The rate of growth of bacteria in a culture i | vedclass

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

Question

(A) Let $B(t)$ be the number of bacteria at time $t$. The rate of growth is given by $\frac{dB}{dt} = \lambda B$.Integrating this,we get $B(t) = B_0 e

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let $B(t)$ be the number of bacteria at time $t$. The rate of growth is given by $\frac{dB}{dt} = \lambda B$.Integrating this,we get $B(t) = B_0 e^{\lambda t}$,where $B_0 = 1000$.Given that at $t = 2$,$B(2) = 1000 + 20\% 	ext{ of } 1000 = 1200$.So,$1200 = 1000 e^{2\lambda} \Rightarrow e^{2\lambda} = \frac{6}{5} \Rightarrow 2\lambda = \log_{e}\left(\frac{6}{5}ight) \Rightarrow \lambda = \frac{1}{2} \log_{e}\left(\frac{6}{5}ight)$.We are given $B(T) = 2000$ where $T = \frac{k}{\log_{e}\left(\frac{6}{5}ight)}$.Using $B(T) = B_0 e^{\lambda T}$,we have $2000 = 1000 e^{\lambda T} \Rightarrow 2 = e^{\lambda T} \Rightarrow \log_{e} 2 = \lambda T$.Substituting $\lambda$ and $T$: $\log_{e} 2 = \left(\frac{1}{2} \log_{e}\left(\frac{6}{5}ight)ight) 	imes \left(\frac{k}{\log_{e}\left(\frac{6}{5}ight)}ight) = \frac{k}{2}$.Thus,$k = 2 \log_{e} 2$.Finally,$\left(\frac{k}{\log_{e} 2}ight)^{2} = \left(\frac{2 \log_{e} 2}{\log_{e} 2}ight)^{2} = 2^{2} = 4$.

Similar Questions

If it is mentioned that for a curve passing through $(3,4)$,the slope of the curve at any point is the reciprocal of twice the ordinate of that point,then that curve is a

The equation to the orthogonal trajectories of the system of parabolas $y = ax^2$ is

The rate of disintegration of a radioactive element at any time is proportional to its mass at that time, where $k (k > 0)$ is the constant of proportionality. The time during which an original mass of $1.5 \text{ gm}$ will disintegrate to a mass of $0.5 \text{ gm}$ is ...

$A$ body cools according to Newton's law of cooling from $100^{\circ} C$ to $60^{\circ} C$ in $15$ minutes. If the temperature of the surrounding is $20^{\circ} C$,then the temperature of the body after cooling down for one hour is (in $^{\circ} C$)

Verify that the given function $y=x \sin 3x$ is a solution of the differential equation $\frac{d^{2}y}{dx^{2}}+9y-6 \cos 3x=0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module