The dimensional formula of 1-0-0 is

The correct option is A [M^0L^2T^ 2] Velocity of EM waves, v = 1√(μ_0ϵ_0) ⇒ = 1μ_0ϵ_0 = v^2 ∴ Dimensional formula of 1μ_0ϵ_0 = ...

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Standard XII

Physics

Dimensional Analysis

The dimension...

Question

The dimensional formula of $\frac{1}{μ_{0} ϵ_{0}}$ is

[M^{0} L T^{- 2}]

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[M^{0} L^{- 2} T^{- 2}]

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[M^{0} L T^{- 1}]

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[M^{0} L^{2} T^{- 2}]

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Solution

The correct option is A $[M^{0} L^{2} T^{- 2}]$
Velocity of $E M$ waves, $v = \frac{1}{\sqrt{μ_{0} ϵ_{0}}}$
$\Rightarrow= \frac{1}{μ_{0} ϵ_{0}} = v^{2}$
$∴$ Dimensional formula of $\frac{1}{μ_{0} ϵ_{0}} = [M^{0} L T^{- 1}]^{2}$
$= [M^{0} L^{2} T^{- 2}]$ .

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The dimensional formula of 1μ0ϵ0 is

The correct option is A [M0L2T−2]Velocity of EM waves, v=1√μ0ϵ0⇒=1μ0ϵ0=v2∴ Dimensional formula of 1μ0ϵ0=[M0LT−1]2=[M0L2T−2].

The dimensional formula of $\frac{1}{μ_{0} ϵ_{0}}$ is

The correct option is A $[M^{0} L^{2} T^{- 2}]$
Velocity of $E M$ waves, $v = \frac{1}{\sqrt{μ_{0} ϵ_{0}}}$
$\Rightarrow= \frac{1}{μ_{0} ϵ_{0}} = v^{2}$
$∴$ Dimensional formula of $\frac{1}{μ_{0} ϵ_{0}} = [M^{0} L T^{- 1}]^{2}$
$= [M^{0} L^{2} T^{- 2}]$ .