The rate of increase of bacteria in a certain | vedclass

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

Question

(C) Let $P_0$ be the initial population and $P$ be the population at time $t$.According to the problem,$\frac{dP}{dt} = kP$,where $k$ is a constant.Se

Vedclass · Accepted Answer

$32$

Vedclass · Answer

(C) Let $P_0$ be the initial population and $P$ be the population at time $t$.According to the problem,$\frac{dP}{dt} = kP$,where $k$ is a constant.Separating the variables,we get $\frac{dP}{P} = k dt$.Integrating both sides,we have $\ln P = kt + C$.At $t = 0$,$P = P_0$,so $\ln P_0 = C$.Thus,$\ln P = kt + \ln P_0$,which simplifies to $\ln \left( \frac{P}{P_0} ight) = kt$.Given that the population doubles in $5$ hours,at $t = 5$,$P = 2P_0$.Substituting these values,$\ln \left( \frac{2P_0}{P_0} ight) = 5k$,so $k = \frac{\ln 2}{5}$.Now,we need to find the population at $t = 25$ hours.$\ln \left( \frac{P}{P_0} ight) = \left( \frac{\ln 2}{5} ight) 	imes 25 = 5 \ln 2 = \ln(2^5) = \ln 32$.Therefore,$\frac{P}{P_0} = 32$,which means $P = 32 P_0$.The number of bacteria will be $32$ times the original.

The rate of increase of bacteria in a certain culture is proportional to the number present. If it doubles in $5$ hours,then in $25$ hours,its number would be ......... times the original.

Similar Questions

If the solution of the differential equation $\frac{dy}{dx} = \frac{1+x}{2y}$ is a conic passing through the point $(1, 1)$,then its eccentricity is:

If the length of the sub-tangent at any point $P$ on a curve is proportional to the abscissa of the point $P$,then the equation of that curve is ($C$ is an arbitrary constant).

The 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 in $12$ hours will be: (in $N$)

Any curve passes through the point $(3, -4)$. If the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2y}{x}$,then the equation of the curve is . . . . . . .

If the surrounding air is kept at $25^{\circ} C$ and a body cools from $80^{\circ} C$ to $50^{\circ} C$ in $30 \text{ minutes}$,then the temperature of the body after one hour will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module