A person’s wound was exposed to some bacteria and then bacterial growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay (r) was found to be proportional with the square of t

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Solution

Q.
A person’s wound was exposed to some bacteria and then bacterial growth started to happen at the same place. The wound was later treated with some antibacterial medicine and the rate of bacterial decay (r) was found to be proportional with the square of the existing number of bacteria at any instance. Which of the following set of graphs correctly represents the ‘before’ and ‘after’ situation of the application of the medicine

Given: N = Number of bacteria, t = time, bacterial growth follows first order kinetics. [2025]

1

2

3

4

Ans.
(2)

Before applying medicine:

Number of bacteria at any instant t = N

Rate of increase of bacteria = $\frac{d N}{d t}$

As bacterial growth is first order reaction:

$\frac{d N}{d t} = k N$

$\frac{d N}{N} = k dt$

Integrating both sides:

$\int_{N_{0}}^{N} \frac{d N}{N} = \int_{0}^{t} k dt$

${[\ln N]}_{N_{0}}^{N} = k {[t]}_{0}^{t}$

$\ln \frac{N}{N_{0}} = k t$

$\frac{N}{N_{0}} = e^{k t}$

After applying medicine:

Rate of decrease of bacteria $(r) = - \frac{d N}{d t}$
As per given data:

$r = - \frac{d N}{d t} = k {[N]}^{2}$

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