GATE 2022 BT – Question 63
A 100 m³ medium initially contains $10^8$ spores/mL. The accepted final contamination is one spore in the vessel and death constant is $2\ \mathrm{min^{-1}}$ at 121°C. Ignoring heating/cooling deaths, holding time (nearest integer min) is ___________.
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Correct answer: 18
Explanation
**Sterilisation kinetics.** The number of viable spores falls as
$$N=N_0\,e^{-k_dt}\;\Rightarrow\;t=\frac1{k_d}\ln\frac{N_0}{N}.$$
**Step 1: initial number of spores.**
$$100\text{ m}^3=100\times10^6\text{ mL}=10^8\text{ mL},\qquad N_0=10^8\times10^8=10^{16}.$$
**Step 2: final number.** One spore, $N=1$.
**Step 3: holding time** with $k_d=2\ \text{min}^{-1}$:
$$t=\frac{\ln(10^{16})}{2}=\frac{36.84}{2}=18.4\approx\mathbf{18\ min}.$$