Directions : In the following question more than one of the answers given may be correct. Select the correct answer and mark it according to the code:
The charge flowing in a conductor varies with time as\[Q=at-b{{t}^{2}},\]then the current changes at the rate of\[(-2b)\] falls to zero after\[T=\left( \frac{a}{2b} \right)\] (3) reaches a maximum and then decreases (4) will remain constantA) 1, 2 and 3 are correct
B) 1 and 2 are correct
C) 2 and 4 are correct
D) 1 and 3 are correct
Correct Answer: B
Solution :
\[Q=at-b{{t}^{2}}\] Current, \[I=\frac{dQ}{dt}\] \[=\frac{d}{dt}(at-b{{t}^{2}})\] \[=a-2bt\] If\[I=0\]then\[0=a-2bt\]or\[t=\frac{a}{2b}\] Also, \[\frac{dI}{dt}=\frac{d}{dt}(a-2bt)\] \[=-2b\] Therefore, current change at the ratio of\[(-2b)\].You need to login to perform this action.
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