Q98 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
The rate of disintegration of a radioactive element at any time is proportional to its mass at that time, where $k$ $(k>0)$ is the constant of proportionality. The time during which an original mass of 1.5 gm will disintegrate to a mass of 0.5 gm is $\ldots$
- A$k\log 3$
- Correct answerB$\dfrac{1}{k}\log 3$
- C$k\log 5$
- D$\dfrac{1}{k}\log 5$
Solution
Let $m$ be the mass of the radioactive element at time $t$.
The rate of disintegration is given by $\dfrac{dm}{dt} = -km$
$\dfrac{dm}{m} = -k dt$
Integrating both sides with appropriate limits:
$\int_{1.5}^{0.5} \dfrac{dm}{m} = -k \int_{0}^{t} dt$
$[\log m]_{1.5}^{0.5} = -kt$
$\log 0.5 - \log 1.5 = -kt$
$\log \left(\dfrac{0.5}{1.5}\right) = -kt$
$\log \left(\dfrac{1}{3}\right) = -kt$
$-\log 3 = -kt$
$t = \dfrac{1}{k} \log 3$
Answer: $\dfrac{1}{k}\log 3$
The rate of disintegration is given by $\dfrac{dm}{dt} = -km$
$\dfrac{dm}{m} = -k dt$
Integrating both sides with appropriate limits:
$\int_{1.5}^{0.5} \dfrac{dm}{m} = -k \int_{0}^{t} dt$
$[\log m]_{1.5}^{0.5} = -kt$
$\log 0.5 - \log 1.5 = -kt$
$\log \left(\dfrac{0.5}{1.5}\right) = -kt$
$\log \left(\dfrac{1}{3}\right) = -kt$
$-\log 3 = -kt$
$t = \dfrac{1}{k} \log 3$
Answer: $\dfrac{1}{k}\log 3$
Hence, the correct answer is option: 'B'.