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  • Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 6

Provide solution for RD Sharma maths class 12 chapter Differential Equations exercise 21.11 question 6

Answers (1)

Answer: 9 times, \frac{5 \log 10}{\log 3}

Given:  The amount of bacteria triples in 5 hours.

To find: The amount of bacteria present after 10 hours and also the time for the number of bacteria to become 10 times.

Hint: Use compound interest i.e. \frac{d P}{d t}=\frac{p r}{100}

Solution: Let A be the amount of bacteria present at time t and A0 be the initial amount of bacteria

\text { Then, } \begin{aligned} \frac{d A}{d t} & \propto A \\ &=\frac{d A}{d t}=\lambda A \end{aligned}

Where lambda is the integrating constant

Integrating on both sides we get,

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

Where C is integral constant

When t=0, we have A=A0

        \begin{aligned} &\left.=>\log A_{0}=0+C \quad \text { [Putting } t=0 \text { and } A=A_{0} \text { in equation }(i)\right]\\\\ &=>C=\log A_{0} \end{aligned}

Putting the value of C in equation (i)

        \begin{aligned} &=>\log A=\lambda t+\log A_{0} \\\\\ &=>\log A-\log A_{0}=\lambda t \\\\ &=>\log \left(\frac{A}{A_{0}}\right)=\lambda t \ldots(i i) \end{aligned}

Since bacteria triples in 5 hours,

\text { So, } A=3 A_{0} \text { and } t=5

Putting in equation (i) we get,

        \begin{aligned} &=\log \left(\frac{4 A_{0}}{A_{0}}\right)=5 \lambda \\\\ &=\log 3=5 \lambda \\\\ &=\lambda=\frac{\log 3}{5} \end{aligned}

Putting \lambda=\frac{\log 3}{5}  in equation (ii) we get,

        =\log \left(\frac{A}{A_{0}}\right)=\frac{\log 3}{5} t \ldots(i i i)

Now let A1 be the amount of bacteria present after 10 hours,

\text { Then, } \log \left(\frac{A_{1}}{A_{0}}\right)=\frac{\log 3}{5} \times 10

        \begin{aligned} &=\log \left(\frac{A_{1}}{A_{0}}\right)=2 \log 3 \\\\ &=\log \left(\frac{A_{1}}{A_{0}}\right)=2 \times 1.0986 \quad[\log \log 3=1.0986] \end{aligned}

        \begin{aligned} &=\log \left(\frac{A_{1}}{A_{0}}\right)=2.1972 \\\\ &=\frac{A_{1}}{A_{0}}=e^{2.1972} \\\\ &=A_{1}=A_{0} e^{2.1972} \\\\ &=A_{1}=9 A_{0} \quad\left[e^{2.1972}=9\right] \end{aligned}

Hence, after 10 hours the number of bacteria will be 9 times the original.

Now, let t1 be the time necessary for the bacteria to be 10 times the amount. So, A=10A0

Then equation (iii) is

        \begin{aligned} &=\log \left(\frac{A}{A_{0}}\right)=\frac{\log 3}{5} \times t \\\\ &=\log \left(\frac{10 A_{0}}{A_{0}}\right)=\frac{\log 3}{5} \times t_{1} \\\\ &=\log \log 10=\frac{\log 3}{5} \times t_{1} \end{aligned}

        \begin{aligned} &=5 \log 10=\log 3 \times t_{1} \\\\ &=t_{1}=\frac{5 \log 10}{\log 3} \end{aligned}

Hence, the time needed for the bacteria to become 10 times the initial amount =\frac{5 \log 10}{\log 3} hours

 

Posted by

infoexpert26

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