In a culture, the 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?

Asked by Sakshi | 1 year ago |  145

##### Solution :-

From the question it is given that,

The bacteria count is 1,00,000

The number is increased by 10% in 2 hours.

So, let us assume C be the surface area of balloon,

$$( \dfrac{dC}{dt})$$ ∝ C

Then,

$$\dfrac{dC}{dt}$$ = λC

$$\dfrac{dC}{dt}$$ = λ dt

Integrating on both side we get,

$$\dfrac{dC}{dt}$$ = λ ∫dt

Log C = λt + log k … [equation (i)]

From question, t = 0 when c = 1,00,000,

log (1,00,000) = λ × 0 + log k … [equation (ii)]

log (1,00,000) = log k … [equation (iii)]

And also form question, given t = 2,

c = 1,00,000 + 1,00,000 × ($$\dfrac{10}{100}$$)

= 110000

So, from equation (i) we have,

log 110000 = λ × 2 + log K … [equation (iv)]

Now, subtracting equation (ii) from equation (iv), we have,

Log 110000 – log 100000 = 2 λ

Then, log 11 × 10000 – log 10 × 10000 = 2 λ

Log $$\dfrac{ ((11 × 10000)}{(10 × 10000))}$$ = 2λ

Log $$\dfrac{11}{10}$$ = 2λ

So, λ = $$\dfrac{1}{2}$$log ($$\dfrac{11}{10}$$) … [equation (v)]

Now we need to find the time ‘t’ in which the count reaches 200000.

Then, substituting the values of λ and k from equation (iii) and (v) in equation (i),

Log 200000 = $$\dfrac{1}{2}$$log ($$\dfrac{11}{10}$$)t + log 100000

$$\dfrac{1}{2}$$log ($$\dfrac{11}{10}$$)t = log 200000 – log 100000

$$\dfrac{1}{2}$$log ($$\dfrac{11}{10}$$)t = log$$\dfrac{ 200000}{100000}$$

$$\dfrac{1}{2}$$log($$\dfrac{11}{10}$$)t = log 2

Therefore, the required time t = $$\dfrac{ 2log2}{log}(\dfrac{11}{10}) hours$$

Answered by Aaryan | 1 year ago

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