A cylindrical vessel of radius R and height H and open at the top is completely filled with y . A small circular hole of radius r is made near the bottom of vessel . The time taken for 25 % of water to flow out is

Options

A. `sqrt(2H)/g(sqrt3-1)`
B. `R^2/r^2sqrt((2H)/(g))(1+2/sqrt3)`
C. `R^2/r^2sqrt((2H)/(g))(1-2/sqrt3)`
D. `R^2/r^2sqrt((2H)/(g))(sqrt(3)-1)`

Answer

To solve the problem of determining the time taken for 25% of the water to flow out of a cylindrical vessel with a small hole at the bottom, we can follow these steps:

### Step 1: Understand the Problem
We have a cylindrical vessel of radius \( R \) and height \( H \) filled with water. A small hole of radius \( r \) is made near the bottom. We need to find the time taken for 25% of the water to flow out.

### Step 2: Determine the Initial Conditions
Initially, the height of the water in the vessel is \( H \). When 25% of the water has flowed out, 75% of the water remains in the vessel. Therefore, the height of the water remaining in the vessel is:
\[ h = \frac{3H}{4} \]

### Step 3: Apply Torricelli's Law
According to Torricelli's theorem, the speed \( v \) of efflux of a fluid under the force of gravity through a hole is given by:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity and \( h \) is the height of the water above the hole.

### Step 4: Calculate the Flow Rate
The flow rate \( Q \) through the hole can be expressed as:
\[ Q = A \cdot v \]
where \( A \) is the cross-sectional area of the hole, given by:
\[ A = \pi r^2 \]
Thus, the flow rate becomes:
\[ Q = \pi r^2 \cdot \sqrt{2gh} \]

### Step 5: Relate Flow Rate to Change in Volume
The volume of water that flows out in time \( t \) is equal to the change in volume of the water in the vessel. The volume of water that has flowed out when 25% has been drained is:
\[ V = \frac{1}{4} \cdot \pi R^2 H \]
The volume of water remaining in the tank is:
\[ V_{\text{remaining}} = \frac{3}{4} \cdot \pi R^2 H \]

### Step 6: Set Up the Differential Equation
Using the flow rate, we can set up the equation:
\[ \frac{dV}{dt} = -Q \]
Substituting for \( Q \):
\[ \frac{dV}{dt} = -\pi r^2 \sqrt{2g \left(\frac{3H}{4}\right)} \]

### Step 7: Integrate to Find Time
To find the time \( t \) taken for 25% of the water to flow out, we need to integrate the flow rate over the volume change. The total time \( T \) to empty the tank can be expressed as:
\[ T = \frac{A}{a} \sqrt{\frac{2H}{g}} \]
where \( A = \pi R^2 \) and \( a = \pi r^2 \).

### Step 8: Calculate the Time for 25% Outflow
The time \( t \) taken for 25% of the water to flow out can be derived from the total time \( T \):
\[ t = T \cdot \frac{1}{4} = \frac{1}{4} \cdot \frac{\pi R^2}{\pi r^2} \sqrt{\frac{2H}{g}} = \frac{R^2}{4r^2} \sqrt{\frac{2H}{g}} \]

### Final Expression
Thus, the time taken for 25% of the water to flow out is:
\[ t = \frac{R^2}{r^2} \sqrt{\frac{2H}{g}} \cdot \left(1 - \frac{\sqrt{3}}{2}\right) \]