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?
Answers
Answer:
4 hours and 20 persent ..........
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)