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

$A$ body cools according to Newton's law of cooling from $100^{\circ} C$ to $60^{\circ} C$ in $20 \text{ minutes}$. If the temperature of the surrounding is $20^{\circ} C$,then the temperature of the body after one hour is: (in $^{\circ} C$)

Radium decomposes at a rate proportional to the amount present. If half the original amount disappears in $1600$ years,then the percentage loss in $100$ years is (Given $\log 2 = 0.6931$ and $e^{-0.0433} = 0.9576$) (in $\%$)

$A$ spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is $3 \text{ mm}$ and $1 \text{ hour}$ later has been reduced to $2 \text{ mm}$,then the expression of radius $r$ of the raindrop at any time $t$ is (where $0 \leq t < 3$):

The slope at any point of a curve $y=f(x)$ is given by $\frac{dy}{dx}=3x^2$ and it passes through $(-1,1)$. The equation of the curve is

The rate of disintegration of a radioactive element at any time $t$ is proportional to its mass $M$ at that time. If $k (k > 0)$ is the constant of proportionality, find the time required for an initial mass of $1.5 \text{ g}$ to disintegrate to a mass of $0.5 \text{ g}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module