The height of the water in a tank is H. The range of the liquid emerging out from a hole in the wall of the tank at a depth `(3H)/4` from the upper surface of water, will be

Options

A. H
B. `H/2`
C. `(3H)/2`
D. `(sqrt3H)/2`

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Understand the Problem
We have a tank filled with water to a height \( H \). There is a hole in the wall of the tank at a depth of \( \frac{3H}{4} \) from the top of the water surface. We need to find the range \( R \) of the liquid emerging from this hole.

### Step 2: Determine the Depth of Water Above the Hole
The depth of water above the hole is:
\[ h = H - \frac{3H}{4} = \frac{H}{4} \]

### Step 3: Calculate the Velocity of Water Emerging from the Hole
Using Torricelli's theorem, the velocity \( V \) of the water emerging from the hole can be calculated using the formula derived from the principle of conservation of energy:
\[ V = \sqrt{2gh} \]
Substituting \( h = \frac{H}{4} \):
\[ V = \sqrt{2g \cdot \frac{H}{4}} = \sqrt{\frac{gH}{2}} \]

### Step 4: Determine the Time of Flight for the Water
The water will fall a vertical distance of \( \frac{H}{4} \) before it hits the ground. We can calculate the time \( t \) it takes to fall this distance using the equation of motion:
\[ h = \frac{1}{2}gt^2 \]
Substituting \( h = \frac{H}{4} \):
\[ \frac{H}{4} = \frac{1}{2}gt^2 \]
Rearranging gives:
\[ t^2 = \frac{H}{2g} \quad \Rightarrow \quad t = \sqrt{\frac{H}{2g}} \]

### Step 5: Calculate the Range of the Water
The range \( R \) of the projectile (the water) can be calculated using the formula:
\[ R = V \cdot t \]
Substituting the values of \( V \) and \( t \):
\[ R = \sqrt{\frac{gH}{2}} \cdot \sqrt{\frac{H}{2g}} = \sqrt{\frac{gH}{2} \cdot \frac{H}{2g}} = \sqrt{\frac{H^2}{4}} = \frac{H}{2} \]

### Final Answer
Thus, the range of the liquid emerging from the hole is:
\[ R = \frac{H}{2} \]