text: #### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 28 maths
  
  Given: Radium decomposes at a rate proportional to the quantity of radium present.
  
  To find: The time taken for half the amount of the original amount of radium to decompose.
  
  Hint: The quantity of radium decomposes at the rate of time t i.e. $\frac{d A}{d t} \propto A$  then solve using integration.
  
  Solution: Let A be the quantity of bacteria present in the culture at any time t and initial quantity of bacteria is A0 be the initial quantity of radium.
  
  \begin{aligned} &=\frac{d A}{d t} \propto A \\\\ &=\frac{d A}{d t}=-\lambda A \end{aligned}        [λ is proportional constant]
  
  $=\frac{d A}{A}=-\lambda d t$
  
  Integrating on both sides
  
  $=\int \frac{d A}{A}=-\lambda \int \mathrm{dt}$
  
  $=\log A=-\lambda t+C \ldots(i)$
  
  Now, $A=A_{0}$ when $t=0$
  
  \begin{aligned} &=\log A_{0}=0+C \\\\ &=C=\log A_{0} \end{aligned}
  
  Substituting in equation (i)
  
  \begin{aligned} &=\log A=-\lambda t+\log A_{0} \\\\ &=\log A-\log A_{0}=-\lambda t \\\\ &=\log \frac{A}{A_{0}}=-\lambda t \ldots(i i) \end{aligned}
  
  Given that in 25 years radium decomposes at 1.1%
  
  So,
  
  $\begin{gathered} A=(100-1.1) \%=98.9 \% \\ A=0.989 A_{0}, t=25 \end{gathered}$
  
  Substituting in equation (ii)
  
  \begin{aligned} &=\log \frac{0.989 A_{0}}{A_{0}}=-25 \lambda \\\\ &=\lambda=-\frac{1}{25} \log (0.989) \end{aligned}
  
  Now equation (ii) becomes
  
  $=\log \frac{A}{A_{0}}=\left[-\frac{1}{25} \log (0.989)\right] t$
  
  Now    $A=\frac{1}{2} A_{0}$
  
  \begin{aligned} &=\log \left(\frac{A}{2 A}\right)=-\frac{1}{25} \log (0.989) t \\\\ &=\log \left(\frac{1}{2}\right)=-\frac{1}{25} \log (0.989) t \end{aligned}
  
  \begin{aligned} &=\log \left(2^{-1}\right)=-\frac{1}{25} \log (0.989) t \\\\ &=-1 \log (2)=-\frac{1}{25} \log (0.989) t \end{aligned}
  
  $=t=\frac{\log 2 \times 25}{\log (0.989)}$
  
  $=t=\frac{0.6931 \times 25}{0.01106} \quad[\log 2=0.6931 \text { And } \log (0.989)=0.01106]$
  
  $=t=1567$
  
  Hence, time taken for radium to decay to half the original amount is 1567 years.
explanation: The document provides a detailed solution to a math problem from RD Sharma class 12, chapter Differential Equations, exercise 21.11, question 28. The problem involves finding the time taken for half the original amount of radium to decompose, given that the rate of decomposition is proportional to the quantity of radium present. The solution includes the setup of the differential equation, integration, and substitution of given values to find the final answer.
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distilabel_metadata.content: #### Explain solution RD Sharma class 12 chapter Differential Equations exercise 21.11 question 28 maths
  
  Given: Radium decomposes at a rate proportional to the quantity of radium present.
  
  To find: The time taken for half the amount of the original amount of radium to decompose.
  
  Hint: The quantity of radium decomposes at the rate of time t i.e. $\frac{d A}{d t} \propto A$  then solve using integration.
  
  Solution: Let A be the quantity of bacteria present in the culture at any time t and initial quantity of bacteria is A0 be the initial quantity of radium.
  
  \begin{aligned} &=\frac{d A}{d t} \propto A \\\\ &=\frac{d A}{d t}=-\lambda A \end{aligned}        [λ is proportional constant]
  
  $=\frac{d A}{A}=-\lambda d t$
  
  Integrating on both sides
  
  $=\int \frac{d A}{A}=-\lambda \int \mathrm{dt}$
  
  $=\log A=-\lambda t+C \ldots(i)$
  
  Now, $A=A_{0}$ when $t=0$
  
  \begin{aligned} &=\log A_{0}=0+C \\\\ &=C=\log A_{0} \end{aligned}
  
  Substituting in equation (i)
  
  \begin{aligned} &=\log A=-\lambda t+\log A_{0} \\\\ &=\log A-\log A_{0}=-\lambda t \\\\ &=\log \frac{A}{A_{0}}=-\lambda t \ldots(i i) \end{aligned}
  
  Given that in 25 years radium decomposes at 1.1%
  
  So,
  
  $\begin{gathered} A=(100-1.1) \%=98.9 \% \\ A=0.989 A_{0}, t=25 \end{gathered}$
  
  Substituting in equation (ii)
  
  \begin{aligned} &=\log \frac{0.989 A_{0}}{A_{0}}=-25 \lambda \\\\ &=\lambda=-\frac{1}{25} \log (0.989) \end{aligned}
  
  Now equation (ii) becomes
  
  $=\log \frac{A}{A_{0}}=\left[-\frac{1}{25} \log (0.989)\right] t$
  
  Now    $A=\frac{1}{2} A_{0}$
  
  \begin{aligned} &=\log \left(\frac{A}{2 A}\right)=-\frac{1}{25} \log (0.989) t \\\\ &=\log \left(\frac{1}{2}\right)=-\frac{1}{25} \log (0.989) t \end{aligned}
  
  \begin{aligned} &=\log \left(2^{-1}\right)=-\frac{1}{25} \log (0.989) t \\\\ &=-1 \log (2)=-\frac{1}{25} \log (0.989) t \end{aligned}
  
  $=t=\frac{\log 2 \times 25}{\log (0.989)}$
  
  $=t=\frac{0.6931 \times 25}{0.01106} \quad[\log 2=0.6931 \text { And } \log (0.989)=0.01106]$
  
  $=t=1567$
  
  Hence, time taken for radium to decay to half the original amount is 1567 years.
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distilabel_metadata.raw_output_q_a_detector_0: The document provides a detailed solution to a math problem from RD Sharma class 12, chapter Differential Equations, exercise 21.11, question 28. The problem involves finding the time taken for half the original amount of radium to decompose, given that the rate of decomposition is proportional to the quantity of radium present. The solution includes the setup of the differential equation, integration, and substitution of given values to find the final answer.
  
  Label: question-and-answer