InterviewSolution
Saved Bookmarks
| 1. |
In aculture, the bacteria count is 1,00,000. The number is increased by 10% in 2 hours.In how many hours will the count reach 2,00,000, if the rate of growth of bacteriais proportional to the number present? |
|
Answer» Let at time `t`, the number of bacteria be `y` then `(dy)/(dt)propyimplies (dy)/(dt)=ky`, where `k` is proportional constant. `implies (dy)/(y)=k dt` `impliesint(dy)/(y)=intk dt` `implies log|y|=kt+C` .........`(1)` Initially at `t=0`, `y=100000` `:. log100000=C` ........`(2)` when `t=2`, `y=110000` `log110000=2k+C` ..........`(3)` `:.` Subtracting equation `(2)` from equation `(3)`, `log110000-log100000=2k` `implies log"(110000)/(100000)=2k` `implies k=(1)/(2)log((11)/(10))` put the value of `k` and `C` in equation `(1)`, `logy=(1)/(2)log((11)/(10))t+log100000` when `y=200000` then `log200000=(1)/(2)((11)/(10))t+log100000` `implies log((200000)/(100000))=(1)/(2)log((11)/(10))t` `implies 2log(2)=log((11)/(10))t` `implies t=(2log2)/(log((11)/(10)))` Therefore, the number of bacteria will increase from `100000` to `200000` in `(2log2)/(log((11)/(10)))` time. |
|