A wet porous subtance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the wind loses half its moisture during the first hour, then the time when it would have lost `99.9%` of its moisture is (weather conditions remaining same)

Options

A. (a) more than 100 h
B. (b) more than 10 h
C. (c) approximately 10 h
D. (d) None of these

Answer

To solve the problem, we will follow these steps:

### Step 1: Set up the differential equation
Let \( m(t) \) be the amount of moisture at time \( t \). According to the problem, the rate of loss of moisture is proportional to the moisture content, which can be expressed as:
\[ \frac{dm}{dt} = -k m \]
where \( k \) is a positive constant.

### Step 2: Separate variables and integrate
We can rearrange the equation to separate the variables:
\[ \frac{dm}{m} = -k \, dt \]
Now, we integrate both sides:
\[ \int \frac{dm}{m} = -k \int dt \]
This gives us:
\[ \ln |m| = -kt + C \]
where \( C \) is the constant of integration.

### Step 3: Solve for \( m(t) \)
Exponentiating both sides, we have:
\[ m = e^{-kt + C} = e^C e^{-kt} \]
Let \( N = e^C \), then:
\[ m(t) = N e^{-kt} \]

### Step 4: Determine the constant \( N \)
At \( t = 0 \), the initial amount of moisture is \( m(0) = N \). Therefore, we can write:
\[ m(0) = N \]

### Step 5: Use the condition of losing half the moisture in the first hour
We know that after 1 hour, the moisture content is half of the initial amount:
\[ m(1) = \frac{N}{2} \]
Substituting into our equation:
\[ \frac{N}{2} = N e^{-k \cdot 1} \]
Dividing both sides by \( N \) (assuming \( N \neq 0 \)):
\[ \frac{1}{2} = e^{-k} \]
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -k \quad \Rightarrow \quad k = -\ln(2) \]

### Step 6: Find the time when 99.9% of moisture is lost
If 99.9% of the moisture is lost, then 0.1% remains:
\[ m(t) = 0.001N \]
Substituting into our equation:
\[ 0.001N = N e^{-kt} \]
Dividing both sides by \( N \):
\[ 0.001 = e^{-kt} \]
Taking the natural logarithm:
\[ \ln(0.001) = -kt \]
Substituting \( k = -\ln(2) \):
\[ \ln(0.001) = \ln(10^{-3}) = -3 \ln(10) = -(-\ln(2))t \]
Thus:
\[ t = \frac{3 \ln(10)}{\ln(2)} \]

### Step 7: Calculate the value of \( t \)
Using approximate values:
- \( \ln(10) \approx 2.302 \)
- \( \ln(2) \approx 0.693 \)

Calculating:
\[ t \approx \frac{3 \times 2.302}{0.693} \approx \frac{6.906}{0.693} \approx 9.95 \text{ hours} \]

### Final Answer
Therefore, the time when the substance would have lost 99.9% of its moisture is approximately:
\[ \boxed{10 \text{ hours}} \]