JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Chemical Kinetics

A bacterial infection in an internal would grows as

N^{'} (t) = N_{0} exp (t)

, where the time t is in hours. A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as

\frac{d N}{d t} = - 5 N^{2}

. What will be the plot of

\frac{N _{0}}{N}

vs, t after 1 hour?

(A)(see structure)
(B)(see structure)
(C)(see structure)
(D)(see structure)
✓ Correct

Answer: (D)

Solution

For the first hour the colony grows as

N (t) = N_{0} e^{t}

, so

N > N_{0}

and the ratio

\frac{N _{0}}{N}

starts below

1

and falls. Once the antibiotic acts, the population obeys

\frac{d N}{d t} = - 5 N^{2}

, a second-order decay faster than exponential, so

N

drops sharply and

\frac{N _{0}}{N}

climbs steeply. The curve that fits dips at first and then rises steeply after the one-hour mark.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Chemical Kinetics

Solution

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