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Physics (SSC, Railway, Police & All State exam)Chapter Unit

Physical Quantities

Introduction to Physical Quantities

Physical quantities are measurable properties of matter or energy used to describe physical phenomena. They are classified into fundamental and derived quantities based on whether they can be expressed in terms of basic units or are formed by their combination.


Classification of Physical Quantities

  1. Fundamental Quantities:

    • Independent and cannot be derived from other quantities.
    • Examples:
      • Length (LL)
      • Mass (MM)
      • Time (TT)
      • Electric Current (II)
      • Temperature (Θ\Theta)
      • Luminous Intensity (JJ)
      • Amount of Substance (NN)
  2. Derived Quantities:

    • Formed by combining fundamental quantities.
    • Examples:
      • Velocity: [LT1][L T^{-1}]
      • Force: [MLT2][M L T^{-2}]
      • Pressure: [ML1T2][M L^{-1} T^{-2}]

System of Units

  1. CGS System:

    • Length: Centimeter (cmcm)
    • Mass: Gram (gg)
    • Time: Second (ss)
  2. MKS System:

    • Length: Meter (mm)
    • Mass: Kilogram (kgkg)
    • Time: Second (ss)
  3. SI System:

    • Length: Meter (mm)
    • Mass: Kilogram (kgkg)
    • Time: Second (ss)
    • Electric Current: Ampere (AA)
    • Temperature: Kelvin (KK)
    • Luminous Intensity: Candela (cdcd)
    • Amount of Substance: Mole (molmol)

Scalars and Vectors

  1. Scalar Quantities:

    • Have only magnitude.
    • Examples: Distance, Speed, Mass, Temperature, Energy.
  2. Vector Quantities:

    • Have both magnitude and direction.
    • Examples: Displacement, Velocity, Force, Acceleration.
PropertyScalarsVectors
MagnitudeYesYes
DirectionNoYes
AdditionSimple arithmeticVector addition rules

Representing Vector Quantities

  1. Graphical Representation:

    • Represented as arrows, where:
      • Length = Magnitude.
      • Arrowhead = Direction.
  2. Vector Addition:

    • Triangle Law: If two vectors are represented by two sides of a triangle in sequence, their resultant is the third side of the triangle taken in reverse order.
    • Parallelogram Law: The resultant vector is the diagonal of a parallelogram formed by the two vectors.

    Formula for resultant magnitude: R=A2+B2+2ABcosθR = \sqrt{A^2 + B^2 + 2AB \cos \theta} where AA and BB are magnitudes, and θ\theta is the angle between them.

  3. Vector Resolution:

    • A vector can be resolved into components along perpendicular axes:
      • Horizontal Component: Ax=AcosθA_x = A \cos \theta
      • Vertical Component: Ay=AsinθA_y = A \sin \theta

Measurement of Physical Quantities

  1. Base Quantities:

    • Measured using standard instruments.
    • Length: Ruler, Vernier Caliper.
    • Mass: Balance.
    • Time: Stopwatch, Atomic Clock.
  2. Derived Quantities:

    • Calculated using formulas.
    • Example: Velocity = DisplacementTime\frac{\text{Displacement}}{\text{Time}}.

Numerical Example

  1. Example: Two forces of 10N10 \, \text{N} and 20N20 \, \text{N} act at an angle of 6060^\circ. Find the resultant force.
    • Using the formula: R=A2+B2+2ABcosθR = \sqrt{A^2 + B^2 + 2AB \cos \theta} R=102+202+2(10)(20)cos60R = \sqrt{10^2 + 20^2 + 2(10)(20)\cos 60^\circ} R=100+400+200×0.5R = \sqrt{100 + 400 + 200 \times 0.5} R=100+400+100=60024.49NR = \sqrt{100 + 400 + 100} = \sqrt{600} \approx 24.49 \, \text{N}

Vector Operations

  1. Addition of Vectors:

    • Analytical Method:

      • Resolve vectors into their components along xx and yy axes.
      • Add corresponding components: Rx=Ax+Bx,Ry=Ay+ByR_x = A_x + B_x, \, R_y = A_y + B_y
      • Resultant magnitude: R=Rx2+Ry2R = \sqrt{R_x^2 + R_y^2}
      • Direction of resultant: θ=tan1(RyRx)\theta = \tan^{-1}\left(\frac{R_y}{R_x}\right)
    • Special Cases:

      • Same Direction: R=A+BR = A + B
      • Opposite Direction: R=ABR = |A - B|
  2. Subtraction of Vectors:

    • Subtract components of one vector from the other: R=AB\vec{R} = \vec{A} - \vec{B}
  3. Multiplication of Vectors:

    • Dot Product: AB=ABcosθ\vec{A} \cdot \vec{B} = AB \cos \theta

      • Produces a scalar quantity.
      • Example: Work done W=FdW = \vec{F} \cdot \vec{d}.
    • Cross Product: A×B=ABsinθn^\vec{A} \times \vec{B} = AB \sin \theta \, \hat{n}

      • Produces a vector quantity.
      • Example: Torque τ=r×F\vec{\tau} = \vec{r} \times \vec{F}.

Derived Quantities and Their SI Units

Derived QuantityFormulaSI UnitDimensional Formula
AreaL×LL \times Lm2m^2[L2][L^2]
VolumeL×L×LL \times L \times Lm3m^3[L3][L^3]
DensityMassVolume\frac{\text{Mass}}{\text{Volume}}kgm3kg \, m^{-3}[ML3][M L^{-3}]
Speed/VelocityDistanceTime\frac{\text{Distance}}{\text{Time}}ms1m \, s^{-1}[LT1][L T^{-1}]
AccelerationVelocityTime\frac{\text{Velocity}}{\text{Time}}ms2m \, s^{-2}[LT2][L T^{-2}]
ForceMass×AccelerationMass \times AccelerationNN (Newton)[MLT2][M L T^{-2}]
PressureForceArea\frac{\text{Force}}{\text{Area}}PaPa (Pascal)[ML1T2][M L^{-1} T^{-2}]
Energy/WorkForce×DisplacementForce \times DisplacementJJ (Joule)[ML2T2][M L^2 T^{-2}]

Physical Constants

  1. Definition:

    • Fundamental quantities with fixed numerical values.
  2. Examples:

    • Speed of Light (cc): 3×108m/s3 \times 10^8 \, m/s.
    • Gravitational Constant (GG): 6.67×1011Nm2kg26.67 \times 10^{-11} \, N \, m^2 \, kg^{-2}.
    • Planck's Constant (hh): 6.626×1034Js6.626 \times 10^{-34} \, J \, s.

Dimensional Consistency

  1. Principle:

    • All terms in a physical equation must have the same dimensions.
  2. Applications:

    • Checking Validity of Equations:
      • Example: Verify v2=u2+2asv^2 = u^2 + 2as.
        • Dimensions of v2v^2, u2u^2: [L2T2][L^2 T^{-2}].
        • Dimensions of 2as2as: [L2T2][L^2 T^{-2}].
        • Equation is dimensionally consistent.
  3. Limitations:

    • Does not confirm numerical constants.
    • Cannot verify equations involving non-dimensional terms like trigonometric functions.

Practical Examples of Physical Quantities

  1. Example 1: A car accelerates uniformly from 5m/s5 \, m/s to 15m/s15 \, m/s over a distance of 50m50 \, m. Find the acceleration.

    • Formula: v2=u2+2asv^2 = u^2 + 2as
    • Rearranging: a=v2u22sa = \frac{v^2 - u^2}{2s} a=152522×50=22525100=2m/s2a = \frac{15^2 - 5^2}{2 \times 50} = \frac{225 - 25}{100} = 2 \, m/s^2
  2. Example 2: Calculate the work done in moving a 10kg10 \, kg object 5m5 \, m with a force of 20N20 \, N at an angle of 3030^\circ to the horizontal.

    • Work done: W=FdcosθW = Fd \cos \theta W=205cos30=10032=503JW = 20 \cdot 5 \cdot \cos 30^\circ = 100 \cdot \frac{\sqrt{3}}{2} = 50\sqrt{3} \, J

Errors in Measurement of Physical Quantities

  1. Definition of Error:

    • The difference between the true value and the measured value of a quantity.
  2. Types of Errors:

    • Systematic Errors:
      • Arise from faulty instruments or consistent biases.
      • Example: Incorrect calibration of a scale.
    • Random Errors:
      • Caused by unpredictable fluctuations in measurement.
      • Example: Variations in stopwatch readings.
    • Gross Errors:
      • Result from human mistakes.
      • Example: Misreading an instrument.
  3. Quantification of Error:

    • Absolute Error: ΔA=AmeasuredAtrue\Delta A = |A_{\text{measured}} - A_{\text{true}}|
    • Relative Error: Relative Error=ΔAAtrue\text{Relative Error} = \frac{\Delta A}{A_{\text{true}}}
    • Percentage Error: Percentage Error=(ΔAAtrue)×100\text{Percentage Error} = \left( \frac{\Delta A}{A_{\text{true}}} \right) \times 100

Propagation of Errors

  1. For Addition/Subtraction:

    • Total absolute error: ΔR=ΔA+ΔB\Delta R = \Delta A + \Delta B
      • Example: If A=5±0.2A = 5 \pm 0.2 and B=3±0.1B = 3 \pm 0.1, then: R=A+B=8±0.3R = A + B = 8 \pm 0.3
  2. For Multiplication/Division:

    • Total relative error: ΔRR=ΔAA+ΔBB\frac{\Delta R}{R} = \frac{\Delta A}{A} + \frac{\Delta B}{B}
      • Example: If A=5±0.2A = 5 \pm 0.2 and B=3±0.1B = 3 \pm 0.1, then: ΔRR=0.25+0.13=0.04+0.03330.0733\frac{\Delta R}{R} = \frac{0.2}{5} + \frac{0.1}{3} = 0.04 + 0.0333 \approx 0.0733 ΔR=0.0733(AB)\Delta R = 0.0733 \cdot (A \cdot B)
  3. For Powers:

    • Error is multiplied by the exponent: ΔRR=nΔAA\frac{\Delta R}{R} = n \cdot \frac{\Delta A}{A}

Significant Figures

  1. Definition:

    • Digits in a measurement that carry meaningful information about precision.
  2. Rules for Significant Figures:

    • All non-zero digits are significant.
    • Zeros between significant digits are significant.
    • Trailing zeros in a decimal are significant.
    • Leading zeros are not significant.
  3. Arithmetic with Significant Figures:

    • Addition/Subtraction:
      • Result has the same number of decimal places as the term with the least decimal places.
    • Multiplication/Division:
      • Result has the same number of significant figures as the term with the least significant figures.

Practical Applications

  1. Dimensional Analysis in Real Life:

    • Verifying equations.
    • Deriving new formulas.
    • Checking the consistency of units in calculations.
  2. Examples of Derived Quantities:

    • Kinetic Energy: KE=12mv2KE = \frac{1}{2} mv^2
      • Dimensional Formula: [ML2T2][M L^2 T^{-2}].
    • Momentum: p=mvp = mv
      • Dimensional Formula: [MLT1][M L T^{-1}].
    • Electric Charge: Q=ItQ = I \cdot t
      • Dimensional Formula: [IT][I T].

Numerical Examples

  1. Example 1: The mass of an object is measured as 5.0±0.1kg5.0 \pm 0.1 \, kg and its volume as 2.0±0.2m32.0 \pm 0.2 \, m^3. Calculate the density and its error.

    • Density: ρ=mV=5.02.0=2.5kg/m3\rho = \frac{m}{V} = \frac{5.0}{2.0} = 2.5 \, kg/m^3
    • Relative Error: Δρρ=Δmm+ΔVV=0.15.0+0.22.0=0.02+0.1=0.12\frac{\Delta \rho}{\rho} = \frac{\Delta m}{m} + \frac{\Delta V}{V} = \frac{0.1}{5.0} + \frac{0.2}{2.0} = 0.02 + 0.1 = 0.12
    • Absolute Error: Δρ=0.122.5=0.3kg/m3\Delta \rho = 0.12 \cdot 2.5 = 0.3 \, kg/m^3
    • Final Result: ρ=2.5±0.3kg/m3\rho = 2.5 \pm 0.3 \, kg/m^3
  2. Example 2: A sphere has a radius of 10.0±0.1cm10.0 \pm 0.1 \, cm. Calculate its volume and error.

    • Volume: V=43πr3V = \frac{4}{3} \pi r^3 V=433.1416(10)3=4188.79cm3V = \frac{4}{3} \cdot 3.1416 \cdot (10)^3 = 4188.79 \, cm^3
    • Relative Error: ΔVV=3Δrr=30.110=0.03\frac{\Delta V}{V} = 3 \cdot \frac{\Delta r}{r} = 3 \cdot \frac{0.1}{10} = 0.03
    • Absolute Error: ΔV=0.034188.79=125.66cm3\Delta V = 0.03 \cdot 4188.79 = 125.66 \, cm^3
    • Final Result: V=4188.79±125.66cm3V = 4188.79 \pm 125.66 \, cm^3

Recap: Key Points to Remember

  • Physical quantities are classified as fundamental or derived.
  • Errors in measurement are unavoidable; they can be minimized but not eliminated.
  • Dimensional analysis is a powerful tool to verify equations and derive relations.
  • Significant figures convey the precision of a measurement.

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