Bacteria increases at the rate proportional to the number of bacteria present. If the original number N doubles in 4 hours, then the number of bacteria will be 4 ~N in

Step-by-step solution

Let N be the number of bacteria present at time t₀ . Let N ₀ be the initial number of bacteria. Here dN dt N dN dt = KN dN N = K dt dN N = K dt N = Kt + C When t =0, ~N = N ₀ N₀=C ( N N₀ )=K t When t=4, N=2 N₀ 4 K= 2 array l K = 1 4 2 ( N N ₀ )= t 4 2 array When N =4 ~N ₀ , we get 4= t 4 2 2( 2)= t 4 ( 2) t=8 hours This problem can also be saved as follows : Number of bacteria doubles in 4 hrs. If initial number of bacteria are N , then After 4 hours number of bacteria become 2 ~N . After 8 hours, number of bacteri