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The rate of growth of bacteria is proportional to the bacteria present. If it is found
that the number doubles in 3 hours, then the number of times the bacteria are
increased in 6 hours is
Options:
that the number doubles in 3 hours, then the number of times the bacteria are
increased in 6 hours is
Solution:
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Verified Answer
The correct answer is:
4 times the original
Let $b$ be the number of bacteria.
We have $\frac{\mathrm{db}}{\mathrm{dt}} \propto \mathrm{b} \Rightarrow \int \frac{\mathrm{db}}{\mathrm{b}}=\int \mathrm{Kdt}$
$\therefore \log \mathrm{b}=\mathrm{Kt}+\mathrm{c}$ ...(1)
Let $b_{0}$ be the initial number of bacteria. At $t=0, b=b_{0}$
$\log \mathrm{b}_{0}=\mathrm{K}(0)+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{b}_{0}$
$\therefore \log \left(\frac{b}{b_{0}}\right)=\mathrm{Kt}$ ...(2)
When $t=3, b=2 b_{0}$
$\therefore \log \left(\frac{2 b_{0}}{b_{0}}\right)=3 K \Rightarrow K=\frac{1}{3}(\log 2)$
Thus $\log b=\frac{1}{3}(\log 2) t+\log b_{0}$
When $\mathrm{t}=6$
$\log \left(\frac{b}{b_{0}}\right)=2 \log 2=\log 4 \Rightarrow \frac{b}{b_{0}}=4 \Rightarrow b=4 b_{0}$
This problem can also be solved as follows :
The number of bacteria doubles in 3 hours.
Let initial number of bacteria $=\mathrm{N}$.
$\therefore$ After 3 hours, number of bacteria $=2 \mathrm{~N}$.
After 6 hours, number of bacteria $=4 \mathrm{~N}$.
We have $\frac{\mathrm{db}}{\mathrm{dt}} \propto \mathrm{b} \Rightarrow \int \frac{\mathrm{db}}{\mathrm{b}}=\int \mathrm{Kdt}$
$\therefore \log \mathrm{b}=\mathrm{Kt}+\mathrm{c}$ ...(1)
Let $b_{0}$ be the initial number of bacteria. At $t=0, b=b_{0}$
$\log \mathrm{b}_{0}=\mathrm{K}(0)+\mathrm{c} \Rightarrow \mathrm{c}=\log \mathrm{b}_{0}$
$\therefore \log \left(\frac{b}{b_{0}}\right)=\mathrm{Kt}$ ...(2)
When $t=3, b=2 b_{0}$
$\therefore \log \left(\frac{2 b_{0}}{b_{0}}\right)=3 K \Rightarrow K=\frac{1}{3}(\log 2)$
Thus $\log b=\frac{1}{3}(\log 2) t+\log b_{0}$
When $\mathrm{t}=6$
$\log \left(\frac{b}{b_{0}}\right)=2 \log 2=\log 4 \Rightarrow \frac{b}{b_{0}}=4 \Rightarrow b=4 b_{0}$
This problem can also be solved as follows :
The number of bacteria doubles in 3 hours.
Let initial number of bacteria $=\mathrm{N}$.
$\therefore$ After 3 hours, number of bacteria $=2 \mathrm{~N}$.
After 6 hours, number of bacteria $=4 \mathrm{~N}$.
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