A vessel of area of cross-section A has liquid to a height H . There is a hole at the bottom of vessel having area of cross-section a . The time taken to decrease the level from `H_(1)` to `H_(2)` will sec

Options

A. `A/asqrt((2)/(g))[sqrt(H_(1)) - sqrt(H_(2))]`
B. `sqrt(2gh)`
C. `sqrt(2gh(H_(1) - H_(2))`
D. `A/asqrt((g)/(2))[sqrt(H_(1)) - sqrt(H_(2))`

Answer

To solve the problem of finding the time taken for the liquid level in a vessel to decrease from height \( H_1 \) to \( H_2 \) through a hole at the bottom, we can use Torricelli's law and the principles of fluid dynamics. Here’s the step-by-step solution:

### Step 1: Understand the flow rate
According to Torricelli's law, 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 liquid above the hole.

### Step 2: Determine the volume flow rate
The volume flow rate \( Q \) through the hole can be expressed as:
\[ Q = a \cdot v = a \cdot \sqrt{2gH} \]
where \( a \) is the area of cross-section of the hole.

### Step 3: Relate the change in height to the volume
The change in volume \( dV \) of the liquid in the vessel as the height decreases by \( dH \) is given by:
\[ dV = A \cdot dH \]
where \( A \) is the area of cross-section of the vessel.

### Step 4: Set up the differential equation
Since the volume flow rate out of the vessel equals the rate of change of volume in the vessel, we can write:
\[ A \frac{dH}{dt} = -a \sqrt{2gH} \]
The negative sign indicates that the height \( H \) is decreasing.

### Step 5: Rearranging the equation
Rearranging gives:
\[ \frac{dH}{\sqrt{H}} = -\frac{a \sqrt{2g}}{A} dt \]

### Step 6: Integrate both sides
Integrating both sides, we have:
\[ \int_{H_1}^{H_2} \frac{dH}{\sqrt{H}} = -\frac{a \sqrt{2g}}{A} \int_{0}^{t} dt \]
This leads to:
\[ 2(\sqrt{H_2} - \sqrt{H_1}) = -\frac{a \sqrt{2g}}{A} t \]

### Step 7: Solve for time \( t \)
Rearranging for \( t \) gives:
\[ t = -\frac{2A}{a \sqrt{2g}} (\sqrt{H_2} - \sqrt{H_1}) \]

### Final Result
Thus, the time taken to decrease the liquid level from \( H_1 \) to \( H_2 \) is:
\[ t = \frac{2A}{a \sqrt{2g}} (\sqrt{H_1} - \sqrt{H_2}) \]