In a culture bacteria count is 1,00,000 initially. The number increases by 10 % in first 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?

Step-by-step solution

The bacterial population N(t) grows exponentially according to dN dt = kN , with solution N(t) = N₀ e^ kt where N₀ = 100 , 000 . In 2 hours, the population increases by 10%, so N(2) = 1.1N₀ = 110 , 000 . Using N(2) = N₀ e^ 2k , we obtain 1.1 = e^ 2k , giving k = (1.1) 2 . To find the time t when N(t) = 200 , 000 = 2N₀ , solve 2 = e^ kt , so t = (2) k . Substituting k yields t = 2 (2) (1.1) = 2 2 (11/10) , which corresponds to option C. Final answer: C