The dimensions of - left- - mu - 0 - varepsilon - 0 - right- - -1-2 - are -L-1-T-L-T-1-2-L-1-2-T-1-2-L-T-1-

Question

Toppr Admin · Accepted Answer

Since expression for speed of light c-1-x221A-x3BC-0-x3B5-0- the given expression has the dimensions of velocity i-e- LT-x2212-1-Hence- option B is correct-

The dimensions of ${(μ_{0} ε_{0})}^{- 1 / 2}$ are :
$[L^{- 1} T]$
$[L T^{1 / 2}]$
$[L^{1 / 2} T^{1 / 2}]$
$[L T^{- 1}]$

Since expression for speed of light $c = \frac{1}{\sqrt{μ_{0} ε_{0}}}$ , the given expression has the dimensions of velocity i.e. $L T^{- 1}$ .
Hence, option B is correct.

The dimensions of (μ0ε0)−1/2 are :[L−1T][LT1/2][L1/2T1/2][LT−1]

Since expression for speed of light c=1√μ0ε0, the given expression has the dimensions of velocity i.e. LT−1.Hence, option B is correct.

The dimensions of ${(μ_{0} ε_{0})}^{- 1 / 2}$ are :
$[L^{- 1} T]$
$[L T^{1 / 2}]$
$[L^{1 / 2} T^{1 / 2}]$
$[L T^{- 1}]$

Since expression for speed of light $c = \frac{1}{\sqrt{μ_{0} ε_{0}}}$ , the given expression has the dimensions of velocity i.e. $L T^{- 1}$ .
Hence, option B is correct.