A) \[8\times {{10}^{-6}}H\]
B) \[6\times {{10}^{-6}}H\]
C) \[4\times {{10}^{-3}}H\]
D) \[2\times {{10}^{-3}}H\]
Correct Answer: D
Solution :
As, \[\left| e \right|=L\frac{dl}{dt}\] \[8\times {{10}^{-3}}=L\frac{d(4+4\,\,t)}{dt}\] \[8\times 1{{0}^{-\,3}}=\,\,L\left( 4 \right)\] \[\operatorname{L}=2\times 1{{0}^{-}}^{3}H\]You need to login to perform this action.
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