Question

The rate of decay of mass of a certain substance at time \( t \) is proportional to the mass at that instant. The time during which the original mass of \( m_0 \) gm will be left to \( m_1 \) gm is

• A. \( K \log \left( \frac{m_1}{m_0} \right) \)
• B. \( \frac{1}{K} \log \left( \frac{m_1}{m_0} \right) \)
• C. \( \frac{1}{K} \log \left( \frac{m_0}{m_1} \right) \)
• D. \( K \log \left( \frac{m_0}{m_1} \right) \)

Hint:

For exponential decay problems, remember the general form \( M(t) = M_0 e^{-Kt} \) and solve for time by using logarithms.

Answer 1

The Correct Option is C

Step 1: Use the decay formula.
The rate of decay is given by \( \frac{dM}{dt} = -K M \), where \( K \) is the constant of proportionality. By solving this differential equation, we get: \[ M(t) = M_0 e^{-Kt} \] Step 2: Solve for time.
To find the time \( t \) when the mass decays from \( M_0 \) to \( M_1 \), we substitute \( M_1 \) for \( M(t) \) and solve for \( t \). The result is: \[ t = \frac{1}{K} \log \left( \frac{m_0}{m_1} \right) \] Step 3: Conclusion.
The correct answer is (C) \( \frac{1}{K} \log \left( \frac{m_0}{m_1} \right) \).