1. Measurement of any physical quantity involves comparison with a certain basic, arbitrary chosen, widely accepted reference standard called
Unit. Mathematically, a measure of a quantity Q = nu, where u is the size of the unit, and n is the numerical value of the given measure.
2. Fundamental quantities: Fundamental quantities are the base quantities. There are 7 fundamental quantities:
(i) Length
(ii) Mass
(iii) Time
(iv) Electric Current
(v) Thermodynamic Temperature
(vi) Amount of substance
(vii) Luminous Intensity.
3. Derived quantities: These quantities are formed using fundamental quantities like density, volume, force, etc.
4. Length: Unit is metre (m). Meter is defined as the length of the path traveled by light in vacuum during a time interval of $ \frac{1}{299792458} $ part of a second.
1 fermi = $1f = 10^{-15} m $
1 angstrom = $ 1A = 10^{-10} m$
1 nano-metre = $1nm = 10^{-10}m$
1 micro-metre = $ 1\mu m = 10^{-6}m$
1 mili-metre = $1mm = 10^{-3} m$
1 Astronomical unit = $ 1AU = 1.496 \times 10^{11}m$
1 light-year = $ 1ly = 9.46 \times 10^{11} m$
1 parsec = $ 3.08 \times 10^{16}m $
5. Mass: Unit is Kilogram(kg). The mass of a cylinder made of platinum-iridium alloy kept at the International Bureau of Weights and Measures is defined as 1 kg.
6. Time: Unit is second(s). One second is the time taken by 9 192 631 770 oscillations of the light (of a specified wavelength) emitted by a cesium-133 atom.
7. Electric Current: Unit is Ampere. If equal currents are maintained in the two wires so that the force between them is $ 2 x 10^{-7} $ newton per meter of the wires, the current in any of the wires is called 1 A
8. Thermodynamic Temperature: Unit is Kelvin(K). The fraction $ \frac{1}{273.16} $ of the thermodynamic temperature of the triple point of water is called 1 K.
9. Amount of the Substance: Unit is mole(mole). The amount of a substance that contains as many
elementary entities as there is the number of atoms in 0.012 kg of carbon-12 is called a mole.
10. Luminous Intensity: Unit is Candela(cd). The SI unit of luminous intensity is 1 cd which is the luminous intensity of a blackbody of surface area $ \frac{1}{600 000} m^{2} $ placed at the temperature of freezing, platinum, and at a pressure of 101,325 $ {N/m^{2}} $, in the direction perpendicular to its surface.
11. Dimensions: Dimensions are the powers to which fundamental quantities are raised to represent that quantity. It is represented by using a square bracket.
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Physical Quantities
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Dimensions
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Distance, Length, Displacement
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$[M^0LT^0]$
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Velocity, Speed
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$[M^0LT^{-1}]$
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Acceleration
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$[M^0LT^{-2}]$
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Force
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$[MLT^{-2}]$
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Linear momentum, Impulse
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$[MLT^{-1}]$
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Torque, Work, Kinetic Energy, Potential Energy, Energy,
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$[ML^2T^2]$
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Power
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$[ML^2T^{-3}]$
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Pressure, Stress, Modulus of Elasticity
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$[ML^{-1}T^{-2}]$ |
12. Principle of homogeneity of Dimensions: A correct dimensional equation must be homogeneous i.e. dimensions on both sides are the same.
13. Use of Dimension: To convert a unit from one system to another system, To find the relation between various physical parameters and to check whether the formula is dimensionally correct or not.
Example1: Find the dimension of the constants a and b in Van Der Wall Equation
i.e. $ (P + \frac{a}{V^2})(V-b) = RT $
Solution: Using principle of homogeneity,
Dimension of b = Dimension of V (volume) = $ [{ L^3 }] $
Dimension of P (pressure) = Dimension of $ (\frac{a}{V^{2}}) $
Dimension of a = dimension of $ PV{^2} $ = $[ML^{-1}T^{-2}] [L^{3}]^{2} $= $ [ML^{5}T^{-2}] $
Example 2: The value of the gravitational constant is G = $ 6.67 * 10^{-11} $ $ Nm{^2}kg^{-2} $. Convert it into a system based on kilometer, tonne and hour as base units.
Solution: Dimnsional formula of G is $ [M^{-1}L^{3}T^{-2}] $
$ n_2 = n_1 [\frac{M_1}{M_2}]^{-1}[\frac{L_2}{L_1}]^{3} [\frac{T_2}{T_1}]^{-2} $
$ n_1 = 6.67 * 10^{-11}, M_1 = 1 kg, M_2 = 1 tonne = 1000kg, $
$T_1 = 1s, T_2 = 3600s, L_1 = 1m and L_2 = 1000m $
$ n_2 = 6.67*10^{-11}[\frac{1}{1000}]^{-1}[\frac{1}{1000}]^{3} [\frac{1}{3600}]^{-2} = 8.64 * 10^{10} $
Example 3: The frequency f of a stretched string depends upon the Tension (T), length (l) and the linear mass density $ /mu $. Find the relation for frequency.
Solution: Let frequency depends on T, l, and $ \mu $ as follow:
$ f = kT^{a}l^{b}{\mu ^{c}} $ where k is constant.
writing dimension formula of both sides,
$ [M^{0}L^{0}T^{-1}] $ = $ [MLT^{-2}]^{a}[L]^{b}[ML^{-1}]^{c} $ = $ [M^{a+c}L^{a+b-c}T^{-2a}] $
Comparing dimensions on both sides,
a + c = 0
a + b - c = 0
-2a = -1
solving these we get, a = $\frac{1}{2} $, b = -1 and c = $ \frac{-1}{2} $
so relation will be, $f = \frac{k}{l} \sqrt {\frac{T}{\mu}} $
Video Lecture:
Fundamental and Derived Quantities, Dimensions, How to find dimension of any physical Quantity, Formula validation by dimensions, Deriving relation between physical quantities Unit conversion Watch video
