VITEEE 2021 Paper

Let y denote the number of bacteria at any instant t . then according to the question
dy‌dt‌α‌y⇒

dy

y

=k‌dt ... (i)
k is the constant of proportionality, taken to be + ve on integrating (i), we get
log‌y=kt+c ...(ii)
c is a parameter.
let y0 be the initial number of bacteria i.e., at t=0 using this in (ii),
c=log‌y0
⇒ log‌y=kt+log‌y0
⇒log‌

y

y0

=kt ... (iii)
y=(y0+

10

100

y0)=

11y0

10

,
when t=2, So, from (iii), we get
log‌

11y0 10

y0

=k (2)
⇒ k=

1

2

‌log‌

11

10

...(iv)
Using (iv) in (iii)
log‌

y

y0

=

1

2

(log‌

11

10

)t ... (v)
let the number of bacteria become 1,00,000 to 2,00,000 in t1 hours. i.e., y=2y0
when t=t1 hours, from (v)
log‌

2y0

y0

=

1

2

(log‌

11

10

)t1⇒t1=

2‌log‌2

log‌ 11 10

Hence, the reqd. no. of hours =

2‌log‌2

log‌ 11 10