Units and Measurement Notes

Units and Measurement

1.1 Introduction

  • Measurement involves comparing a physical quantity to a reference standard called a unit.
  • A measurement's result is expressed as a number with a unit.
Fundamental and Derived Quantities
  • Physical quantities can be measured directly or indirectly.
  • Two types of physical quantities:
    • Fundamental quantities (Base quantities): Independent of each other and cannot be expressed in terms of other physical quantities (e.g., length, mass, time).
    • Derived quantities: Can be expressed in terms of fundamental quantities (e.g., volume, velocity, force).
Fundamental and Derived Units
  • Fundamental or base units: Units for fundamental quantities.
  • Derived units: Combinations of base units used to express units of derived quantities.

1.2 The International System of Units

  • A system of units is a complete set of base and derived units.
  • Earlier systems included CGS, FPS (British), and MKS.
    • CGS: centimetre, gram, second.
    • FPS: foot, pound, second.
    • MKS: metre, kilogram, second.
  • In 1971, the General Conference on Weights and Measures developed the Système Internationale d’ Unites (SI system).
  • The SI system is used internationally in scientific, technical, industrial, and commercial work.
  • The SI system includes seven base units and two supplementary units.

1.3 Significant Figures

  • The result of a measurement includes all reliable digits plus the first uncertain digit.
  • Significant digits (or figures): Reliable digits plus the first uncertain digit in a measurement.
  • Example: If the period of oscillation of a simple pendulum is 1.62 s, 1 and 6 are reliable, and 2 is uncertain.
Rule 6
  • The power of 10 in scientific notation is irrelevant to significant figures.
Rounding off the Uncertain Digits
  1. If the insignificant digit to be dropped is more than 5, the preceding digit is raised by 1.

    • Example: 2.746 rounded to three significant figures is 2.75 (6 > 5, so 4 becomes 5).
  2. If the insignificant digit to be dropped is less than 5, the preceding digit is left unchanged.

    • Example: 2.743 rounded to three significant figures is 2.74 (3 < 5, so 4 remains 4).
  3. If the insignificant digit to be dropped is 5:

    • Case i) If the preceding digit is even, the insignificant digit is simply dropped.
      • Example: 2.745 rounded to three significant figures is 2.74 (4 is even, so 5 is dropped).
    • Case ii) If the preceding digit is odd, the preceding digit is raised by 1.
      • Example: 2.735 rounded to three significant figures is 2.74 (3 is odd, so 3 becomes 4).
Rules for Arithmetic Operations with Significant Figures
  1. In multiplication or division, the final result should retain as many significant figures as the original number with the least significant figures.
    • Example: Density calculation with mass = 4.237 g (4 significant figures) and volume = 2.51 cm³ (3 significant figures): Density=massvolume=4.237g2.51cm3=1.688047Density = \frac{mass}{volume} = \frac{4.237 g}{2.51 cm^3} = 1.688047
      • Rounded to 3 significant figures: 1.69 g/cm³.
  2. In addition or subtraction, the final result should retain as many decimal places as the number with the least decimal places.
    • Example: Sum of 436.32 g (2 decimal places), 227.2 g (1 decimal place), and 0.301 g (3 decimal places): 436.32g+227.2g+0.301g=663.821g436.32 g + 227.2 g + 0.301 g = 663.821 g
      • Rounded to 1 decimal place: 663.8 g.

1.4 Dimensions of Physical Quantities

  • The nature of a physical quantity is described by its dimensions.
  • Derived units can be expressed in terms of fundamental quantities.
  • Base quantities are the seven dimensions of the physical world, denoted with square brackets [ ]. Length [L], mass [M], time [T], electric current [A], thermodynamic temperature [K], luminous intensity [cd], and amount of substance [mol].
  • The dimensions of a physical quantity are the powers to which the base quantities are raised to represent that quantity.
  • Example: Volume = length × breadth × thickness; Dimensions of volume [V].

1.5 Dimensional Formulae and Dimensional Equations

  • Volume has zero dimension in mass, zero dimension in time, and three dimensions in length.
  • Unit of density = kg m⁻³
  • Speed:
    • [S]=Distancetime=[L][T]=[LT1][S] = \frac{Distance}{time} = \frac{[L]}{[T]} = [LT^{-1}]
    • Unit of speed = m/s
  • Velocity:
    • [v]=Displacementtime=[L][T]=[LT1][v] = \frac{Displacement}{time} = \frac{[L]}{[T]} = [LT^{-1}]
    • Unit of velocity = m/s
    • Speed and Velocity have the same dimensional formula.
  • Momentum:
    • [p]=Mass×Velocity=[M]×[LT1]=[MLT1][p] = Mass \times Velocity = [M] \times [LT^{-1}] = [MLT^{-1}]
    • Unit of momentum = kg m/s
  • Angular Momentum:
    • [L]=momentum×Distance=[MLT1]×[L]=[ML2T1][L] = momentum \times Distance = [MLT^{-1}] \times [L] = [ML^2T^{-1}]
    • Unit of angular momentum = kg m²/s
  • Acceleration:
    • [a]=Change in velocitytime=[LT1][T]=[LT2][a] = \frac{Change \space in \space velocity}{time} = \frac{[LT^{-1}]}{[T]} = [LT^{-2}]
    • Unit of acceleration = m/s²
  • Force:
    • [F]=Mass×Acceleration=[M]×[LT2]=[MLT2][F] = Mass \times Acceleration = [M] \times [LT^{-2}] = [MLT^{-2}]
    • Unit of force = kg m/s² or newton (N); 1 kg m/s² = 1 N
  • Impulse:
    • [I]=Force×Time=[MLT2]×[T]=[MLT1][I] = Force \times Time = [MLT^{-2}] \times [T] = [MLT^{-1}]
    • Unit of Impulse = kg m/s
  • Work:
    • [W]=Force×Displacement=[MLT2]×[L]=[ML2T2][W] = Force \times Displacement = [MLT^{-2}] \times [L] = [ML^2T^{-2}]
    • Unit of work = kg m²/s² or joule (J); 1 kg m²/s² = 1 J
  • Energy:
    • [E]=Workdone=[ML2T2][E] = Workdone = [ML^2T^{-2}]
    • Unit of energy = kg m²/s² or joule (J)
  • Torque:
    • [τ]=Force×perpendicular Distance=[MLT2]×[L]=[ML2T2][\tau] = Force \times perpendicular \space Distance = [MLT^{-2}] \times [L] = [ML^2T^{-2}]
    • Unit of torque = kg m²/s²
    • Work, Energy, and Torque have the same dimensional formula.
  • Pressure:
    • [P]=ForceArea=[MLT2][L2]=[ML1T2][P] = \frac{Force}{Area} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]
    • Unit of pressure = kg m⁻¹ s⁻² or pascal (Pa)
  • Stress:
    • [stress]=ForceArea=[MLT2][L2]=[ML1T2][stress] = \frac{Force}{Area} = \frac{[MLT^{-2}]}{[L^2]} = [ML^{-1}T^{-2}]
    • Unit of stress = kg m⁻¹ s⁻²
    • Pressure and Stress have the same dimensional formula.
  • Power:
    • [P]=WorkTime=[ML2T2][T]=[ML2T3][P] = \frac{Work}{Time} = \frac{[ML^2T^{-2}]}{[T]} = [ML^2T^{-3}]
    • Unit of power = kg m² s⁻³ or watt (W)
  • Strain:
    • Change in dimensionOriginal dimension=[L][L]=[1]\frac{Change \space in \space dimension}{Original \space dimension} = \frac{[L]}{[L]} = [1]
  • Relative Density:
    • Density of substanceDensity of water=[ML3][ML3]=[L0]\frac{Density \space of \space substance}{Density \space of \space water} = \frac{[ML^{-3}]}{[ML^{-3}]} = [L^0]
  • Physical quantities having units but no dimension:
    • Plane angle
    • Solid Angle
    • Angular Displacement
  • Gravitational constant (G):
    • Using F=Gm<em>1m</em>2r2F = G \frac{m<em>1m</em>2}{r^2}, [G]=[F][r2][m<em>1][m</em>2]=[MLT2][L2][M][M]=[M1L3T2][G] = \frac{[F][r^2]}{[m<em>1][m</em>2]} = \frac{[MLT^{-2}][L^2]}{[M][M]} = [M^{-1}L^3T^{-2}]
  • Planck's constant (h):
    • Using λ=hmv\lambda = \frac{h}{mv}, h=λmvh = \lambda mv, [h]=[λ][m][v]=[L][M][LT1]=[ML2T1][h] = [\lambda][m][v] = [L][M][LT^{-1}] = [ML^2T^{-1}]

1.6 Dimensional Analysis and its Applications

  1. Checking the dimensional consistency (correctness) of equations.
  2. Deducing relations among physical quantities.
Examples of Checking Dimensional Consistency
  1. Equation: s=ut+ats = ut + at
    • [s]=L[s] = L
    • [ut]=LT1×T=L[ut] = LT^{-1} \times T = L
    • [at]=LT2×T=LT1[at] = LT^{-2} \times T = LT^{-1}
    • Since the dimensions of all terms are not the same, this equation is wrong.
  2. Equation: s=ut+at2s = ut + at^2
    • [s]=L[s] = L
    • [ut]=LT1×T=L[ut] = LT^{-1} \times T = L
    • [at2]=LT2×T2=L[at^2] = LT^{-2} \times T^2 = L
    • Since each term has the same dimension, this equation is dimensionally correct.
    • However, dimensional correctness does not guarantee an exact equation.
  3. Equation: E=12mv2E = \frac{1}{2}mv^2
    • [mv2]=M[LT1]2=ML2T2[mv^2] = M [LT^{-1}]^2 = ML^2T^{-2}
    • [mgh]=MLT2L=ML2T2[mgh] = M LT^{-2} L = ML^2T^{-2}
    • The dimensions of LHS and RHS are the same, hence the equation is dimensionally correct.
  4. Equation: E=mc2E = mc^2
    • [E]=ML2T2[E] = ML^2T^{-2}
    • [mc2]=M[LT1]2=ML2T2[mc^2] = M [LT^{-1}]^2 = ML^2T^{-2}
    • The dimensions of LHS and RHS are the same, hence the equation is dimensionally correct.
  5. Given equation: v=x+atv = x + at, find the dimensions of x.
    • [v]=[x]=[at][v] = [x] = [at]
    • [x]=[v]=LT1[x] = [v] = LT^{-1}
  6. Given equation: x=a+bt+ct2x = a + bt + ct^2, find the dimensions of a, b, and c (where x is in meters and t in seconds).
    • [x]=[a]=[bt]=[ct2][x] = [a] = [bt] = [ct^2]
    • [a]=[x]=L[a] = [x] = L
    • [bt]=[x]    [b]×T=L    [b]=LT1[bt] = [x] \implies [b] \times T = L \implies [b] = LT^{-1}
    • [ct2]=[x]    [c]×T2=L    [c]=LT2[ct^2] = [x] \implies [c] \times T^2 = L \implies [c] = LT^{-2}
  7. Van der Waals equation: (P+aV2)(Vb)=nRT(P + \frac{a}{V^2})(V - b) = nRT. Find the dimensional formula for a and b.
    • [P]=[aV2]    [a]=[PV2]=[ML1T2]×[L6]=[ML5T2][P] = [\frac{a}{V^2}] \implies [a] = [PV^2] = [ML^{-1}T^{-2}] \times [L^6] = [ML^5T^{-2}]
    • [b]=[V]=[L3][b] = [V] = [L^3]
Examples of Deducing Relations Among Physical Quantities
  1. Kinetic energy (E) of a body of mass m moving with velocity v:
    • Emxvy    E=kmxvyE \propto m^x v^y \implies E = k m^x v^y
    • ML2T2=Mx(LT1)y=MxLyTyML^2 T^{-2} = M^x (LT^{-1})^y = M^x L^y T^{-y}
    • Equating dimensions: x = 1, y = 2
    • E=km1v2=kmv2E = k m^1 v^2 = k mv^2
  2. Time period (T) of a simple pendulum depends on mass (m), length (l), and acceleration due to gravity (g):
    • Tmxlygz    T=kmxlygzT \propto m^x l^y g^z \implies T = k m^x l^y g^z
    • M0L0T1=MxLy(LT2)z=MxLy+zT2zM^0 L^0 T^1 = M^x L^y (LT^{-2})^z = M^x L^{y+z} T^{-2z}
    • Equating dimensions: x = 0, y + z = 0, -2z = 1 => z = -1/2, y = 1/2
    • T=km0l1/2g1/2=klgT = k m^0 l^{1/2} g^{-1/2} = k \sqrt{\frac{l}{g}}

Limitations of Dimensional Analysis

  1. Checks only dimensional correctness, not exact correctness.
  2. Dimensionless constants cannot be obtained.
  3. Cannot deduce relations if a quantity depends on more than three physical quantities.
  4. Cannot derive equations involving more than one term.
  5. Formulas containing trigonometric, exponential, and logarithmic functions cannot be derived.
  6. Does not distinguish between physical quantities with the same dimensions.