Answer :

Let the count of bacteria be C at any time t.

According to question,

where k is a constant

Integrating both sides, we have

= k∫dt

log|C| = kt + a……(1)

Given, we have C = C0 when t = 0 sec

Putting the value in equation (1)

log|C| = kt + a

log| C0| = 0 + a

a = log| C0| ……(2)

Putting the value of a in equation (1) we have,

log|C| = kt + log|100000|

log|C| – log| C0| = k t []

log ( = kt ……(3)

Also, at t = 5 years, C = 3C0

From equation(3),we have

kt = log (

k×5 = log (

k = ……(4)

Now, equation (3) becomes,

Now, let C1 be the number of bacteria present in 10 hours, as

Let the time be t1 for bacteria to be 10 times

The time required for = hours

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