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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