Question

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?

• A. \(\frac{2\mathrm{log}(\frac{11}{10})}{\mathrm{log}2}\)
• B. \(\frac{\mathrm{log}(\frac{11}{10})}{\mathrm{log}2}\)
• C. \(\frac{2\mathrm{log}2}{\mathrm{log}(\frac{11}{10})}\)
• D. \(\frac{\mathrm{log}(2)}{\mathrm{log}(\frac{11}{10})}\)

Hint:

Use dN/dt = kN to get N = N0 e^(kt). Find k from the 10 percent rise in 2 hours, then the doubling time.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Growth proportional to the number present gives \(\dfrac{dN}{dt}=kN\), so \(N=N_0e^{kt}\).

Step 2: Find k
After 2 hours, \(N=1.1N_0\), so \(e^{2k}=\dfrac{11}{10}\Rightarrow k=\dfrac{1}{2}\log\dfrac{11}{10}\).

Step 3: Doubling Time
\(N=2N_0\Rightarrow e^{kt}=2\Rightarrow t=\dfrac{\log2}{k}\)

Step 4: Result
\(t=\dfrac{2\log2}{\log\left(\frac{11}{10}\right)}\) hours, the third option.

Final Answer:
The count doubles after 2 log 2 divided by log (11/10) hours. \[ \boxed{\text{(C) }\dfrac{2\log 2}{\log\left(\dfrac{11}{10}\right)}} \]