Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • 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 . The number of bacteria is in

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}{log_{e}\left ( \frac{6}{5} \right )} hours, then \left ( \frac{k}{log_{e}2} \right )^{2} is equal to :
 
Option: 1 2
Option: 2 4
Option: 3 8
Option: 4 16

Answers (1)

best_answer

Initial bacteria count = 1000

20% bacteria increased in 2 hours = 1200

\\\frac{\mathrm{dB}}{\mathrm{dt}}=\lambda \mathrm{B}\\ \Rightarrow \int_{1000}^{1200} \frac{\mathrm{dB}}{\mathrm{B}}=\lambda \int_{0}^{2} \mathrm{dt} \\\Rightarrow \lambda=\frac{1}{2} \ell \mathrm{n}\left(\frac{6}{5}\right)

\\\int_{1000}^{2000} \frac{\mathrm{dB}}{\mathrm{B}}=\frac{1}{2} \ln \left(\frac{6}{5}\right) \int_{0}^{\mathrm{T}} \mathrm{dt}\\ \Rightarrow \mathrm{T}=\frac{2 \ell \mathrm{n} 2}{\ln \left(\frac{6}{5}\right)} \\ \Rightarrow \mathrm{k}=2 \ln 2

\\\left(\frac{\mathrm{k}}{\log _{\mathrm{e}} 2}\right)^{2}=\left(\frac{\mathrm{2\ln 2}}{\log _{\mathrm{e}} 2}\right)^{2}=4\\\ln 2a\text{ and }\log_e 2\;\text{is same thing}

Posted by

himanshu.meshram

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

BTech aspirant from north-east or UT? Apply for CSAB NEUT counselling 2025 today
anam-khan

JoSAA round 2 cut-off 2025 out; BTech closing ranks for top engineering courses at IITs
anam-khan

Goa Engineering Admission 2025: 1,415 eligible for BE seats; merit list on June 27

Latest Question