A bacterial infection in an internal wound 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 `(dN)/(dt)= -5 N^(2)`. What will be the plot of `N_(0)/N` vs. t after 1 hour?

Options

A.
B.
C.
D.

Explanation

To solve the problem step by step, we need to analyze the growth of the bacterial infection and the effect of the antibiotic over time.

### Step 1: Understand the Growth of Bacteria
The bacterial population grows according to the equation:
\[ N'(t) = N_0 e^t \]
where \( N_0 \) is the initial bacterial population and \( t \) is the time in hours.

### Step 2: Calculate Bacterial Population Before Antibiotic Effect
For the first hour (from \( t = 0 \) to \( t = 1 \)):
At \( t = 1 \):
\[ N(1) = N_0 e^1 = N_0 e \]

### Step 3: Effect of Antibiotic After 1 Hour
After 1 hour, the antibiotic reaches the wound, and the bacterial population starts to decrease according to the equation:
\[ \frac{dN}{dt} = -5N^2 \]

### Step 4: Solve the Differential Equation
We need to solve the differential equation:
\[ \frac{dN}{dt} = -5N^2 \]

Rearranging gives:
\[ \frac{dN}{N^2} = -5 dt \]

Now, we integrate both sides. The left side integrates to:
\[ \int \frac{dN}{N^2} = -\frac{1}{N} \]

The right side integrates to:
\[ -5 \int dt = -5t + C \]

### Step 5: Apply Initial Conditions
At \( t = 1 \), \( N = N_0 e \):
\[ -\frac{1}{N_0 e} = -5(1) + C \]
Thus,
\[ C = -\frac{1}{N_0 e} + 5 \]

### Step 6: Substitute Back to Find N
Substituting \( C \) back into the equation gives:
\[ -\frac{1}{N} = -5t - \frac{1}{N_0 e} + 5 \]

Rearranging gives:
\[ \frac{1}{N} = 5t - 5 + \frac{1}{N_0 e} \]

### Step 7: Express \( \frac{N_0}{N} \)
To find \( \frac{N_0}{N} \):
\[ \frac{N_0}{N} = \frac{N_0}{\frac{1}{5t - 5 + \frac{1}{N_0 e}}} \]
This simplifies to:
\[ \frac{N_0}{N} = N_0 \left( 5t - 5 + \frac{1}{N_0 e} \right) \]

### Step 8: Analyze the Plot of \( \frac{N_0}{N} \) vs. \( t \)
The equation for \( \frac{N_0}{N} \) is linear in \( t \):
\[ \frac{N_0}{N} = 5N_0 t - 5N_0 + \frac{1}{e} \]
This indicates that the plot of \( \frac{N_0}{N} \) vs. \( t \) will be a straight line with a slope of \( 5N_0 \) and a y-intercept of \( -5N_0 + \frac{1}{e} \).

### Conclusion
The plot of \( \frac{N_0}{N} \) vs. \( t \) after 1 hour will be a straight line with a positive slope.

---