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 \( m_0 \) gram will be left to \( m_1 \) gram is
\[ k \text{ is constant of proportionality} \]

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

Hint:

In decay problems, remember the general formula for exponential decay: \( m(t) = m_0 e^{-kt} \).

Answer 1

The Correct Option is D

Step 1: Using the formula for exponential decay.
The rate of decay is given by the differential equation: \[ \frac{dm}{dt} = -km \] Integrating this equation gives: \[ \ln m = -kt + C \] Step 2: Finding the time for decay.
We use the initial condition \( m = m_0 \) at \( t = 0 \) to find the constant \( C \), which gives: \[ \ln m_0 = C \] Now, solving for the time when the mass decays from \( m_0 \) to \( m_1 \), we get: \[ t = \frac{1}{k} \log \left( \frac{m_0}{m_1} \right) \] Step 3: Conclusion.
Thus, the time during which the original mass of \( m_0 \) gram will decay to \( m_1 \) gram is \( \boxed{\frac{1}{k} \log \left( \frac{m_0}{m_1} \right)} \).