A) \[\text{weber/}{{\text{m}}^{\text{2}}}\]
B) Henry(H)
C) \[H/{{m}^{2}}\]
D) weber
Correct Answer: B
Solution :
(b) Henry(H) Dimension of Henry \[=[M{{L}^{2}}{{T}^{-2}}{{A}^{-2}}]\] Dimension of charge = [AT] ...(1) \[\therefore \] Dimension of Henry \[=[M{{L}^{2}}{{(AT)}^{-2}}]\] \[=[M{{L}^{2}}{{Q}^{-2}}]\] [from(1)] \[\left[ \frac{M{{L}^{2}}}{{{Q}^{2}}} \right]\]You need to login to perform this action.
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