Physics And Measurement Question 87

Question: $ {\mu _{0}} $ and $ {\varepsilon _{0}} $ denote the permeability and permittivity of free space, the dimensions of $ {\mu _{0}}{\varepsilon _{0}} $ are

Options:

A) $ L{{T}^{-1}} $

B) $ {{L}^{-2}}T^{2} $

C) $ {{M}^{-1}}{{L}^{-3}}Q^{2}T^{2} $

D) $ {{M}^{-1}}{{L}^{-3}}I^{2}T^{2} $

Show Answer

Answer:

Correct Answer: B

Solution:

$ C=\frac{1}{\sqrt{{\mu _{0}}{\varepsilon _{0}}}} $

therefore $ {\mu _{0}}{\varepsilon _{0}}=( \frac{1}{C^{2}} ) $ (where C = velocity of light)

$ [{\mu _{0}}{\varepsilon _{0}}]={{L}^{-2}}T^{2} $



NCERT Chapter Video Solution

Dual Pane