Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimensions of permeability are :A. `[M^(-1)LT^(-2)A]`B. `[ML^(-2)T^(-2)A^(-1)]`C. `[MLT^(-2)A^(-2)]`D. `[MLT^(-1)A^(-1)]`

Using mass (M), length (L), time (T) and current (A) as fundamental qu

1.	Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimensions of permeability are :A. `[M^(-1)LT^(-2)A]`B. `[ML^(-2)T^(-2)A^(-1)]`C. `[MLT^(-2)A^(-2)]`D. `[MLT^(-1)A^(-1)]`
Answer» Correct Answer - C 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)]`

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Using mass (M), length (L), time (T) and current (A) as fundamental quantities, the dimensions of permeability are :A. `[M^(-1)LT^(-2)A]`B. `[ML^(-2)T^(-2)A^(-1)]`C. `[MLT^(-2)A^(-2)]`D. `[MLT^(-1)A^(-1)]`

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