Question

A bacterial colony grows via cell division where each mother bacterium independently produces two daughter cells in 20 minutes. If the concentration of bacteria is \( 10^4 \, \text{cm}^{-3} \), the colony becomes harmful. Starting from a colony with an initial concentration of 5 cm$^{-3}$, the time taken (in minutes) for the colony to become harmful is ......... (Round off to nearest integer)
 

Hint:

Exponential growth is common in bacterial colonies, and the doubling time can be used to calculate the time for a population to reach a certain threshold.

Answer 1

Correct Answer: 210 - 225

The growth of the bacterial colony follows an exponential pattern since each bacterium divides to form two daughter cells every 20 minutes. We aim to determine the time required for the concentration to increase from its initial value to a harmful level. Let's break this down:

Step 1: Understand Exponential Growth
Exponential growth can be expressed by the formula: \( N = N_0 \times 2^{t/T} \), where \( N \) is the final concentration, \( N_0 \) is the initial concentration, \( t \) is the time in minutes, and \( T \) is the doubling time (20 minutes here).

Step 2: Define Parameters
Initial concentration, \( N_0 = 5 \, \text{cm}^{-3} \).
Final harmful concentration, \( N = 10^4 \, \text{cm}^{-3} \).
Doubling time, \( T = 20 \, \text{minutes} \).

Step 3: Formulate the Equation
\( 10^4 = 5 \times 2^{t/20} \)

Step 4: Solve for \( t \)
First, divide both sides by 5:
\( 2^{t/20} = \frac{10^4}{5} = 2000 \).
Take the logarithm on both sides:
\( \log_{10}(2^{t/20}) = \log_{10}(2000) \).
Using the logarithmic identity, \( \log_{10}(a^b) = b \log_{10}(a) \):
\( \frac{t}{20} \log_{10}(2) = \log_{10}(2000) \).
Solve for \( t \):
\( t = 20 \frac{\log_{10}(2000)}{\log_{10}(2)} \approx 20 \times \frac{3.3010}{0.3010} \approx 219.09 \).

Step 5: Round to Nearest Integer
Rounding 219.09 gives us 219 minutes.