$\begin{array}{1 1}\large\frac{2.\log 2}{\log(\large\frac{11}{10})} \\ \large\frac{\log 2}{\log(\large\frac{11}{10})} \\\large\frac{2.\log 2}{\log(\large\frac{10}{11})} \\ \large\frac{\log 2}{\log(\large\frac{10}{11})}\end{array} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- $\large\frac{dy}{dt}$ is proportional to y where y is the number of bacteria at any instant $t$

Step 1:

From the given information we know that dy/dt is proportional to $y$

$\large\frac{dy}{dt}=$$ky$

on seperating the variables

we get $\large\frac{dy}{y}$$=k.dt$

Step 2:

Integrating on both sides

we get $\int \large\frac{dy}{y}$$=k \int dt$

$\log y=kt+c$ ----(1)

Let $y_0$ be the number of bacteria when $t=0$

Hence $\log y_0=C$

Substituting this value in equation (1) we get $\log y =kt+\log y_0$

$\log y- \log y_0=kt$

$\log(\large\frac{y}{y_0})$$=kt$ ---(2)

Step 3:

It is also given that the number of bacteria increases 10% in 2 hours

Hence $y=110\large\frac{y_0}{100}$

$\large\frac{y}{y_0}=\frac{11}{10}$

Substituting this in (2)

we get $2k=\log (\large\frac{ 11}{10})$

or $k=\large\frac{1}{2}$$\log \large\frac{11}{10}$

Therefore $\large\frac{1}{2} $$\log (\large\frac{11}{10})$$t = \log (\large\frac{ y}{y_0})$

$t=\large\frac{2\log (\Large\frac{y}{y_0})}{ \log(\Large\frac{11}{10})}$

Step 4:

Let the time when the number of bacteria increases from 1,00,000 to 2,00,000 be $t_1$

$y=2.y_0$ at $t=t_1$

Hence $t_1=\large\frac{2 \log (\large\frac{y}{y_0})}{\log (\large\frac{11}{10})} = \large\frac{2.\log 2}{\log(\large\frac{11}{10})}$

Hence in $\large\frac{2.\log 2}{\log(\large\frac{11}{10})}$ the number of bacteria increases from 1,00,000 to 2,00,000

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...