Discussion Forum

 In a culture the bacteria count is 1,00,000. the number is increased by 10% in 2 hours. In how many hours will be the count reach 2,00,000 if the rate of growth of bacteria is proportional to the number present

Answer:

Explanation:

 Let y denote the number of bacteria at any instant t - then according to the question

 $\frac{dy}{dt}\alpha y\Rightarrow \frac{dy}{y}=kdt$  .....(i)

  k is the constant of proportionality , taken  to be +ve integrating (i) , we get

log y=kt+c   ....... (ii)

c is a parameter , let y₀ be the initial number of bacteria

 i.e,  at t=0 using thus in (ii) , c = log y₀

$\Rightarrow \log y=kt+\log y_{0}$

$\Rightarrow \log\frac{y}{y_{0}}=kt$  ....(iii)

$y= \left(y_{0}+\frac{10}{100}y_{0}\right)=\frac{11y_{0}}{10},$, when t=2

 so , from (iii)   , we get   $\log \frac{\frac{11y_{0}}{10}}{y_{0}}=k(2)$

 $\Rightarrow k=\frac{1}{2}\log \frac{11}{10}$    .........(iv)

 Using (iv) in (iii)       $\log\frac{y}{y_{0}}=\frac{1}{2}\left(\log \frac{11}{10}\right)t$   .....(v)

 let the number of bacteria become 1,00,000  to 2,00,000 in t ₁  hours .i.e, y=2y₀ when t= t₁   hours , from (v)

$\log\frac{2y_{0}}{y_{0}}=\frac{1}{2}\left(\log \frac{11}{10}\right)t_{1}\Rightarrow t_{1}=\frac{2\log 2}{\log \frac{11}{10}}$

 Hence , the reqd. no of hours =  $\frac{2\log 2}{\log \frac{11}{10}}$