Rate of evaporation (volume evaporated per unit time) of water kept in an earthen pot is proportional to the volume of water present in the pot. It is observed that it takes 48 hrs for 75% of the water kept in the pot to evaporate. The empty pot was placed below a tap from which water leaks in small drops each of volume `v_(0)`. The drops fall at a uniform rate of n drops per hour. Calculate the volume of water in the pot 24hrs after it was kept below the tap.

Answer

To solve the problem, we need to analyze the evaporation process and the addition of water from the leaking tap. Here’s a step-by-step solution:

### Step 1: Understand the Rate of Evaporation
The rate of evaporation \( R \) is proportional to the volume of water \( V \) in the pot:
\[ R = kV \]
where \( k \) is a constant of proportionality.

### Step 2: Determine the Initial Conditions
Let the initial volume of water in the pot be \( V_0 \). According to the problem, 75% of the water evaporates in 48 hours. Therefore, the volume of water evaporated is:
\[ 0.75V_0 \]
The remaining volume after 48 hours is:
\[ V = V_0 - 0.75V_0 = 0.25V_0 \]

### Step 3: Set Up the Differential Equation
The change in volume over time can be described by the differential equation:
\[ \frac{dV}{dt} = -kV \]
This is a separable differential equation. We can separate variables and integrate.

### Step 4: Integrate the Differential Equation
Integrating both sides gives:
\[ \int \frac{dV}{V} = -k \int dt \]
This results in:
\[ \ln V = -kt + C \]
Exponentiating both sides, we have:
\[ V = e^{-kt + C} = Ce^{-kt} \]
where \( C \) is a constant determined by initial conditions.

### Step 5: Apply Initial Conditions
At \( t = 0 \), \( V = V_0 \):
\[ V_0 = Ce^{0} \implies C = V_0 \]
Thus, the volume as a function of time is:
\[ V(t) = V_0 e^{-kt} \]

### Step 6: Find the Constant \( k \)
At \( t = 48 \) hours, \( V(48) = 0.25V_0 \):
\[ 0.25V_0 = V_0 e^{-48k} \]
Dividing both sides by \( V_0 \):
\[ 0.25 = e^{-48k} \]
Taking the natural logarithm:
\[ \ln(0.25) = -48k \implies k = -\frac{\ln(0.25)}{48} \]

### Step 7: Calculate the Volume After 24 Hours
Now, we need to find \( V(24) \):
\[ V(24) = V_0 e^{-24k} \]
Substituting for \( k \):
\[ V(24) = V_0 e^{-24 \left(-\frac{\ln(0.25)}{48}\right)} = V_0 e^{\frac{24 \ln(0.25)}{48}} = V_0 e^{\frac{1}{2} \ln(0.25)} = V_0 \sqrt{0.25} = V_0 \cdot 0.5 = 0.5V_0 \]

### Step 8: Account for Water from the Tap
In 24 hours, the tap adds water. The volume added from the tap is:
\[ \text{Volume added} = n \cdot v_0 \cdot 24 \]
Thus, the total volume of water in the pot after 24 hours is:
\[ V_{\text{total}} = 0.5V_0 + n \cdot v_0 \cdot 24 \]

### Final Answer
The volume of water in the pot 24 hours after it was kept below the tap is:
\[ V_{\text{total}} = 0.5V_0 + 24nv_0 \]