4 examples of dimension of physical quantity

1 of 4 examples of dimension of physical quantity

 Example1.  Physicist Fermi once pointed out that a standard lecture period (50 min) is close to 1 micro century. Find the percentage difference from Fermi’s approximation?

Solution:   Here, 1 micro century = 10^-6 × 100 years = 10^-4 years

= 10^-4  × 365 × 60 min  = 52.56 min.

Percentage difference = (actual – approximation) 100/actual

=(50-52.56)100/50 = -5.12%

So percentage difference from Fermi’s approximation will  -5.12%

2 of 4 examples of dimension of physical quantity

 Example2. Iron has a mass of 7.87 g per cubic centimeter of volume and the mass of an iron atom is 9.27×10^-26 kg. If the atoms are spherical and tightly packed, (a) What is the volume of an iron atom and (b) What is the distance between the centers of adjacent atom?

Solution:  Here, mass/volume = density (Ρ)

                                                 =7.87 g/c.c. = 7.87 × 10³   kg/m³

Mass of an iron atom,  m = 9.27  ×   10^-26  kg

Volume of an iron atom,  V=mass/density=m/ Ρ = (9.27 ×10^-26 )/(7.87  × 10³  )=1.18×   10^-29m³

If r is radius of an iron atom, then V = 4/3πr³

                                                              r = (3V/4π)^1/3 = [(3×1.18×10^-29)/(4×22/7)]^1/3

                                                               r =1.41 × 10^-10m

The centre to centre distance between two adjacent atoms = 2r = 2×1.41 × 10^-26m

                                            = 2.82 × 10^-10 m

Hence, the volume of an iron atom is 1.18×   10^-29m³

And distance between the centers of adjacent atom = 2.82 × 10^-10 m

3 of 4 examples of dimension of physical quantity

Example 3.  If a composite physical quantity in terms of moment of inertia I, force F, volicity v, work W and length  L is defined as Q = IFv³/WL³ find the dimensions of Q and identify it.

Solution : As I =[ML²],        F=[MLT-²]

  V= [LT^-1],    W = [ML²T^-2], therefore,

Q = IFv²/WL³ ==[ML²] [MLT-²] [LT^-1]²/[ML²T^-2]L³

Q = [MT-²]

As [MT-²] is the dimensional formula of surface tension, force constant, surface energy (energy per unit area) therefore, the physical quantity Q can be any one of these

4 of 4 examples of dimension of physical quantity

Example 4. Find the dimension of (a) electric field E (b)magnetic field B (c) magnetic permeability µ.

The relevant equations are F = qE; F = q v B and B = µI/2πa

Where F is the force, q is the charge, v is speed, I is current and a is distance.

Solution. (a) From F= qE

E= F/q = [MLT-²]/[AT]

E=[MLT^-3 A^-1]

Here, A stand for ampere, the S.I unit of current.

(b) F=q v B

B =F/qv = [MLT-²]/[AT][LT^-1]=[ML^o T^-2 A ^-1]

(C)  From B = µI/2πa=2πaB/I=[L] [ML^o T^-2 A ^-1]/[A]

=[MLT^-2 A ^-1]