MHT CET · Maths · Differential Equations
If the population grows at the rate \(5 \%\) peryear, then the time taken for the population
to become double is \(\quad\) (Given \(\log 2=0.6912\) )
- A \(13.8275\) years
- B \(13.624\) years
- C 13.725 years
- D \(13.8240\) years
Answer & Solution
Correct Answer
(D) \(13.8240\) years
Step-by-step Solution
Detailed explanation
Let initial population \(=P_{0}\) and let \(P\) be the population at time \(t\) We have \(\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{5 \mathrm{P}}{100}\)
\(\therefore \frac{d P}{d t}=\frac{P}{20} \Rightarrow \int \frac{d P}{P}=\int \frac{1}{20} d t\)
\(\therefore \log P=\frac{1}{20} t+C\)....(1)
When \(t=0, P=P_{0}\)
\(\therefore \log \mathrm{P}_{0}=0+\mathrm{C} \Rightarrow \mathrm{C}=\log \mathrm{P}_{0}\)
From (1), \(\log \left(\frac{P}{P_{0}}\right)=\frac{t}{20}\)....(2)
When \(P=2 P_{0}\), we get
\(\log 2=\frac{t}{20} \Rightarrow t=20(\log 2)=20 \times 0.6912=13.8240\) years
\(\therefore \frac{d P}{d t}=\frac{P}{20} \Rightarrow \int \frac{d P}{P}=\int \frac{1}{20} d t\)
\(\therefore \log P=\frac{1}{20} t+C\)....(1)
When \(t=0, P=P_{0}\)
\(\therefore \log \mathrm{P}_{0}=0+\mathrm{C} \Rightarrow \mathrm{C}=\log \mathrm{P}_{0}\)
From (1), \(\log \left(\frac{P}{P_{0}}\right)=\frac{t}{20}\)....(2)
When \(P=2 P_{0}\), we get
\(\log 2=\frac{t}{20} \Rightarrow t=20(\log 2)=20 \times 0.6912=13.8240\) years
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