First slide

Dimensions of physical quantities

Question

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

Moderate

A

$h e^{- 2} μ^{- 1} G^{0}$
B

$h^{2} e G^{0} μ$
C

$h^{0} e^{2} G^{- 1} μ$
D

$h G e^{- 2} μ^{0}$

Solution

$\begin{array}{l} Here v = e^{a} h^{b} μ^{c} G^{d} \\ . Taking the dimensions, \\ [M^{0} L T^{- 1} A^{0}] = {[A T^{1}]}^{a} {[M L^{2} T^{- 1}]}^{b} {[M L T^{- 2} A^{- 2}]}^{c} {[M^{- 1} L^{3} T^{- 2}]}^{d} \\ [M^{0} L T^{- 1} A^{0}] = [M^{b + c - d} L^{2 b + c + 3 d} T^{a - b - 2 c - 2 d} A^{a - 2 c}] \\ b + c - d = 0 - - - (1) \\ 2 b + c + 3 d = 1 - - - (2) \\ a - b - 2 c - 2 d = - 1 - - - (3) \\ a - 2 c = 0 - - - (4) \end{array}$
$\begin{array}{l} Solving for a, b, c and d, we get \\ a = - 2, b = 1, c = - 1, d = 0 \\ Thus, v = e^{- 2} h μ^{- 1} G^{0} \end{array}$

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