The rate of growth of bacteria is proportiona | vedclass

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

Question

(C) The rate of growth is proportional to the number present.$\frac{dN}{dt} = kN \Rightarrow \frac{dN}{N} = k dt$.Integrating both sides,we get $\ln N

Vedclass · Accepted Answer

$5656$ approximately

Vedclass · Answer

(C) The rate of growth is proportional to the number present.$\frac{dN}{dt} = kN \Rightarrow \frac{dN}{N} = k dt$.Integrating both sides,we get $\ln N = kt + C$.At $t = 0$,$N = 1000$,so $C = \ln 1000$.Thus,$\ln N = kt + \ln 1000 \Rightarrow \ln(\frac{N}{1000}) = kt \Rightarrow N = 1000 e^{kt}$.Given that at $t = 1$,$N = 2000$,we have $2000 = 1000 e^k$,which implies $e^k = 2$.Substituting this into the equation,$N = 1000 	imes (e^k)^t = 1000 	imes 2^t$.For $t = 2 \frac{1}{2} = 2.5$ hours:$N = 1000 	imes 2^{2.5} = 1000 	imes 2^2 	imes 2^{0.5} = 1000 	imes 4 	imes \sqrt{2}$.Given $\sqrt{2} = 1.414$,we get $N = 1000 	imes 4 	imes 1.414 = 5656$.

The rate of growth of bacteria is proportional to the number present. If initially there were $1000$ bacteria and the number doubles in $1$ hour,then the number of bacteria after $2 \frac{1}{2}$ hours is (Given $\sqrt{2} = 1.414$):

Similar Questions

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

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 $\%$)

An ice ball melts at a rate proportional to the amount of ice present at that instant. Half of the initial quantity of ice melts in $15 \text{ minutes}$. Let $x_0$ be the initial quantity of ice. If after $30 \text{ minutes}$ the amount of ice left is $k x_0$,then the value of $k$ is:

The equation of the curve passing through $\left(2, \frac{9}{2}\right)$ and having the slope $\left(1-\frac{1}{x^2}\right)$ at $(x, y)$ is

Water flows from the base of a rectangular tank of depth $16 \ m$. The rate of flow of the water is proportional to the square root of the depth at any time $t$. If the depth is $4 \ m$ when $t = 2 \ hours$,then after $3.5 \ hours$,the depth (in meters) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module