A bacterium divides every 35 minutes. If a cu | vedclass

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(C) The number of generations $(n)$ is calculated by dividing the total time by the time taken for one division.$n = \frac{	ext{Total time}}{	ext{Ge

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$32 	imes 10^5$ cells per $ml$

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(C) The number of generations $(n)$ is calculated by dividing the total time by the time taken for one division.$n = \frac{	ext{Total time}}{	ext{Generation time}} = \frac{175 	ext{ min}}{35 	ext{ min}} = 5$.The final number of cells $(N_f)$ is calculated using the formula $N_f = N_i 	imes 2^n$,where $N_i$ is the initial number of cells.$N_f = 10^5 	imes 2^5$.Since $2^5 = 32$,the final concentration is $N_f = 10^5 	imes 32 = 32 	imes 10^5$ cells per $ml$.

$A$ bacterium divides every $35$ minutes. If a culture containing $10^5$ cells per $ml$ is grown for $175$ minutes,what will be the cell concentration per $ml$ after $175$ minutes?

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