The rate of growth of bacteria is proportional to the bacteria present. If it is found that the number…
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
6 times the original
4 times the original
8 times the original
5 times the original
Solution
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 $=4 \mathrm{~N}$.