Question:
In a certain culture of bacteria, the rate of increase is proportional to the number present.
It is found that the number doubles in 4 hours.
Then the number of times the bacteria are increased in 12 hours is
In a certain culture of bacteria, the rate of increase is proportional to the number present.
It is found that the number doubles in 4 hours.
Then the number of times the bacteria are increased in 12 hours is
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In exponential growth problems, if the quantity doubles in time \(T\), then in \(nT\) it becomes \(2^n\) times.
Updated On: Mar 28, 2026
- \( 6 \)
- \( 8 \)
- \( 12 \)
- \( 4 \)
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The Correct Option is B
Solution and Explanation
Step 1: Write the law of growth.
Since the rate of increase is proportional to the number present, \[ \frac{dN}{dt} = kN \] whose solution is \[ N = N_0 e^{kt} \]
Step 2: Use the doubling condition.
Given that the population doubles in 4 hours, \[ 2N_0 = N_0 e^{4k} \Rightarrow e^{4k} = 2 \]
Step 3: Find the increase in 12 hours.
\[ N = N_0 e^{12k} = N_0 (e^{4k})^3 = N_0 (2)^3 \] \[ N = 8N_0 \]
Step 4: Conclusion.
The number of bacteria increases \(\boxed{8}\) times in 12 hours.
Since the rate of increase is proportional to the number present, \[ \frac{dN}{dt} = kN \] whose solution is \[ N = N_0 e^{kt} \]
Step 2: Use the doubling condition.
Given that the population doubles in 4 hours, \[ 2N_0 = N_0 e^{4k} \Rightarrow e^{4k} = 2 \]
Step 3: Find the increase in 12 hours.
\[ N = N_0 e^{12k} = N_0 (e^{4k})^3 = N_0 (2)^3 \] \[ N = 8N_0 \]
Step 4: Conclusion.
The number of bacteria increases \(\boxed{8}\) times in 12 hours.
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