Question

Which of the following product of e , h , mu , G ( where mu is permeability ) be taken so that the dimensions of the product are same as that of the speed of light ?

Answer 1

Answer:

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Explanation:

e is denoted electron charge .

∴ dimension of e = dimension of { current × time} = [AT]

[∵ dimension of current = [A] and dimension of time = [T]

h is denoted Plank's constant .

∵ E = hν , here E is energy of photons , h is plank's constant and ν is frequency of light .

∴ dimension of h = dimension of { E/v} = dimension of {E}/dimension of v

= [ML²T⁻²]/[T⁻¹] = [ML²T⁻¹]

μ is denoted permeability of medium

we know, F = Bil and B = μidlsinΘ/4πr² [ both equations are General formula ]

after solving we get

dimension of μ = dimension of {F/i²} = dimension of F/dimension of i²

= [MLT⁻²]/[A²] = [MLT⁻²A⁻²]

G is denoted gravitational constant

∵ F = GmM/r² [ Gravitational force ]

∴ dimension of G = dimension of {Fr²/mM} = dimension of Fr²/dimension of {mM}

= [MLT⁻²][L²]/[M²]

= [M⁻¹L³T⁻²]

Now, speed of light in term of all above :

dimension of speed = [LT⁻¹]

Let, [LT⁻¹] = k[AT]^a [ML²T⁻¹]^b [MLT⁻²A⁻²]^c [M⁻¹L³T⁻²]^d

= [M]^(b+c -d) [L]^(2b+c+3d) [T]^(a-b-2c-2d) [A]^(a -2c)

Compare both sides,

b + c - d = 0⇒b + c = d

2b + c + 3d = 1

a - b - 2c - 2d = -1

a - 2c = 0 ⇒ a = 2c

after solving above equations ,

d = 0 , b = 1 , c = -1 and a = -2

Now, dimension of speed of light = h/(e² × μ )

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