# A bacterial infection in an internal wound grows as $N{}'(t)=N_{o}exp(t)$,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 $\frac{dN}{dt}=-5N^{2}$.What will be the plot of $\frac{N_{o}}{N}$ vs. t after 1 hour ? Option 1)Option 2)Option 3)Option 4)

From o to 1 hour => $N^{t}=N_{o}e^{t}$

After 1 hour, it is given : $\frac{dN}{dt}=-5N^{2}$

$\therefore$ at t = 1 hour , $N^{'}=eN_{o}$

Now,

$\int_{eN_o}^{N}N^{-2}dN=-5\int_{1}^{t}dt$

$\frac{1}{N}-\frac{1}{eN_o}=5(t-1)$

$\frac{N_o}{N}=5N_o(t-1)+\frac{1}{e}$

$\frac{N_o}{N}=5N_ot+(\frac{1}{e}-5N_o)$

This equation is similar to the straight line equation (Y=mx+C)

So, the curve will be straight line.

$\therefore$ option (3) is correct.

Option 1)

Option 2)

Option 3)

Option 4)

Exams
Articles
Questions