Question

If no of 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 bacteria is proportional to the number present?

Answer 1

Answer:

4 hours and 20 persent ..........

Answer 2

Let the number of bacteria at time t be y

Given that rate growth of bacteria is proportional to the number present

dy/dt ∝ y

dy/dt = ky

dy/y = kdt

Integrating sides,

∫dy/y = k∫dt

⇒ logy=kt+C ---(1)

Now according to question,

The bacteria court is 1,00,000. The number is increased by 10% in 2 hours, in how many hours will the count reach 2,00,000.

Putting t=0 and y=1,00,000 in ( 1 )

log1,00,000=k×0+C

C=log1,00,000

Putting value of C in ( 1 )

logy=kt+C

⇒ logy=kt+log1,00,000 ---(2)

Now,

Putting t=2 and y=1,10,000 in ( 2 )

log1,10,000=2k+log1,00,000

log1,10,000−log1,00,000=2k

log(1,10,000/1,00,000)=2k

½log(11/10)=k

Putting value of k in ( 2 )

logy=kt+log1,00,000

⇒ logy= ½ log (11/10)t+log1,00,000 ---(3)

Now, if bacterial =2,00,000 in ( 3 )

log2,00,000= ½ log(11/10)t+log(1,00,000)

log2,00,000−log1,00,000= ½log(11/10)t

log(2,00,000/1,00,000)½log(11/10)t

log2= ½ log(11/10)t

⇒t= 2log2/log(11/10)