A) \[3s\]
B) \[4s\]
C) \[1s\]
D) \[2s\]
Correct Answer: D
Solution :
Emf across an inductor is given by \[emf=L\frac{di}{dt}\] \[=L\frac{d}{dt}({{t}^{2}}{{e}^{-t}})=0\] \[\Rightarrow \] \[\frac{d}{dt}({{t}^{2}}{{e}^{-1}})=0\] \[\Rightarrow \] \[{{e}^{-t}}\times 2t+{{t}^{2}}\times (-1){{e}^{-1}}=0\] \[\Rightarrow \] \[t{{e}^{-1}}(2-t)=0\] \[\Rightarrow \] \[2-t=0\] \[\Rightarrow \] \[t=2s\]
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