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  • Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 4 maths

Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 4 maths

Answers (1)

Answer: \frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \text { hours }

Given:  Present bacteria count=1,00,000

To find: The amount of hours taken for the count to reach 2,00,000

Hint: Use \frac{d C}{d t}=\lambda C and then find the value of lambda using integration.

Solution: The bacteria count is 1,00,000 and rate of increase =10% in 2 hours

Let C be the bacteria count at any time t,

Then \frac{d C}{d t} \propto C

        =>\frac{d C}{d t}=\lambda C           [Where \lambda is the constant]

Integrating on both sides

        \begin{aligned} &=>\int \frac{d C}{d t}=\int \lambda C \\\\ &=>\int \frac{d C}{C}=\lambda \int d t \\\\ &=>\log C=\lambda t+\log k \ldots(i) \end{aligned}

Where \log \log k  is integral constant

At t=0 we have C=1,00,000

Then \log (1,00,000)=\lambda \times 0+\log k         [putting t=0 and C=1,00,000 in equation (i)]

        =\log (1,00,000)=\log k \ldots(i i)

\begin{aligned} &\text { At } t=2 \\ &\qquad \begin{aligned} C &=1,00,000+1,00,000\left(\frac{10}{100}\right) \\ &=1,10,000 \end{aligned} \end{aligned}

From equation (i) we have,

        \begin{aligned} &=\log 1,10,000=\lambda \times 2+\log k \quad \text { [Putting } C=1,10,000 \text { and } t=2] \\\\ &=\log 1,10,000=2 \lambda+\log k \ldots(i i i) \end{aligned}

Now, subtracting equation (ii) from (iii) we have

        \begin{aligned} &=\log 1,10,000-\log 1,00,000=2 \lambda+\log k-\log k \\\\ &=\log \left(\frac{1,10,000}{1,00,000}\right)=2 \lambda \\\\ &=\log \frac{11}{10}=2 \lambda \\\\ &=\lambda=\frac{1}{2} \log \left(\frac{11}{10}\right) \end{aligned}

Now, we have to find time t in which the count reaches 2,00,000

        \begin{aligned} &\text { Now } C=2,00,000 \text { in equation }(i)\\\\ &=\log (2,00,000)=\lambda t+\log k \end{aligned}

Putting value of lambda and  \log k, we get

        \begin{aligned} &=\log (2,00,000)=\frac{1}{2} \log \left(\frac{11}{10}\right) t+\log 1,00,000 \\\\ &=\frac{1}{2} \log \left(\frac{11}{10}\right) t=\log 2,00,000-\log 1,00,000 \end{aligned}

        \begin{aligned} &=\frac{1}{2} \log \left(\frac{11}{10}\right) t=\log \left(\frac{2,00,000}{1,00,000}\right) \\\\ &=\frac{1}{2} \log \left(\frac{11}{10}\right) t=\log 2 \\\\ &=t=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)} \end{aligned}

Hence, the time required =\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}  hours

Posted by

infoexpert26

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