Physics Study Module - Past Year NEET Trend PDF

1 Units and Measurements Past Years NEET Trend 6 5 4 No. of MCQs 3 2 1 0 2...

1 Units and Measurements Past Years NEET Trend 6 5 4 No. of MCQs 3 2 1 0 2021 2020 2019 2018 2017 2016 2015 2014 2013 2012 Investigation Report TARGET EXAM PREDICTED NO. OF MCQs CRITICAL CONCEPTS NEET 1-2 SI units, Dimensional Analysis, Errors Perfect Practice Plan Topicwise Questions Learning Plus Multiconcept MCQs NEET Past 10 Years Questions Total MCQs 59 26 11 19 115 PHYSICAL QUANTITIES Systems of Units: All quantities that can be measured are called physical There are four systems of units quantities. e.g. length, mass, force, work done, etc. In physics F.P.S C.G.S we study about physical quantities and their inter relationship. M.K.S SI There are two types of physical quantities FPS or British Engineering system: In this system length, (i) Fundamental quantities mass and time are taken as fundamental quantities and their (ii) Derived quantities base units are foot (ft), pound (lb) and second (s) respectively. Fundamental Quantity: Physical quantities which cannot be CGS or Gaussian system: In this system the fundamental expressed in terms of any other physical quantities are called quantities are length, mass and time and their respective units fundamental physical quantities. are centimetre (cm), gram (g) and second (s). E.g. length, mass, time, temperature etc. MKS system: In this system also the fundamental quantities Derived Quantity: Physical Quantities which are derived are length, mass and time but their fundamental units are from fundamental quantities are called derived quantities. metre (m), kilogram (kg) and second (s) respectively. E.g. Area, density, force etc. Units of some fundamental physical quantities in different systems MEASUREMENT Type of Physical System Measurement is the comparison of a physical quantity with a physical Quantity standard of the same physical quantity. Quantity Different countries followed different standards. CGS MKS FPS Units of Measurement: Length cm m ft A fixed measurement chosen as a standard of measurement to Fundamental Mass g kg lb measure a physical quantity is called a Unit. Time s s s To measure a physical quantity means to determine the number International system (SI) of units: This system is modification of times its standard unit is contained in that physical quantity. over the MKS system. Besides the three base units of MKS system four fundamental and two supplementary units are also A standard Unit is necessary for the sake of included in this system. (i) Accuracy, Based on SI there are three categories of physical quantities. (ii) Convenience, 7 fundamental quantities (iii) Uniformity and 2 supplementary quantities (iv) Equal justice to all. and derived quantities The standard unit chosen should have the following characteristics. SET OF FUNDAMENTAL QUANTITIES (i) Consistency (or) invariability A set of physical quantities which are completely independent (ii) Availability (or) reproducibility of each other and all other physical quantities can be (iii) Imperishability (Permanency) expressed in terms of these physical quantities is called Set of (iv) Convenience and acceptability. Fundamental Quantities. The numerical value obtained on measuring a physical Fundamental Quantities and Their S.I. Units quantity is inversely proportional to the magnitude of the unit There are seven fundamental quantities and two supplementary chosen. quantities in S.I. system. The names and units with symbols are given below: 1 nα ⇒ n1U1 = n2U 2 SI Base Quantities and Their Units U S. No. Physical quantity Unit Symbol Where n1 and n2 are the numerical values U1 and U2 are the units of same physical quantity in different systems. 1 Length Metre m Fundamental unit: The unit used to measure the 2 Mass Kilogram kg fundamental quantity is called fundamental unit. 3 Time Second s e.g., Metre for length, kilogram for mass etc. 4 Temperature Kelvin kg Derived unit: The unit used to measure the derived 5 Electric current Ampere A quantity is called derived unit. 6 Luminous Intensity Candela cd e.g., m2 for area, gm cm–3 for density etc... 7 Amount of substance Mole mol 6 Dropper NEET  Supplementary quantities Such an expression for a physical quantity in terms of base quantities is called dimensional formula. 1. Plane angle Radian Rad Physical quantity can be further of four types : 1. Dimension less constant i.e. 1,2,3, π etc. 2. Solid angle Steradian sr 2. Dimension less variable i.e. angle q etc. 3. Dimensional constant i.e. G, h etc. Some Special Units for Length: 4. Dimensional variable i.e. F, v, etc. angstrom (Å) = 10–10 m = 10–8 cm Dimension nanometre (nm) = 10–9 m = 10 Å Dimensions of a physical quantity are the powers to which the Fermi = 10–15 m fundamental units are to be raised to obtain one unit of that Micron = 10–6 m quantity X-ray unit = 10–13 m DIMENSIONAL EQUATION 1 A.U. = Distance between sun & earth Whenever the dimension of a physical quantity is equated with its dimensional formula, we get a dimensional equation. = 1.496×1011 m Dimensional Formula Light year = 9.46 × 1015 m An expression showing the powers to which the fundamental Par sec = 3.26 light years = 30.84 × 1015 m units are to be raised to obtain one unit of the derived quantity is called Dimensional formula of that quantity. Bohr radius = 0.5 × 10–10 m In general the dimensional formula of a quantity can be written Mile = 1.6 km as [MxLyTz]. Here x,y,z are dimensions. Some Special Units for Mass: Dimensional Constants Quintal = 100 kg The physical quantities which have dimensions and have a Metric ton = 1000 kg fixed value are called dimensional constants. Eg: Gravitational Constant (G), Planck’s Constant (h), Atomic mass unit (a.m.u) = 1.67 × 10–27 kg Universal gas constant (R), Velocity of light in vacuum (c) etc., Some Special Units for Time: Dimensionless Quantities One day = 86400 second Dimensionless quantities are those which do not have Shake = 10–8 second dimensions but have a fixed value. One light year is distance travelled by light in one year in vacuum or air. This unit is used in astronomy. (a) Dimensionless quantities without units. Astronomical unit is the mean distance of the earth from the Eg: Pure numbers, etc., sun. This unit is used in astronomy. (b) Dimensionless quantities with units. DIMENSIONS AND DIMENSIONAL FORMULA Eg: Angular displacement-radian, angle Joule’s constant-joule/calorie,etc., All the physical quantities of interest can be derived from the base quantities. “The power (exponent) of base quantity PRINCIPLE OF HOMOGENEITY that enters into the expression of a physical quantity, is called The magnitude of a physical quantity may be added or the dimension of the quantity in that base. To make it clear, subtracted from each other only if they have the same consider the physical quantity force. dimension, also the dimension on both sides of an equation Force = mass × acceleration must be same. This is called as principle of homogeneity. length/time Dimensional Variables mass × time Dimensional variables are those physical quantities which = mass × length × (time)–2 have dimensions and do not have fixed value. So the dimensions of force are 1 in mass, 1 in length and –2 Eg: velocity, acceleration, force, work, power... etc. in time. Thus [Force] = MLT–2 Dimensionless Variables Similarly energy has dimensional formula given by Dimensionless variables are those physical quantities which [Energy ] = ML2T–2 do not have dimensions and do not have fixed value. i.e. energy has dimensions, 1 in mass, 2 in length and –2 in Eg: Specific gravity, refractive index, Coefficient of friction, time. Poisson’s Ratio etc., Units and Measurements 7 USES OF DIMENSIONAL ANALYSIS Sol. The dimensional equation of force is [F] = [M1 L1T–2] (i) To check the dimensional correctness of a given Therefore if n1, u1 and n2, u2 corresponds to SI & CGS physical relation unit respectively, then It is based on principle of homogeneity, which 1 1 −2  M1   L1   T1  states that a given physical relation is dimensionally = n2u2 = n1u1 ⇒ n2 n= 1      1 correct if the dimensions of the various terms on  M 2   L2   T2  −2 either side of the relation are the same.  kg   m   s   g   cm   s  = 1× 1000 × 100 × 1 = 105  KEY NOTE     Š Powers are dimensionless ⇒ 105 Dyne = 1 N Š sin q, eq, cos q, log q gives dimensionless value and in above expression q is dimensionless Q. Let us find an expression for the time period t of a simple Š We can add or subtract quantity having same dimensions. pendulum. The time period t may possibly depend upon (i) mass m of the bob of the pendulum, (ii) length l of pendulum, (iii) acceleration due to gravity g at the place TRAIN YOUR BRAIN where the pendulum is suspended. Q. The position of a particle at time t, is given by the v Sol. Let (i) t ∝ ma (ii) t ∝ lb (iii) t ∝ gc equation, x(t) = 0 (1 – e–at), where v0 is a constant and α a > 0. The dimensions of v0 & a are respectively. Combining all the three factors, we get (a) M0L1 T0 & T–1 (b) M0L1 T–1 & T t ∝ malbgc or t = Kmalbgc (c) M0 L1 T–1 & T–1 (c) M1 L1 T–1 & LT–2 Where K is a dimensionless constant of proportionality. Sol. (c) [v0] = [x] [a] = M 0L1T –1 & [a] [t] = M0L0T0 Writing down the dimensions on either side of equation, ⇒ [a] = M0L0T–1 we get L Q. Check the accuracy of the relation T = 2π for a [T] = [Ma][Lb][LT–2]c = [MaLb + c T–2c] g simple pendulum using dimensional analysis.  Sol. The dimensions of LHS = the dimension of T = [M0L0T1] Comparing dimensions we can get t = K g 1/2  dim.of length  Q. A calorie is a unit of heat or energy and it equals about The dimensions of RHS =  n   dim.of acc  4.2 J, where 1 J = 1 kg m2/s2. Suppose we employ a (Q 2p is a dimensionless const.) system of units in which the unit of mass equals a kg, 1/2 the unit of length equals b metre, the unit of time is g  L  2 1/2 =  = (T )= (T =) [ M 0 L0T 1 ] second. Show that a calorie has a magnitude 4.2 a–1b–2g2  LT −2  in terms of the new units. (ii) To establish a relation between different physical quantities Sol. 1 cal = 4.2 kg m2s–2 If we know the various factors on which a physical quantity depends, then we can find a relation among SI New system different factors by using principle of homogeneity. (iii) To convert units of a physical quantity from one n1 = 4.2 n2 = ? system of units to another M1 = 1 kg M2 = a kg It is based on the fact that, L1 = 1 m L2 = b metre Numerical value × unit = constant T1 = 1 s T2 = g second So on changing unit, numerical value will also get changed. If n1 and n2 are the numerical values of a given physical quantity and u1 and u2 be the units Dimensional formula of energy is [ML2T–2] respectively in two different systems of units, then 1 2 −2  M   L  T  n1u1 = n2u2 Now, n2 = n1  1   1   1   M 2   L2   T2  TRAIN YOUR BRAIN 1 2 −2 Q. Convert 1 newton (SI unit of force) into dyne (CGS unit  1 kg  1 m  1 s  −1 −2 2 = 4.2    β m   γ s  = 4.2 α β γ of force)  α kg      8 Dropper NEET  TRAIN YOUR BRAIN Thus, [a] = length = L Q. The distance covered by a particle in time t is given by x [bt] = L, or [b] = LT–1 = a + bt + ct2 + dt3 ; find the dimensions of a, b, c and d. [ct2] = L, or [c] = LT–2 Sol. The equation contains five terms. All of them should have the same dimensions. Since [x] = length, each of and [dt3] = L or [d] = LT–3 the remaining four must have the dimension of length. The Following is the list of some Physical Quantities with their Formula and Dimensional Formula Dimensional S. No. Physical Quantity Explanation or Formulae S.I.Unit Formulae Distance, Displacement, Wave Length, 1. Radius of gyration, Circumference, [M0 L1 T 0] m Perimeter, Light year, Par-sec. 2. Mass [M1 L0 T 0] kg total time T= 3. Period of oscillation, Time, time constant No.of oscillations [M0 L0 T1] s T = Capacity × Resistance 4. Frequency 1 [M 0 L0 T –1] Hertz (Hz) Reciprocal of time period n = T 5. Area A = length × breadth [M0 L2 T 0] m2 6. Volume V = Length × breadth × height [M0 L3 T 0] m3 Mass 7. Density D= [M1 L–3 T 0] kgm–3 Volume Mass 8. Linear density λ= [M1 L–1 T 0] kgm–1 Length displacement 9. Speed, Velocity v= [M0 L1 T –1] ms–1 time Change in Velocity 10. Acceleration a= [M 0 L1 T –2] ms–2 time 11. Linear Momentum P = mass × velocity [M1 L1 T –1] kgms–1 12. Impulse J = Force × time [M1 L1 T –1] Ns 13. Force F = Mass × acceleration [M1 L1 T –2] N W = Force × displacement 1 1 14. Work,Energy,PE, KE, Strain Energy P.E = mgh; KE = MV 2 SE = [M1 L2 T –2] J(or) N.m 2 2 × Stress × Strain × Volume Work 15. Power P= [M1 L2 T–3] JS–1 (or) watt time Units and Measurements 9  The Following is the list of some Physical Quantities with their Formula and Dimensional Formula Dimensional S. No. Physical Quantity Explanation or Formulae S.I.Unit Formulae Force pressure = Pressure , Stress, Modulus of Elasticity Area 16. [M1 L–1 T–2] Nm–2 (or) Pascal (Y, n, k) Stress Y= Strain Change in dimension 17. Strain [M0 L0 T0] No units Original dimensions Work 18. Strain energy density E= [M1 L–1 T–2] Jm–3 Volume length of arc 19. Angular displacement θ= [M0 L0 T0] rad radius angular displacement 20. Angular Velocity ω= [M0 L0 T–1] rads–1 time Change in angular velocity 21. Angular acceleration α= [M0 L0 T–2] rads–2 time 22. Angular momentum L = Linear momentum × arm [M1 L2 T–1] Js energy 23. Planck’s constant h= [M1 L2 T–1] Js frequency 24. Torque =τ force × ⊥ dis tan ce [M1 L2 T–2] Nm Acceleration due to gravity(g) weight 25. g= [M0 L T–2] ms–2 or Nkg–1 = gravitational field strength mass F.d 2 26. Universal gravitational Constant G= [M–1 L3 T–2] Nm2 kg–2 M1.M 2 27. Moment of inertia I = MK2 [M1 L2 T0] kgm2 dv 28. Velocity gradient [M0 L0 T–1] S–1 dx tangential stress 29. Coefficient of Viscosity η= [M1 L–1 T–1] Pa s (or) Ns m–2 Velocity gradient 30. Heat energy msq [M1 L2 T–2] Joule 31. Temperature q [M0 L0 T0. q1] Kelvin (K) heat energy 32. Specific heat Capacity S (or)C = [M0 L2 T–2. q–1] JKg–1 K–1 mass×temp. 33. Water Equivalent W = MC [M1 L0 T0] kg 34. Coefficient of Thermal expansion a or b or g ask [q–1] k–1 PV [M1 L2 T–2 q–1 35. Universal gas constant R= Jmol–1 K–1 nT mol–1] 10 Dropper NEET  The Following is the list of some Physical Quantities with their Formula and Dimensional Formula Dimensional S. No. Physical Quantity Explanation or Formulae S.I.Unit Formulae R [M0 L2 T–2 q–1 36. Gas constant ( for 1 gram) r= Jkg–1 K–1 Mol.wt mol–1] R 37. Boltzman constant (for 1 Molecule) k= [M1 L2 T–2 q–1] JK–1 molecule–1 AvagadroNo. W 38. Mechanical equivalent of heat J= [M0 L0 T0] no S.I. units H Q.d Js–1 m–1 K–1 (or) 39. Coeff of Thermal Conductivity K= [M1 L1 T–3 q–1] A ∆θ.t Wm–1 K–1 dQ heat energy 40. Entropy = [M1 L2 T–2 q–1] JK–1 T temperature ∆E Js–1 m–2 K–4 (or) 41. Stefan’s Constant σ= [M1 L0 T–3 q–4] ∆A.∆T.θ 4 Wm–2 K–4 dθ temp × time =R =  dQ  Heat 42. Thermal resistance   [M–1 L–2 T3 q1] KsJ–1  dt  d or R = K.A Change in temp d θ 43. Temperature gradient = [q L–1] Km–1 length dl Change in pressure dp 44. Pressure gradient = [M1 L–2 T–2] Pascal m–1 length dl 45. Enthalpy Heat. (DQ) [M1 L2 T–2] Joule M = 2l × m = pole strength × length of 46. Magnetic Moment [M0 L2 T0 A] Am2 magnet φ= B × A = (Magnetic induction × 47. Magnetic flux [M1 L2 T–2 A–1] Wb area) φ Magnetic flux F Tesla (or) Wbm–2 48. Magnetic induction field strength B = = = [M1 L0 T–2 A–1] A area il (or) Na–1 m–1 4π.Fd 2 49. Magnetic permeability of free space µ0 = [M1 L1 T–2 A–2] Hm–1 m1.m2 50. Electric current I [M0 L0 T0 A.] A 51. Charge Q = Current × time [M0 L0 T.A] C 52. Electric dipole moment P = Charge × distance [M0 L0 T.A] Cm Electric field strength (or) Electric field force 53. E= [M1 L T–3 A–1] NC–1 Intensity Charge 54. Electrical flux (fE) Electrical Intensity × area [M1 L3 T–3 A–1] Nm2C–1 Units and Measurements 11  The Following is the list of some Physical Quantities with their Formula and Dimensional Formula Dimensional S. No. Physical Quantity Explanation or Formulae S.I.Unit Formulae Work 55. Electric potential (or) Potential difference V= [M1 L2 T–3 A–1] V Charge Pot.diff 56. Electrical resistance R= [M1 L2 T–3 A–2] W Current 1 1 57. Electrical conductance C= = [M–1 L–2 T3 A2] mho (or) siemen (S) R Resistance R. A 58. Specific resistances (or) Resistivity r (or) ρ= [M1 L3 T–3 A–2] Ohm m l 1 s= Electrical conductivity Resistivity [M–1 L–3 T3 A2] Ohm–1 m–1 (or) 59. Current density (Current per unit area of J = Electrical Intensity siemen m–1 70 cross section)  Current  0 –2 [M L T A]0 Am–2 × Conductivity or    area  Q Charge 60. Capacitance C = = [M–1 L–2 T4 A2] F V Potential dE Voltage × time =L = 61. Self (or) Mutual Inductance dI   Current [M1 L2 T–2 A–2] H (or) Wb/amp    dt  q.q 62. Electrical permittivity of free space ε0 = 1 2 2 [M–1 L–3 T4 A2] farad/m 4πFd Charge 63. Surface charge density [M0 L–2 T1 A1] Cm–2 area 64. Focal Power 1 [M0 L–1 T0] Dioptre P= focal length Physical Quantities Having Same Dimensional Angular momentum, Angular impulse, Planck’s constant Formulas: [ML2T–1] Distance, Displacement, radius, light year wavelength, radius Angular velocity, Frequency, Velocity gradient, Decay of gyration [L] constant, rate of disintegration [T–1] Speed, Velocity, Velocity of light [LT–1] Stress, Pressure, Modulus of Elasticity, Energy density [ML–1T–2] Acceleration, acceleration due to gravity, intensity of gravitational field, centripetal acceleration [LT–2] Specific heat, Specific gas constant [L2T–1q–1] Impulse, Change in momentum [MLT–1] Thermal capacity, Entropy, Boltzmann constant, Molar thermal capacity, [ML2T–2q–1] Force, Weight, Tension, Thrust [MLT–2] Intensity of magnetic field, Intensity of magnetization [IL–1] Work, Energy, Moment of force or Torque, Moment of couple [ML2T–2] Frequency, angular frequency, angular velocity, Disintegration constant and velocity gradient have same dimensional formula Force constant, Surface Tension, Spring constant, Energy per [M0L0T–1] unit area [MT–2] 12 Dropper NEET Pressure, stress, coefficient of elasticity, energy density have Unit Definition and Conversion same dimensional formula [ML–1T–2] 1 fermi 1f = 10–15 m Electric field and potential gradient have same dimensional formula [MLT–3A–1] 1 angstrom 1Å = 10–10 m 1 astronomical unit 1 AU = (average distance of the sum Surface tension, surface energy, force gradient and spring constant have same dimensional formula [ML0T–2] from the earth) = 1.496 × 1011 m Force, weight and Energy gradient have same dimensions [MLT–2] 1 light year 1 1y = 9.46 × 1015 m Light year, wave length and radius of gyration have same dimensional formula [M0LT0] (Distance that light travels with velocity of 3 × 108m/ s) in 1 year Strain, Poisson’s ratio, refractive index, dielectric constant, coefficient of friction, relative permeability, Magnetic 1 par sec 3.08 × 106 m susceptibility, Electric susceptibility, angle, solid angle, Parsec is the distance at which average Trigonometric ratios, exponential constant are all radius of earth’s orbit subtends an dimensionless. angle of 1 arc second TRAIN YOUR BRAIN Accuracy and Precision The numerical values obtained on measuring physical Q. If P is the pressure of a gas and r is its density, then find quantities depend upon the measuring instruments, methods the dimension of velocity of measurement. (a) P1/2r – 1/2 (b) P1/2r1/2 A unit of measurement of a physical quantity is the standard (c) P–1/2r1/2 (d) P–1/2r–1/2 reference of the same physical quantity which is used for comparison of the given physical quantity. Sol. (a) Method - I [P] = [ML–1T–2]...(1) Accuracy refers to how closely a measured value agrees with the true values. [r] = [ML–3]...(2) Precision refers to what limit or resolution the given physical Dividing eq. (1) by (2) quantity can be measured [Pr–1] = [L2T–2] Accuracy refers to the closeness of observed values to its ⇒ [LT–1] = [P1/2r–1/2] ⇒ [V] = [P1/2r–1/2] true value of the quantity while precision refer's to closeness Method - II between the different observed values of the same quantity. V ∝ Pa rb High precision does not mean high accuracy. The difference V = kPa rb between accuracy and precision can be understand by the following example: Suppose three students are asked to find [LT–1] = [ML–1T–2]a [ML–3]b the length of a rod whose length is known to be 2.250cm. The 1 1 observations are given in the table. a= ,b=– ⇒ [V] = [P1/2r–1/2] 2 2 Stu- Measure- Measure- Measure- Average LIMITATIONS OF DIMENSIONAL ANALYSIS dent ment-1 ment-2 ment-3 length METHOD: Dimensionless quantities cannot be determined by this A 2.25cm 2.27cm 2.26cm 2.26cm method. B 2.252cm 2.250cm 2.251cm 2.251cm Constant of proportionality cannot be determined by this C 2.250cm 2.250cm 2.251cm 2.250cm method. It is clear from the above table, that the observation taken by a This method is not applicable to trigonometric, logarthmic student A are neither precise nor accurate. The observations of and exponential functions. student B are more precise. The observations of student C are In the case of physical quantities which are dependent precise as will as accurate upon more than three physical quantities, this method will be difficult. Types of Errors In some cases, the constant of proportionality also possesses Uncertainity in measurement of a physical quantity is called dimensions. In such cases we cannot use this system. the error in measurement. If one side of equation contains addition or subtraction of The difference between the measured value and true value as physical quantities, we cannot use this method. per standard method without mistakes is called the error. We use certain special length units for short and large lengths. Error = True value - Measured value Correction = – error These are True value means, standard value free of mistakes. Units and Measurements 13 Errors are broadly classified into 3 types: (iii) Gross Errors (i) Systematic errors The cause for gross errors are improper recording, neglecting (ii) Random errors the sources of the error, reading the instrument incorrectly, (iii) Gross errors sheer carelessness (i) Systematic Errors/Controllable Errors Ex: In a tangent galvanometer experiment, the coil is to be The errors due to a definite cause and which follow a particular placed exactly in the magnetic meridian and care should be rule are called systematic errors. They always occur in one taken to see that no other magnetic material are present in the direction. Following are some systematic errors vicinity. Constant Error No correction can be applied to these gross errors. Systematic errors with a constant magnitude are called When the errors are minimized, the accuracy increases. constant errors. The systematic errors can be estimated and observations can The constant arised due to imperfect design, zero error in the be corrected. instrument or any other such defects. These are also called Random errors are compensating type. A physical quantity is instrumental errors. measured number of times and these values lie on either side of mean value-with random errors. These errors are estimated Zero Error: by statistical methods and accuracy is achieved. The error due to improper designing and construction. Ex: If a screw gauge has a zero error of -4 head scale divisions,  KEY NOTE then every reading will be 0.004cm less than the true value. Š Personal errors like parallax error can be avoided by Environmental Error: taking proper care. Š The instrumental errors are avoided by calibrating the The error arised due to external conditions like changes in environment, changes in temperature, pressure, humidity etc. instrument with a standard value and by applying proper corrections. Ex: Due to rise in temperature, a scale gets expanded and this results in error in measuring length. True value and Errors Imperfection in Experimental Technique or True Value Procedure: In the measurement of a physical quantity the arithmetic The error due to experimental arrangement, procedure followed mean of all readings which is found to be very close to the and experimental technique is called Imperfection error. most accurate reading is to be taken as True value of the Ex: In calorimetric experiments, the loss of heat due to quantities. If a1, a2, a3............ an are readings then true value radiation, the effect on weighing due to buoyancy of air cannot 1 n be avoided. amean = ∑ ai n i =1 Personal Errors or Observational Errors: Absolute Errors These errors are entirely due to personal pecularities like The magnitude of the difference between the true value of individual bias, lack of proper settings of the aparatus, the measured physical quantity and the value of individual carelessness in taking observations. measurement is called absolute error. 1 Probable error ∝ |True value - measured values| no. of readings Ex: Parallax error Dai = |amean – ai| The absolute error is always positive. (ii) Random Errors They are due to uncontrolled disturbances which influence Mean Absolute Error the physical quantity and the instrument. These errors are The arithmetic mean of all the absolute errors is considered as estimated by statistical methods the mean absolute error or final absolute error of the value of 1 the physical concerned. Random error ∝ no. of observations ∆a1 + ∆a2 + − − − ∆an 1 ∞ Ex: The errors due to line voltage changes and backlash error. ∆amean = n ∑ ∆ai = n i =1 Backlash errors are due to screw and nut. The mean absolute error is always positive. 14 Dropper NEET  Relative Error  ∆A   ∆B  = × 100  +  × 100  The relative error of a measured physical quantity is the ratio  A   B  of the mean absolute error to the mean value of the quantity Here Percentage error is the sum of individual percentage measured. errors. ∆amean Error due to Division: Relative error = amean A It is a pure number having no units. Z= B Percentage Error Maximum possible relative error  ∆a  ∆Z ∆A ∆B δa  mean × 100  % = = + Z A B  amean  ∆Z ∆A ∆B Max. Percentage error in division = +  KEY NOTE Z A B Error due to Power Relative error and percentage error give a measure If Z = An of accuracy i.e., percentage error increases accuracy ∆Z ∆A decreases =n Z A Combination of Errors In more general form Error due to addition A p Bq If Z = then maximum fractional error in Z is If Z = A + B Cr DZ = DA + DB (Max. Possible error) ∆Z ∆A ∆B ∆C = p +q +r Z ± DZ = (A ± DA) + (B ± DB) Z A B C ∆A + ∆B As we check for maximum error a + ve sign is to be taken Relative error = ∆C A+ B for the term r ∆A + ∆B C Percentage= error × 100 Maximum Percentage error in A+ B ∆Z ∆A ∆B ∆C Error due to subtraction Z= × 100 = p × 100 + q × 100 + r × 100 Z A B C If Z = A – B DZ = DA + DB (Max. Possible error) TRAIN YOUR BRAIN Q. In an experiment, two capacities measured are (1.3 ± 0.1) Z ± DZ = (A ± DA) – (B ± DB) mF and (2.4 ± 0.2) mF. Calculate the total capacity if the ∆A + ∆B two capacitors are connected in parallel with percentage Relative error = error A− B Sol. Here, C1 = (1.3 ± 0.1) mF and C2 = (2.4 ± 0.2) mF. ∆A + ∆B In parallel, Cp = C1 + C2 = 1.3 + 2.4 = 3.7 mF Percentage = error × 100 A− B DCp = ± (DC1 + DC2) = ± (0.1 + 0.2) = ± 0.3 % error Whether it is addition or subtraction, absolute error is 0.3 same. =± × 100 =±8.1% 3.7 More admissible error ∆U =± ( ∆x )2 + ( ∆y )2 Hence, Cp = (3.7 ± 0.3) mF = 3.7 mF ± 8.1% Significant Figures: In subtraction the percentage error increases. A significant figure is defined as the figure, which is considered Error due to Multiplication reasonably, trust worthy in number. ∆Z ∆A ∆B Eg: p = 3.141592654 If Z = AB then = + Z A B (upto 10 digits) ∆Z is called fractional error or relative error. = 3.14 (with 3 figures ) Z ∆Z = 3.1416 (upto 5 digits ) Percentage error= × 100 Z Units and Measurements 15  (b) If the preceding digit is odd, it is to be raised by 1  KEY NOTE Ex: 4.7153 is to be rounded off to two decimal places. The significant figures indicate the extent to which the As the preceding digit 1 is odd, it is to be raised by 1 readings are reliable. as 2. 4.7153 = 4.72 Rules for Determining the Number of Significant Figures: Rules for Arithmetic Operations with Significant All the non-zero digits in a given number are significant Figures: without any regard to the location of the decimal point if any. In multiplication or division, the final result should retain only Ex: 194,52 has five significant digits. that many significant figures as are there in the original number 1945.2 or 194.52 all have the same number of significant with the least number of significant figures. digits,that is 5. Ex: But the result should be limited to the least number of All zeros accruing between two non zero digits are significant digits-that is two digits only. So final answer is 9.9. significant without any regard to the location of decimal point if any. In addition or subtraction the final result should retain only Ex: 107008 has six significant digits. that many decimal places as are there in the number with the 107.008 or 1.07008 has also got six significant digits. least decimal places. If the number is less than one, all the zeros to the right Ex: 2.2 + 4.08 + 3.12 + 6.38 = 15.78. Finally we should have only of the decimal point but to the first non-zero digit are not one decimal place and hence 15.78 is to be rounded off as 15.8. significant. Ex: 0.000408 VERNIER CALIPERS In this example all zeros before 3 are non-significant. It is a device used for accurate measurement. There are two (i) All zeros to the right of a decimal point are significant scales in the vernier calipers, vernier scale and main scale. The if they are not followed by a non-zero digit. main scale is fixed whereas the vernier scale is movable along Ex: 40.00 has 4 significant digits the main scale. (ii) All zeros to the right of the last non-zero digit after the decimal point are significant. Determination of Least Count (Vernier Constant) Ex: 0.05700 has 4 significant digits Note the value of the main scale division and count the number All zeros to the right of the last non-zero digit in a number n of vernier scale divisions. Slide the movable jaw till the zero having no decimal point are not significant. of vernier scale coincides with any of the mark of the main Ex: 4030 has 3 significant digits scale and find the number of divisions (n – 1) on the main scale coinciding with n divisions of vernier scale. Then Rounding off Numbers nV.S.D. = (n – 1) M.S.D. or 1 V.S.D. = M.S.D.  n –1  The process of omitting the non significant digits and retaining  n  only the desired number of significant digits, incorporating  n –1  1 the required modifications to the last significant digit is called or V.C. = 1 M.S.D. – 1 V.S.D. = 1 –  M.S.D = M.S.D  n  n rounding off the number. Rules for Rounding off Numbers: Determination of Zero error and Zero Correction The preceding digit is raised by 1 if the immediate For this purpose, movable jaw B is brought in contact with insignificant digit to the dropped is more than 5. fixed jaw A. One of the following situations will arise. Ex: 4727 is rounded off to three significant figures as 4730. (i) Zero of Vernier scale coincides with zero of main scale The preceding digit is to be left unchanged if the immediate (see figure) insignificant digit to be dropped is less than 5. Ex: 4722 is rounded off to three significant figures as 4720 0 0.5 M 1 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. 0 5 V 10 Ex: 4.7252 is to be rounded off to two decimal places. In this case, zero error and zero correction, both are nil. The digit to be dropped here is 5 (along with 3) and the Actual length = observed (measured) length. preceding digit 2 is even and hence to be retained as (ii) Zero of vernier scale lies on the right of zero of main two only scale (see figure) 4.7252 = 4.72 16 Dropper NEET  TRAIN YOUR BRAIN 0 0.5 M 1 Q. The least count of vernier caliper is 0.1 mm. The main scale reading before the zero of the vernier scale is 10 and the zeroth division of the vernier scale coincides 0 5 V 10 with the main scale division. Given that each main scale division is 1 mm, what is the measured value? Here 5th vernier scale division is coinciding with any Sol. Length measured with vernier caliper main sale division. Hence, N = 0, n = 5, L.C. = 0.01 cm. = reading before the zero of vernier scale + number of Zero error N = n × (L.C.) = 0 + 5 × 0.01 = + 0.05 cm vernier divisions coinciding Zero correction = – 0.05 cm. with any main scale division × least count Actual length will be 0.05 cm less than the observed = 10 mm + 0 × 0.1 mm = 10 mm = 1.00 cm (measured) length. (iii) zero of the vernier scale lies left of the main scale. SCREW GAUGE, SPHEROMETER Linear scale 0 0.5 M 1 Circulars head scale S N H AB E K 0 0 5 V 10 Stud Sleeve Ratchet R Here, 5th vernier scale division is coinciding with any Thimble main scale division. In this case, zero of vernier scale lies on the right of –0.1 Frame cm reading on main scale. M Hence, N = – 0.1 cm, n = 5, L.C. = 0.01 cm Screw gauge Zero error = N + n × (L.C.) = – 0.1 + 5 × 0.01 = –0.05 cm. Zero correction = +0.05 cm. Actual length will be 0.05 cm more than the observed (measured) length. Experiment Aim: To measure the diameter of a small spherical/cylindrical body, using a vernier caliper. Apparatus: Vernier caliper, a spherical (pendulum bob) or a cylinder. S C D Main Scale 3 E 0 1 5 6 7 8 9 10 V M P Determination of Least Count of Screw Gauge or A B Sphero-meter SPHERE Note the value of linear (pitch) scale division. Rotate screw Theory: If with the body between the jaws, the zero of vernier to bring zero mark on circular (head) scale on reference line. scale lies ahead of Nth division of main scale, then main scale Note linear scale reading i.e. number of divisions of linear scale uncovered by the cap. reading (M.S.R.) = N. If nth division of vernier scale coincides with any division of Now give the screw a few known number of rotations. (one main scale, then vernier scale reading (V.S.R.) rotation completed when zero of circular scale again arrives on the reference line). Again note the linear scale reading. = n × (L.C.) (L.C. is least count of vernier caliper) Find difference of two readings on linear scale to find distance = n × (V.C.) (V.C. is vernier constant of vernier caliper) moved by the screw. Total reading, T.R. = M.S.R. + V.S.R. = N + n × (V.C.) Then, pitch of the screw Units and Measurements 17  Distance moved by in n rotation Sol. (c) 1 main scale div = 0.1 cm = 10V = 8S No. of full rotation (n) 8 Now count the total number of divisions on circular (head) V= S scale. 10 Then, least count 8 2 1 Pitch S − V =S − S = S. = S = 10 10 5 Total number of divisions on the circular scale But 1S = 0.1 cm The least count is generally 0.001 cm. 0.1 = = 0.02 cm TRAIN YOUR BRAIN 5 Least count = 0.02 cm Q. A vernier caliper has its main scale of 10 cm equally divided into 200 equal parts. Its vernier scale of 25 Q. In four complete revolutions of the cap, the distance divisions coincides with 12 mm on the main scale. The travelled on the pitch scale is 2mm. If there are fifty least count of the instrument is– divisions on the circular scale, then (a) Calculate the pitch of the screw gauge (a) 0.020 cm (b) 0.002 cm (b) Calculate the least count of the screw gauge (c) 0.010 cm (d) 0.001 cm Sol. Pitch of screw = Linear distance travelled in one Sol. (b) In vernier caliper main scale 10 cm. 10 Revolution 10 cm divided in 200 divisions. 1 div. = 200 2mm = P = 0.5= mm 0.05 cm = 0.05 cm. 4 12 mm Least count 12 mm on main scale = = 24 MSD 0.0.5cm Pitch 25 V = 24S. = no. of divisions in circular scale 24 1 0.05 S − V =S − S = S = = 0.001 cm 25 25 50 1S = 0.05 cm Q. The pitch of a screw gauge 0.5 mm and there are 50 Least count = 0.002 cm divisions on the circular scale. In measuring the thickness Q. One centimetre on the main scale of vernier caliper is of a metal plate, there are five divisions on the pitch scale divided into ten equal parts. If 10 divisions of vernier (or main scale) and thirty fourth division coincides with the scale coincide with 8 small divisions of the main scale, reference line. Calculate the thickness of the metal plate. the least count of the caliper is – Sol. Pitch of screw = 0.5 mm. (a) 0.005 cm 0.5 L.C = = 0.01 mm. (b) 0.05 cm 50 (c) 0.02 cm Thickness = (5 × 0.5 + 34 × 0.01) mm (d) 0.01 cm = (2.5 + 0.34) = 2.84 mm ILLUSTRATIONS Learning Plus 1. A new system of units is proposed in which, unit of mass Sol. (b) Joule is a unit of energy. is α kg, unit of length is β m and unit of time is γ s. What n1 = 5 n2 = ? will be value of 5 J in this new system? M1 = 1 kg M2 = α kg (a) 5αβ2γ–2 L1 = 1 m L2 = β m (b) 5α–1β–2γ2 T1 = 1 s T2 = γ s (c) 5α–2β–1γ–2 Dimensional formula of energy is [ML2T–2]. Comparing with [MaLbTc], we get (d) 5α–1β2γ2 18 Dropper NEET  a = 1, b = 2, c = –2 5. By what percentage should the pressure of a given mass M  L  T  a b c of gas be increased so as the decrease in its volume is As, n 2 = n1  1   1   1  10% at a constant temperature?  M 2   L 2   T2  1 2 −2 (a) 5% (b) 7.2%  1kg   1m   1s  5γ 2 (c) 12.5% (d) 11.1% = 5      = = 5α −1β−2 γ 2  α kg   β m   γ s  αβ2 Sol. (d) PV = constant  2π  2. Given = that y A sin  ( ct − x )   where y and x are  10V    λ   P ' V    PV measured in meter. Which of the following statements  100  are true?  100V  10V  P '   PV (a) The unit of λ is same as that of x and A  100  (b) The unit of λ is same as of x but not of A  90V  10 2π P '   PV  P '  P (c) The unit of c is same as that of  100  9 λ 2π Percentage increase in volume (d) The unit of (ct – x) is same as that of λ  2π    P  P  Sol. (a) f  A sin    ct  x . Since Angle has [MoLoTo]   100%  λ   P  10   2π    P  P   100% dimensions.   ct  x  needs to be dimensionless 9  λ  P  11.1% 6. Percentage error in the measurement of mass and speed x = λ =  M L T  , A = 0 1 0 [M0 L1 T0] are 2% and 3% respectively. The error in the estimation of kinetic energy obtained by measuring mass and speed 3. If the unit of force and length are doubled, the unit of will be: energy will be: (a) 12% (b) 10% (a) 1/4 times the original (b) 1/2 times the original (c) 2% (d) 8% (c) 2 times the original (d) 4 times the original Sol. (d) Work = Energy = F’. L’ = 2 F × 2 L, 4 times the original Sol. (d) K = 1 mV 2 2 4. The speed of light c, gravitational constant G and Planck’s constant h are taken as fundamental units in ∆K ∆m 2 ∆V = + a system. The dimensions of time in this new system K m V should be: (a) [G1/2h1/2c–5/2] (b) [G–1/2h1/2c–1/2] ΔK  2  3 2 (c) [G–1/2 h1/2c–3/2] (d) [G–1/2 h1/2c1/2] K Sol. (a) c = [ M0 L1 T–1], G=[ M–1 L3 T–2], h = [ M1 L2 T–1] ∆K = 8% T = ca Gb hc K [ T1] = [ M0 L1 T–1]a [ M–1 L3 T–2]b [ M1 L2 T–1]c 7. If E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively, the quantity [ T1] = M–b + c La+3 b+2c T–a –2b–c ∴ b = c  El 2  a + 3b + 2c = 0, a + 2b + c = –1  5 2  has the dimensions of: m G  Solving the three equations (a) Mass 1 a = –5/2, b = , c = 1/2 (b) Length 2 (c) Time T = [ c–5/2 G1/2 h1/2] (d) Angle Units and Measurements 19   El 2  [ M 1 L2T −2 ][ M L2 T −1 ]2 12. What are dimensions of E/B? Sol. (d)  5 2  = (a) [LT–1] (b) [LT–2] m G  [ M 5 ][ M −2 L6T −4 ] (c) [MLT ]–1 (d) [ML2T–1]  M 3L6T 4  E MLT −3 A −1 E   3 6 4    M 0 L0T 0  Sol. (a) = = LT −1  M L T  B MT −2 A −1 B 2nd solutions Mass = [M1] Length = [L1] Ratio E/B is the speed of EM wave. Time = [T1] ∴ Dimension of E/B is the dimension of speed. (Angle dimension less) 13. A physical quantity is given by X = [MaLbTc]. The 8. In an experiment four quantities a, b, c and d are measured percentage error in measurements of M, L and T are α, β, with percentage error 1%, 2%, 3% and 4% respectively. γ. Then, the maximum % error in the quantity X is: a 3b2 (a) aα + bβ + c γ (b) aα + bβ − c γ Quantity P is calculated as follows: P = % error in P is: cd a b c (a) 4% (b) 14% (c) + + (d) None of these α β γ (c) 10% (d) 7% Sol. (a) X = M a LbT c a 3b2 P  a b c d  dX dM dL dT Sol. (b) P    3 2    a b c cd P  a b c d  X M L T    3  1  2  2  3  4   14% dX  a  b  c X 9. The respective number of significant figure for the 14. In dimension of critical velocity, of liquid flowing number 23.023, 0.0003 and 2.1 × 10–3 are: through a tube are expressed as νc∝[ηxρyrz] where η, ρ (a) 5, 1, 2 (b) 5, 1, 5 and r are the coefficient of viscosity of liquid, density of liquid and radius of the tube respectively, then the values (c) 5, 5, 2 (d) 4, 4, 2 of x, y and z are given by: Sol. (a) 23.023 → 5, 0.0003 = 1, 2.1 × 10–3 = 2 (a) 1, 1, 1 (b) 1, –1, –1 10. The dimensions of coefficient of viscosity and self (c) –1, –1, 1 (d) –1, –1, –1 inductance are: Sol. (b)  c    r  x y z (a) [M L–1 T–1], [M L2 T–2 I–2] x y z  L1T 1    M1L1T 1   M1L3   L1  (b) [M2 L–1 T–1], [M L2 T–1 I–2] (c) [M L–1 T–1], [M L3 T–1 I–2]  L1T 1    M x  y   L x 3 y z  T  x  (d) [M1 L1 T–2], [M1 L3 T2 I–3] Taking comparison on both size x + y = 0, –x – 3y + z = 1, –x = –1 Sol. (a)   F dimensions of force  x = 1, y = –1, z = –1 A  V / L  D. of area  D. of velocity gradient [ MLT −2 ] 15. The absolute error in density of a sphere of radius 10.01 = = [ ML−1T −1 ] cm and mass 4.692 kg is: [L2 ][T −1 ] (a) 3.59 kg m–3 (b) 4.692 kg m–3 dimensions of e [ ML2T −3 I −1 ] L= = (c) 0 (d) 1.12 kg m–3 D. of I / dimension of t [ I /T ] L = [M L2 T–2= I–2]f m Sol. (a) = ρ 4.692 f =×m3 4.692 × 3 = 4 4 −×6 3.14 × ( 10.01 ) × 10 −6 3 4 2 4 × 3.14 × ( 10.01πr) 23 × 10 1 πr 3 11. St = U + a ( 2t − 1) is: 3 2 ρ = 1.12 × 103 kg /m3 (a) Only numerically correct abs. errors: (b) Only dimensionally correct (c) Both numerically and dimensionally correct Δm = 1gm = 0.001 kg (d) Neither numerically nor dimensionally correct ΔR = 0.01 cm Sol. (d) St = distance traveled, U = velocity so, it is not  m r  3 dimensionally correct.  m r 20 Dropper NEET  ∆P  0.001 3 × 0.01  18. Dimensions of resistance in an electrical circuit, in terms 3 ∆ρ=  +  × 1.12 × 10 of dimension of mass M, length L, time T and current I,  4.692 10.01  would be: = 3.59 kg /m3 (a) [ML2T-2] 16. The dimension of Planck’s constant equals to that of: (b) [ML2T-1I-1] (a) Energy (b) Momentum (c) Angular momentum (d) Power (c) [ML2T-3I-2] Sol. (c) Dimensions of Planck’s constant, h = Energy (d) [ML2T-3I-1] Frequency Sol. (c) According to Ohm’s law,  ML2 T 2   V = RI or R = V T 1  I W  ML T  2 2   ML2 T 1  Dimensions of V   Dimensions of angular momentum L q  IT  = Moment of inertia I × Angular velocity  ML2 T 2 / IT   R      ML2 T 3 I 2  = [ML2][T-1] = [ML2T-1]  I 17. In the following dimensionally constant equations, we have 19. If A = (1.0 m ± 0.2) gm and B = (2.0 ± 0.2) m, then AB is: X =F + Y where F = force. The dimensional (a) 1.4 m ± 0.4 m Linear density (b) 1.41 m ± 0.15 m formula of X are Y are: (a) [M L T–2], [M2 L0 T–2] (c) 1.4 m ± 0.3 m (b) [M2 L0 T–2], [M L T–2] (d) 1.4 m ± 0.2 m (c) [M L2 T–4], [M2 L–2 T–2] Sol. (d) x  AB  1.0  2.0  1.414 m (d) None of these Rounding off x = 1.4 m  X  Sol. (b) = [F ]   + [Y ] x 1  A B  1  0.2 0.2   L.D         x 2 A B  2  1.0 2.0  ∴ [Y] = [F] = [MLT–2] = 0.2 m (round figure)  X   MLT    2  X   M 2 L0 T 2  AB  1.4  0.2  m  ML  1 Units and Measurements 21  Topicwise Questions DIMENSIONS & DIMENSIONAL FORMULA 9. Surface tension has the same dimensions as that of: (a) Coefficient of viscosity (b) Impulse 1. The dimensions of magnetic moment are: (a) L2A–1 (b) L2A1 (c) Momentum (d) Spring constant (c) LA 2 (d) L2A–3 10. The dimensions of RC, where R is resistance and C is 2. The velocity “V” of a particle is given in terms of time t capacitance are same as that of: b (a) Inverse time (b) Time as V = at +. t +C (c) Square of time (d) Square root of time The dimensions of a, b, C are: