Bacteria increases at a rate proportional to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

$8$

Vedclass · Answer

(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 $hours$)

Similar Questions

$A$ piece of steel is heated to $100^{\circ}C$ and allowed to cool in a room. Which graph correctly represents the cooling process?

If the population grows at the rate of $8 \%$ per year,then the time taken for the population to be doubled,is (Given $\log 2 = 0 \cdot 6912$)

If the slope of the tangent to a curve at any point $(x, y)$ is given by $3x^2 + 2x + 5$ and the curve passes through the point $(0, 1)$,then the equation of the curve is:

The slope of the tangent to a curve $C : y = y(x)$ at any point $(x, y)$ on it is $\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$. If $C$ passes through the points $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$ and $(\alpha, \frac{1}{2}e^{2\alpha})$,then $e^{\alpha}$ is equal to:

The solution of the differential equation $\frac{dy}{dx} + \frac{x}{y} \cdot \frac{x^2+y^2-1}{2(x^2+y^2)+1} = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module