JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Differential Equations

The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time

t = 0

. The number of bacteria is increased by

20%

in 2 hours. If the population of bacteria is 2000 after

\frac{k}{l o g _{e} ( \frac{6}{5} )}

hours, then

(\frac{k}{l o g _{e} 2})^{2}

is equal to :

Answer: (A)

Solution

With

N

the bacteria count,

\frac{d N}{d t} = c N

integrates to

N = 1000 e^{c t}

.
A

20%

rise in

2

hours means

N (2) = 1200

, so

e^{2 c} = \frac{6}{5}

and

c = \frac{1}{2} lo g_{e} \frac{6}{5}

.
The count reaches

2000

when

e^{c t} = 2

, that is at

t = \frac{lo g _{e} 2}{c} = \frac{2 lo g _{e} 2}{lo g _{e} \frac{6}{5}}

Comparing this with

t = \frac{k}{lo g _{e} \frac{6}{5}}

gives

k = 2 lo g_{e} 2

, so

(\frac{k}{lo g _{e} 2})^{2} = 2^{2} = 4

The required value is

4

JEE Main · 2021 · 26 Feb 2021 · Shift 1 · Single correct option