A cylindrical vessel of area of cross-section A is filled with water to a height H. It has capillary tube of length l and radius r fitted horizontally at its bottom. If the coeffiecient of viscosity of water is `eta`, then time required in which level will fall to a height `(H)/(2)` is (density of water is `rho`)

Options

A. `(etalr^(2))/(4Arho)1n (2)`
B. `(4 etalr^(4))/(pi g A rho) ln ((1)/(2))`
C. `(8 etalA)/(rho pi g r^(4)) ln (2)`
D. `(4 Hetalrho)/(pi g r^(4))ln (2)`

Answer

To solve the problem step by step, we will use the principles of fluid dynamics and the properties of viscosity.

### Step 1: Understand the Setup
We have a cylindrical vessel filled with water to a height \( H \). A capillary tube of length \( l \) and radius \( r \) is fitted horizontally at the bottom of the vessel. We need to find the time required for the water level to fall from \( H \) to \( \frac{H}{2} \).

### Step 2: Define Variables
- Let \( A \) be the area of the cross-section of the cylindrical vessel.
- Let \( \eta \) be the coefficient of viscosity of water.
- Let \( \rho \) be the density of water.
- Let \( g \) be the acceleration due to gravity.
- Let \( h \) be the height of the water level at any time \( t \).

### Step 3: Rate of Flow from the Vessel
The rate of flow of water out of the vessel can be expressed as:
\[ V = -A \frac{dh}{dt} \]
This equation indicates that as the height \( h \) decreases, the volume flow rate \( V \) is negative.

### Step 4: Apply Poiseuille's Law
According to Poiseuille's equation, the flow rate \( V \) through the capillary tube can be expressed as:
\[ V = \frac{\pi r^4}{8 \eta l} \Delta P \]
Where \( \Delta P \) is the pressure difference driving the flow. The pressure difference can be given by:
\[ \Delta P = \rho g h \]
Thus, we can rewrite the flow rate as:
\[ V = \frac{\pi r^4}{8 \eta l} \rho g h \]

### Step 5: Set the Two Expressions for Flow Rate Equal
Equating the two expressions for flow rate, we have:
\[ -A \frac{dh}{dt} = \frac{\pi r^4}{8 \eta l} \rho g h \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ \frac{dh}{h} = -\frac{\pi \rho g r^4}{8 \eta l A} dt \]

### Step 7: Integrate Both Sides
Integrating both sides, we have:
\[ \int \frac{dh}{h} = -\frac{\pi \rho g r^4}{8 \eta l A} \int dt \]
The limits for \( h \) will change from \( H \) to \( \frac{H}{2} \) and for \( t \) from \( 0 \) to \( T \):
\[ \left[ \ln h \right]_{H}^{\frac{H}{2}} = -\frac{\pi \rho g r^4}{8 \eta l A} T \]

### Step 8: Evaluate the Integral
Evaluating the left side:
\[ \ln\left(\frac{H}{2}\right) - \ln(H) = \ln\left(\frac{H/2}{H}\right) = \ln\left(\frac{1}{2}\right) = -\ln(2) \]
Thus, we have:
\[ -\ln(2) = -\frac{\pi \rho g r^4}{8 \eta l A} T \]

### Step 9: Solve for Time \( T \)
Rearranging gives:
\[ T = \frac{8 \eta l A \ln(2)}{\pi \rho g r^4} \]

### Final Answer
The time required for the water level to fall from \( H \) to \( \frac{H}{2} \) is:
\[ T = \frac{8 \eta l A \ln(2)}{\pi \rho g r^4} \]