × 2% = 8% F l P max Problem 57. The relative density of material of a body is found by weighing it first in air and then in ID water. If the weight in air is (5.00 0.05 ) Newton and weight in water is (4.00 0.05) Newton. Then the relative density along with the maximum permissible percentage error is (b) 5.0 1% (a) 5.0 11% (d) 1.25 5% U Solution : (a) Weight in air (5.00 0.05 ) N (c) 5.0 6% Weight in water (4.00 0.05 ) N D YG Loss of weight in water (1.00 0.1) N Now relative density weight in air weight loss in water i.e. R. D 5.00 0.05 1.00 0.1 Now relative density with max permissible error 5.00 0.05 0.1 100 5.0 (1 10)% 1.00 5.00 1.00 5.0 11 % Problem 58. The resistance R = V where V= 100 5 volts and i = 10 0.2 amperes. What is the total i U error in R ST (a) 5% Solution : (b) R V I (b) 7% (c) 5.2% (d) 5 % 2 V I 5 0.2 R 100 100 100 100 100 (5 2)% = 7% V I 100 10 R max Problem 59. The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s respectively. The average absolute error is (a) 0.1 s Solution : (b) Average value (b) 0.11 s (c) 0.01 s 2.63 2.56 2.42 2.71 2.80 2.62 sec 5 Now | T1 | 2.63 2.62 0.01 | T2 | 2.62 2.56 0.06 (d) 1.0 s Units, Dimensions and Measurement 81 | T3 | 2.62 2.42 0.20 | T4 | 2.71 2.62 0.09 | T5 | 2.80 2.62 0.18 | T1 | | T2 | | T3 | | T4 | | T5 | 0.54 0.108 0.11 sec 5 5 60 Mean absolute error T Problem 60. The length of a cylinder is measured with a meter rod having least count 0.1 cm. Its (a) 1% E3 diameter is measured with venier calipers having least count 0.01 cm. Given that length is 5.0 cm. and radius is 2.0 cm. The percentage error in the calculated value of the volume will be (b) 2% (c) 3% Solution : (c) Volume of cylinder V r 2 l V 2r l 100 100 100 V r l ID Percentage error in volume (d) 4% U 0.01 0.1 2 100 100 (1 2)% = 3 % 2. 0 5. 0 Problem 61. In an experiment, the following observation's were recorded : L = 2.820 m, M = 3.00 kg, l = D YG 0.087 cm, Diameter D = 0.041 cm Taking g = 9.81 m / s 2 using the formula , Y= 4 Mg D 2 l , the maximum permissible error in Y is (a) 7.96% Solution : (c) Y 4 MgL D l 2 (b) 4.56% (c) 6.50% so maximum permissible error in Y = (d) 8.42% M g L 2D l Y 100 100 Y g L D l M 1 1 1 1 1 2 100 41 87 300 9.81 9820 U 0.065 100 6.5 % Problem 62. According to Joule's law of heating, heat produced H I 2 Rt, where I is current, R is ST resistance and t is time. If the errors in the measurement of I, R and t are 3%, 4% and 6% respectively then error in the measurement of H is (a) 17% (b) 16% (c) 19% (d) 25% Solution : (b) H I 2 R t H 2I R t 100 100 (2 3 4 6)% 16 % H R t I Problem 63. If there is a positive error of 50% in the measurement of velocity of a body, then the error in the measurement of kinetic energy is (a) 25% (b) 50% (c) 100% (d) 125% 82 Units, Dimensions and Measurement Solution : (c) Kinetic energy E E m 2v 100 100 E v m Here m 0 and v 100 50 % v 60 1 mv 2 2 E 100 2 50 100 % E 1 3 C 4 D 2 percentage error in P is (a) A (b) B. The quantity which brings in the maximum E3 Problem 64. A physical quantity P is given by P= A3B 2 (c) C ST U D YG U ID Solution : (c) Quantity C has maximum power. So it brings maximum error in P. (d) D
