GATE 2026 BT Q.59: solution and answer

GATE 2026 · BTQ.59 · 2 marks · NAT

A batch sterilizer is being operated at 121 °C for sterilizing a medium containing microbial cells. Assume that the thermal deactivation of cells is a first order process with a death rate constant of 0.69 min $^{- 1}$ at 121 °C. If the initial concentration of microbes in the medium is $1 0^{10}$ cells m $^{- 3}$ , the time taken to reduce the microbial load to a final concentration of 10 cells m $^{- 3}$ is ________ min. (rounded off to the nearest integer)

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Answer and solution

ANSWER

The time needed to bring the load down from $1 0^{10}$ to 10 cells m $^{- 3}$ is $t = ln (1 0^{9}) /0.69 \approx 30$ min.

Official answer: 30Source: Official answer key, GATE 2026 BT (IIT Guwahati)

Thermal death is first order: $\frac{d N}{d t} = - k_{d} N$ , where $N$ is the viable cell concentration and $k_{d}$ is the specific death rate constant ( $k_{d} = 0.69 min^{- 1}$ at 121 °C).
Integrate at constant temperature (batch holding at 121 °C): $ln \frac{N _{0}}{N} = k_{d} t$ , so $t = \frac{1}{k _{d}} ln \frac{N _{0}}{N}$ .
Ratio of concentrations (the medium volume cancels, so concentrations can be used directly): $\frac{N _{0}}{N} = \frac{1 0 ^{10} cells m ^{- 3}}{10 cells m ^{- 3}} = 1 0^{9}$ .
Del factor: $\nabla = ln \frac{N _{0}}{N} = ln (1 0^{9}) = 9 ln 10 = 9 \times 2.3026 = 20.723$ .
$t = \frac{\nabla}{k _{d}} = \frac{20.723}{0.69 min ^{- 1}} = 30.03 min$
Rounded off to the nearest integer: $t = 30 min$ .

KEY IDEA

First-order death gives $ln (N_{0} / N) = k_{d} t$ at constant temperature. The time depends only on the ratio $N_{0} / N$ (here $1 0^{9}$ , a 9-log reduction), so the medium volume does not matter. A quick check: the decimal reduction time is $D = ln 10/ k_{d} = 2.3026/0.69 = 3.34$ min, and 9 log cycles take $9 \times 3.34 \approx 30$ min.

WATCH OUT FOR

Using $lo g_{10}$ where the natural log belongs: $lo g_{10} (1 0^{9}) /0.69 = 13$ min is wrong, because $k_{d}$ is defined through $e^{- k_{d} t}$ . Another slip is taking the ratio as $1 0^{10} /1$ , or mixing up which concentration goes on top, which changes the number of log cycles.

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GATE 2026 BT Q.59: solution and answer

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