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| 1. |
Using mass `(M)` , length `(L)` , time `(T)` , and electric current `(A)` as fundamental quantities , the dimensions of permitivity will beA. `[M^(-1)LT^(-2)A]`B. `[ML^(-2)T^(-2)A^(-1)]`C. `[MLT^(-2)A^(-2)]`D. `[MLT^(-1)A^(-1)]` |
| Answer» Units of permeability are equivalent to `N/(Amp)^(2)` thus dimensions are `=(M^(1)L^(1)T^(-2))/(A^(2))=M^(1)L^(1)T^(-2)A^(-2)` | |