Question

In a culture, the bacteria count is 1,00,000 initially. The number increases by 10% in the 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. \( \frac{2 \log 11}{\log 2} \)
• B. \( \frac{\log(11/10)}{\log 2} \)
• C. \( \frac{2 \log 2}{\log 11} \)
• D. \( \frac{\log 2}{\log(11/10)} \)

Hint:

When solving problems involving exponential growth, use the formula \( N(t) = N_0 e^{kt} \) and solve for the unknowns step by step, taking logarithms where necessary.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The rate of growth \( \frac{dN}{dt} \) is proportional to the current population \( N \). This is a standard first-order differential equation model for growth.
Step 2: Key Formula or Approach:
\[ \frac{dN}{dt} = kN \Rightarrow N(t) = N_0 e^{kt} \]
Step 3: Detailed Explanation:
Initially, \( N_0 = 1,00,000 \).
At \( t = 2 \), the count increases by \( 10% \): \( N(2) = 1.1 N_0 \).
\[ 1.1 N_0 = N_0 e^{2k} \Rightarrow 1.1 = e^{2k} \Rightarrow 2k = \log(1.1) = \log\left(\frac{11}{10}\right) \Rightarrow k = \frac{1}{2} \log\left(\frac{11}{10}\right) \]
We want to find \( t \) when \( N(t) = 2,00,000 = 2 N_0 \):
\[ 2 N_0 = N_0 e^{kt} \Rightarrow 2 = e^{kt} \]
Taking log:
\[ \log 2 = kt \]
\[ t = \frac{\log 2}{k} = \frac{\log 2}{\frac{1}{2} \log(11/10)} = \frac{2 \log 2}{\log(11/10)} \]
Step 4: Final Answer:
The required time is \( \frac{2 \log 2}{\log(11/10)} \) hours.