#### Question

The rate of growth of bacteria is proportional to the number present. If, initially, there were

1000 bacteria and the number doubles in one hour, find the number of bacteria after 2½

hours.

[Take `sqrt2` = 1.414]

#### Solution

Let ‘N’ be the number of bacteria at time’t ’

`therefore "dN"/dt ooN`

`"dN"/dt=kN`

`"dN"/N=kdt`

Integrating on both sides, we get

`int"dN"/N=kintdt`

`logN=kt+c`

`when t=0,N=1000`

`c=log1000`

`logN=kt+log1000`

`log(N/1000)=kt`

`N=1000e^(kt)` ..........(1)

when t = 1 hour, N = 2000

`e^k=2`

`N=1000xx(2)^t` ....................from (1)

when `t=2 1/2` hours, we get

`N=1000xx(2)^(5/2)`

`=1000xx4xxsqrt2=4000xx1.414`

`N=5656`

Number of bacteria present after `2 1/2` hours is 5656

Is there an error in this question or solution?

#### APPEARS IN

Solution The rate of growth of bacteria is proportional to the number present. If, initially, there were 1000 bacteria and the number doubles in one hour, find the number of bacteria after 2½ hours. Concept: Rate of Change of Bodies Or Quantities.