MHT CET · Maths · Application of Derivatives
The rate of disintegration of a radio active element at time \(t\) is proportional to its mass at that time. Then the time during which the original mass of \(6 \mathrm{gm}\) will disintegrate into its mass of \(3 \mathrm{gm}\) is proportional to
- A \(\log 4\)
- B \(\log 3\)
- C \(\log 5\)
- D \(\log 2\)
Answer & Solution
Correct Answer
(D) \(\log 2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{d(m)}{d t} \propto m \\ & \Rightarrow \frac{d m}{d t}=-k m \\ & \int \frac{d m}{m}=-\int k d t \\ & \Rightarrow \log _e m=-k t+c \\ & \Rightarrow m=e^{-k t+c}=e^c \cdot e^{-k t} \\ & \text { at } t=0, m=6 \Rightarrow e^c=6 \\ & \text { i.e., } m=6 \cdot e^{-k t}\end{aligned}\)
at \(t=t, m=3\)
Hence \(3=6 \cdot e^{-k t}\)
\(\begin{aligned} & \Rightarrow \frac{1}{2}=e^{-k t} \\ & \Rightarrow \log \left(\frac{1}{2}\right)=-k t \\ & \Rightarrow-\log 2=-k t \\ & \left.\Rightarrow t=\frac{\log 2}{k} \text { [Propotional to } \log 2\right]\end{aligned}\)
at \(t=t, m=3\)
Hence \(3=6 \cdot e^{-k t}\)
\(\begin{aligned} & \Rightarrow \frac{1}{2}=e^{-k t} \\ & \Rightarrow \log \left(\frac{1}{2}\right)=-k t \\ & \Rightarrow-\log 2=-k t \\ & \left.\Rightarrow t=\frac{\log 2}{k} \text { [Propotional to } \log 2\right]\end{aligned}\)
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