Liquid is filled in a container upto a height of `H`. A small hole is made at the bottom of the tank. Time taken to empty from `H` to `(H)/(3)` is `t_(0)`. Find the time taken to empty tank from `(H)/(3)` to zero.

Explanation

To solve the problem of finding the time taken to empty the tank from a height of \( \frac{H}{3} \) to 0, we can follow these steps:

### Step 1: Understand the flow of liquid
When a small hole is made at the bottom of a tank filled with liquid, the speed of the liquid flowing out can be determined using Torricelli's theorem, which states that the speed \( v \) of efflux of a fluid under the force of gravity through an orifice is given by:
\[ v = \sqrt{2gh} \]
where \( h \) is the height of the liquid above the hole.

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

### Step 3: Relate flow rate to change in height
The change in volume \( dV \) of the liquid in the tank can also be expressed in terms of the cross-sectional area \( A \) of the tank and the change in height \( dh \):
\[ dV = A \, dh \]
Thus, we can equate the two expressions for volume flow rate:
\[ A \, dh = -a \sqrt{2gh} \, dt \]
The negative sign indicates that the height \( h \) is decreasing over time.

### Step 4: Rearranging the equation
Rearranging gives us:
\[ dt = -\frac{A}{a} \frac{dh}{\sqrt{2gh}} \]

### Step 5: Integrate to find time
To find the time taken to empty the tank from height \( H \) to \( \frac{H}{3} \), we integrate:
\[ t_0 = \int_{H}^{\frac{H}{3}} -\frac{A}{a} \frac{dh}{\sqrt{2gh}} \]
This can be simplified to:
\[ t_0 = \frac{A}{a} \sqrt{\frac{1}{2g}} \int_{H}^{\frac{H}{3}} h^{-1/2} \, dh \]
The integral evaluates to:
\[ \int h^{-1/2} \, dh = 2\sqrt{h} \]
Thus:
\[ t_0 = \frac{A}{a} \sqrt{\frac{1}{2g}} \left[ 2\sqrt{h} \right]_{H}^{\frac{H}{3}} = \frac{A}{a} \sqrt{\frac{1}{2g}} \left( 2\sqrt{\frac{H}{3}} - 2\sqrt{H} \right) \]
This simplifies to:
\[ t_0 = \frac{2A}{a} \sqrt{\frac{H}{2g}} \left( \sqrt{\frac{1}{3}} - 1 \right) \]

### Step 6: Find time to empty from \( \frac{H}{3} \) to 0
Now, we need to find the time \( t_1 \) to empty from \( \frac{H}{3} \) to 0:
\[ t_1 = \int_{\frac{H}{3}}^{0} -\frac{A}{a} \frac{dh}{\sqrt{2gh}} = \frac{A}{a} \sqrt{\frac{1}{2g}} \left[ 2\sqrt{h} \right]_{\frac{H}{3}}^{0} \]
This evaluates to:
\[ t_1 = \frac{A}{a} \sqrt{\frac{1}{2g}} \left( 0 - 2\sqrt{\frac{H}{3}} \right) = -\frac{2A}{a} \sqrt{\frac{H}{2g}} \sqrt{\frac{1}{3}} \]
Thus:
\[ t_1 = \frac{2A}{a} \sqrt{\frac{H}{2g}} \left( \sqrt{\frac{1}{3}} \right) \]

### Step 7: Relate \( t_1 \) to \( t_0 \)
From the expressions for \( t_0 \) and \( t_1 \), we can see that:
\[ t_1 = \frac{t_0}{\sqrt{3} - 1} \]

### Final Answer
The time taken to empty the tank from \( \frac{H}{3} \) to 0 is:
\[ t_1 = \frac{t_0}{\sqrt{3} - 1} \]