JEE Mains · Maths · STD 12 - 9. differential equations
The rate of growth of bacteria in a culture is proportional to the number of bacteris 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 \(\left(\frac{ k }{\log _{ c } 2}\right)^{2}\) is equal to
- A \(4\)
- B \(8\)
- C \(2\)
- D \(16\)
Answer & Solution
Correct Answer
(A) \(4\)
Step-by-step Solution
Detailed explanation
\(\frac{ dB }{ dt }=\lambda B \Rightarrow \int_{1000}^{1200} \frac{ dB }{ B }=\lambda \int_{0}^{2} dt \Rightarrow \lambda=\frac{1}{2} \ell n \left(\frac{6}{5}\right)\)…
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