A) 8 times the original
B) 16 times the original
C) 32 times the original
D) 64 times the original
Correct Answer: C
Solution :
| [c] Let N be the no. of bacteria at time t. |
| Let \[{{N}_{0}}\] be the initial original no. of bacteria. |
| Then, \[\frac{d}{dt}N\infty N\frac{d}{dt}N=kN\Rightarrow \frac{dN}{N}=kdt\] |
| \[\Rightarrow \int{\frac{dN}{N}=k}\int{dt\Rightarrow \log N=kt+c}\] |
| At \[t=0,N={{N}_{0}}\Rightarrow \log {{N}_{0}}=0+c\Rightarrow c=\log {{N}_{0}}\] |
| \[\therefore \log N=kt+\log {{N}_{0}}\Rightarrow \log \frac{N}{{{N}_{0}}}=kt\] |
| When t = 5 hrs. \[N=2{{N}_{0}}\therefore \log \left( \frac{2{{N}_{0}}}{{{N}_{0}}} \right)=5k\] |
| \[\Rightarrow k=\frac{\log 2}{5}\therefore \log \frac{N}{{{N}_{0}}}=\frac{\log 2}{5}t\] |
| When \[t=25hrs,\log \frac{N}{{{N}_{0}}}=\frac{\log 2}{5}\times 25\] |
| \[\Rightarrow \log \frac{N}{{{N}_{0}}}=5\log 2\log \frac{N}{{{N}_{0}}}=\log {{2}^{5}}\] |
| \[\Rightarrow \frac{N}{{{N}_{0}}}=32\therefore N=32{{N}_{0}}\]. |
| Therefore, 32 times the original. |
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