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(D) Let $N$ be the number of bacteria present at time $t$. Let $N_{0}$ be the initial number of bacteria. Here $\frac{dN}{dt} \propto N \Rightarrow \f

Vedclass · Accepted Answer

$8$

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(D) Let $N$ be the number of bacteria present at time $t$. Let $N_{0}$ be the initial number of bacteria. Here $\frac{dN}{dt} \propto N \Rightarrow \frac{dN}{dt}=KN \Rightarrow \frac{dN}{N}=K dt$.Integrating both sides,we get $\int \frac{dN}{N} = \int K dt \Rightarrow \log N = Kt + C$.When $t=0$,$N=N_{0}$,so $\log N_{0} = C$.Thus,$\log N - \log N_{0} = Kt \Rightarrow \log \left(\frac{N}{N_{0}}ight) = Kt$.When $t=4$ hours,$N=2N_{0}$,so $\log(2) = 4K \Rightarrow K = \frac{\log 2}{4}$.Now,we want to find $t$ when $N=4N_{0}$.Substituting the values: $\log \left(\frac{4N_{0}}{N_{0}}ight) = \left(\frac{\log 2}{4}ight)t$.$\log 4 = \frac{t}{4} \log 2 \Rightarrow 2 \log 2 = \frac{t}{4} \log 2$.Dividing by $\log 2$,we get $2 = \frac{t}{4} \Rightarrow t = 8$ hours.Alternatively,since the population doubles every $4$ hours,after $4$ hours it is $2N$,and after another $4$ hours (total $8$ hours),it becomes $2 	imes (2N) = 4N$.

Bacteria increases at a rate proportional to the number of bacteria present. If the original number $N$ doubles in $4$ hours,then the number of bacteria will be $4N$ in (in $hours$)

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