The dimensions of frac-1-mu -0- in -0- is -where symbols have their usual menaing-ML-2-T-2-left- -L-2-T- - 2- right-A-ML-2T-2-

Question

Toppr Admin · Accepted Answer

Velocity of light in vacuum is 1-x221A-x3BC-0-x2208-0-LT-x2212-1-1-x221A-x3BC-0-x2208-0-L2T-x2212-2-1-x3BC-0-x2208-0-Hence-option -A- is correct answer-

The dimensions of $\frac{1}{μ_{0} \in_{0}}$ is (where symbols have their usual menaing)
$[M L^{- 2} T^{- 2}$
$[L^{2} T^{- 2}]$
$[A]$
$[M L^{2} T^{- 2}]$

Velocity of light in vacuum is $\frac{1}{\sqrt{(μ_{0} \in_{0})}}$
$[L T^{- 1}] = [\frac{1}{\sqrt{(μ_{0} \in_{0})}}]$
$[L^{2} T^{- 2}] = [\frac{1}{(μ_{0} \in_{0})}]$
Hence,
option $(A)$ is correct answer.

The dimensions of 1μ0∈0 is (where symbols have their usual menaing)[ML−2T−2[L2T−2][A][ML2T−2]

Velocity of light in vacuum is 1√(μ0∈0)[LT−1]=[1√(μ0∈0)][L2T−2]=[1(μ0∈0)]Hence,option (A) is correct answer.

The dimensions of $\frac{1}{μ_{0} \in_{0}}$ is (where symbols have their usual menaing)
$[M L^{- 2} T^{- 2}$
$[L^{2} T^{- 2}]$
$[A]$
$[M L^{2} T^{- 2}]$

Velocity of light in vacuum is $\frac{1}{\sqrt{(μ_{0} \in_{0})}}$
$[L T^{- 1}] = [\frac{1}{\sqrt{(μ_{0} \in_{0})}}]$
$[L^{2} T^{- 2}] = [\frac{1}{(μ_{0} \in_{0})}]$
Hence,
option $(A)$ is correct answer.