The rate of growth of bacteria in a culture i | vedclass

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(D) Let $x$ be the number of bacteria present at time $t$.$	herefore \quad \frac{dx}{dt} \propto x$$	herefore \quad \frac{dx}{dt} = \lambda x$,where

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Let $x$ be the number of bacteria present at time $t$.$	herefore \quad \frac{dx}{dt} \propto x$$	herefore \quad \frac{dx}{dt} = \lambda x$,where $\lambda$ is the constant of proportionality.Integrating on both sides,we get$\log x = \lambda t + c$When $t = 0, x = 1000$$	herefore \quad \log 1000 = 0 + c \Rightarrow c = \log (1000)$$	herefore \quad \log x = \lambda t + \log (1000)$ . . . . . . $(i)$When $t = 2$$x = 1000 + (20 \% 	ext{ of } 1000) = 1000 + 200 = 1200$$	herefore \quad \log 1200 = 2 \lambda + \log 1000$$\Rightarrow \lambda = \frac{1}{2} \log \left(\frac{1200}{1000}ight) = \frac{1}{2} \log \left(\frac{6}{5}ight)$$	herefore \quad \log x = \frac{t}{2} \log \left(\frac{6}{5}ight) + \log (1000)$ . . . . . . $[	ext{From } (i)]$When $t = \frac{k}{\log \left(\frac{6}{5}ight)}, x = 2000$$	herefore \quad \log 2000 = \frac{k}{\log \left(\frac{6}{5}ight)} 	imes \frac{1}{2} \log \left(\frac{6}{5}ight) + \log (1000)$$\Rightarrow \log \left(\frac{2000}{1000}ight) = \frac{k}{2} \Rightarrow \log 2 = \frac{k}{2}$$\Rightarrow \frac{k}{\log 2} = 2$$\Rightarrow \left(\frac{k}{\log 2}ight)^2 = 2^2 = 4$

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