1.

Dimension of `(1)/(mu_(0)epsilon_(0))`, where symbols have usual meaning, areA. `L^-1T`B. `L^2T^2`C. `L^2T^-2`D. `LT^-1`

Answer» Correct Answer - c
As `c=1/(sqrt(in_0mu_0)), c^2=1/(mu_0in_0),`
so `[LT^-1]^2=1/(mu_0in_0) or [(mu_0in_0)]=L^2T^-2`


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