1. Use radioactive decay law for activity

   Activity decays in the same way as the number of undecayed nuclei: $A = A_0\left(\frac{1}{2}\right)^{t/T_{1/2}}$ where:

   • $A_0$ = initial activity
   • $A$ = activity after time $t$
   • $T_{1/2} = 30$ days

2. Interpret the safety condition

   Initially, the activity is 10 times the permissible value.

   Let the permissible activity be $A_p$. Then initially, $A_0 = 10A_p$

   The room becomes safe when: $A = A_p$

3. Substitute into the decay equation

   $A_p = 10A_p\left(\frac{1}{2}\right)^{t/30}$

   Cancel $A_p$ from both sides: $1 = 10\left(\frac{1}{2}\right)^{t/30}$

   So, $\left(\frac{1}{2}\right)^{t/30} = \frac{1}{10}$

4. Solve for $t$

   Taking logarithms: $\frac{t}{30}\log\left(\frac{1}{2}\right) = \log\left(\frac{1}{10}\right)$

   Hence, $\frac{t}{30} = \frac{\log(1/10)}{\log(1/2)} = \frac{\log 10}{\log 2} \approx \frac{1}{0.3010} \approx 3.32$

   Therefore, $t \approx 30 \times 3.32 = 99.6 \text{ days}$

   So the required time is approximately: $t \approx 100 \text{ days}$

5. Check options

   • A: $1000$ days — too large
   • B: $300$ days — too large
   • C: $10$ days — too small
   • D: $100$ days — correct

Final Answer: Option D