Chemical Kinetics - Result Question 4
4. A bacterial infection in an internal wound grows as $N^{\prime}(t)=N _0 \exp (t)$, where the time $t$ is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as $\dfrac{d N}{d t}=-5 N^{2}$. What will be the plot of $\dfrac{N _0}{N} v s t$ after 1 hour?
(2019 Main, 10 April I)
(a)
(c)
(b)
(d)
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Answer:
Correct Answer: 4. (a)
Solution:
- The expression for bacterial growth is
$ \begin{aligned} N & =N _0 e^{t} \\ \dfrac{N _0}{N} & =e^{-t} \end{aligned} $
From 0 to 1 hour $N^{\prime}(t)=N _0 e^{t}$
From 1 hour onwards, $\dfrac{d N}{d t}=-5 N^{2}$
On differentiating the above equation from $N^{\prime}$ to $N$ we get.
$ \begin{aligned} & \int _{e N _0}^{N} N^{-2} d N=-5 \int _1^{t} d t \quad\left[\because \text { At } 1 \text { hour, } N^{\prime}=e N _0\right] \\ & {\left[\dfrac{1}{N}-\dfrac{1}{e N _0}\right]=5(t-1)} \end{aligned} $
Multiply both sides by $N _0$, we get
$ \begin{aligned} \dfrac{N _0}{N}-\dfrac{1}{e} & =5 N _0(t-1) \text { or, } \dfrac{N _0}{N}=5 N _0(t-1)+\dfrac{1}{e} \\ \dfrac{N _0}{N} & =5 N _0 t+\left[\dfrac{1}{e}-5 N _0\right] \end{aligned} $
On comparing the above equation with equation of straight line, $y=m x+c$
We get $m=5 N _0, c=\dfrac{1}{e}-5 N _0$
$\therefore$ Plot of $\dfrac{N _0}{N} v _s t$ is shown aside.
BETA
