Solveeit Logo
Login

Question

Question: Dimensions of \(\dfrac{1}{{{\mu }_{0}}\in_0}\), where symbols have their usual meaning, are: A. \...

Dimensions of 1μ0∈0\dfrac{1}{{{\mu }_{0}}\in_0}, where symbols have their usual meaning, are:
A. [L−1T][{{L}^{-1}}T]
B. [L−2T2][{{L}^{-2}}{{T}^{2}}]
C. [L2T−2][{{L}^{2}}{{T}^{-2}}]
D. [LT−1][L{{T}^{-1}}]

Explanation

Solution

We are supposed to find the dimensional formula of 1μ0∈0\dfrac{1}{{{\mu }_{0}}\in_0}. For that, we have to identify what μ0{{\mu }_{0}} and ∈0\in_0 denotes. Further, we can deduce the value from the dimensional formula of these individual elements by finding the degree of dependence of a physical quantity on another. The principle of consistency of two expressions can be used to find the equation relating these two quantities.
We know that μ0{{\mu }_{0}} is the permeability and ∈0\in_0 is the permittivity of the free space.

Formulas used:
Velocity of light in vacuum=1μ0∈0=\dfrac{1}{\sqrt{{{\mu }_{0}}\in_0}}, where μ0{{\mu }_{0}} is the permeability and ∈0\in_0 is the permittivity of the free space.
Dimensional formula for velocity = [LT−1][L{{T}^{-1}}], where LL denotes the length or distance and TT denotes the time. Here, the time is taken in reverse. This implies the basic formula of velocity, i.e. displacementtime\dfrac{displacement}{time}.

Complete step by step answer:
We know that the Velocity of light in vacuum=1μ0∈0=\dfrac{1}{\sqrt{{{\mu }_{0}}\in_0}}.
i.e. v=1μ0∈0v=\dfrac{1}{\sqrt{{{\mu }_{0}}\in_0}}.
Upon squaring, we get v2=1μ0∈0{{v}^{2}}=\dfrac{1}{{{\mu }_{0}}\in_0}
⇒[L2T−2]=1μ0∈0\Rightarrow [{{L}^{2}}{{T}^{-2}}]=\dfrac{1}{{{\mu }_{0}}\in_0}
⇒[LT−1]2=1μ0∈0\Rightarrow {{[L{{T}^{-1}}]}^{2}}=\dfrac{1}{{{\mu }_{0}}\in_0}
∴[L2T−2]=1μ0∈0\therefore [{{L}^{2}}{{T}^{-2}}]=\dfrac{1}{{{\mu }_{0}}\in_0}
Therefore, the dimensions of 1μ0∈0\dfrac{1}{{{\mu }_{0}}\in_0} = [L2T−2][{{L}^{2}}{{T}^{-2}}]
Hence, option C is the right choice.

Note: Though this method is widely used, there are few drawbacks for this method. Dimensionless quantities cannot be determined by this method. Constant proportionality cannot be determined by this method. It is not applicable to trigonometric, logarithmic and exponential functions physical quantities which are dependent upon more than three physical quantities, this method will be difficult.