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The chrage flowing in a conductor varies with time Q=5 t-\frac{1}{2} t^2+\frac{1}{6} t^3 . Then the current -

Option: 1

Has an initial value 5.


Option: 2

Reaches a minimum value after time 1A.


Option: 3

Has either a maximum or a minimum value of \frac{9}{2} A .


Option: 4

Option (i), (ii), (iii)


Answers (1)

best_answer

Given Q=5 t-\frac{1}{2} t^2+\frac{1}{6} t^3 .

We know that

\begin{aligned} I=\frac{d Q}{d t} & =\frac{d\left(s t-\frac{1}{2} t^2+\frac{1}{6} t^3\right)}{d t} \\ & =\left(5-t+\frac{1}{2} t^2\right) \end{aligned}

For maximum and minimum value of current

\begin{aligned} \frac{d I}{d t}=0 & \Rightarrow \frac{d\left(5-t+\frac{1}{2} t^2\right)}{d t}=0 \\ & \Rightarrow\left(0-1+\frac{1}{2} \cdot 2 t\right)=0 \\ & \Rightarrow-1+t=0 \Rightarrow t=1 \end{aligned}

dt \quad t=1, I=5-t+\frac{1}{2} t^2=5-1+\frac{1}{2}(1)^2=\frac{9}{2} \mathrm{~A}

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Divya Prakash Singh

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