Question

Answer the question which is given in attachment.

Answer 1

We are asked to find the dimension of:
$(\mu_{\circ}\varepsilon_{\circ})^{-1/2}$

Here, $\mu_{\circ}$ is the Absolute Permeability. And $\varepsilon_{\circ}$ is the Absolute Permittivity.


These both are related to the speed of light in vacuum (c) as follows:

$\boxed{c = \frac{1}{\sqrt{\mu_{\circ}\varepsilon_{\circ}}}}$


This relation is very useful, and must be remembered.

We have:

$c = \frac{1}{\sqrt{\mu_{\circ}\varepsilon_{\circ}}} \\ \\ \\ \implies c = (\mu_{\circ}\varepsilon_{\circ})^{-1/2}$

This means that the dimensions of $(\mu_{\circ}\varepsilon_{\circ})^{-1/2}$ are the same as dimensions of $c$

Now, c is the Speed of light in vacuum. It's unit is that of speed,which is metres per second. 


So, the dimensions are:

$\left[ c \right] = \left [ L \, T^{-1} \right] \\ \\ \\ \implies \boxed{\left[ \mu_{\circ}\varepsilon_{\circ} \right] = \left [ L \, T^{-1} \right]}$

So, the answer is Option (4).




Hope it helps
Purva
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