A wide tank of cross-section area A, containing a liquid to a height H, has an orifice at its base of area`A/3` . The initial velocity of top surface of liquid is

Options

A. `sqrt((gH)/4)`
B. `3sqrt((gH)/4)`
C. `sqrt((gH)/2)`
D. `1/3sqrt((gH)/2)`

Explanation

To solve the problem, we will use Bernoulli's equation and the principle of continuity. Here’s the step-by-step solution:

### Step 1: Understand the Setup
We have a tank with a cross-sectional area \( A \) filled with liquid to a height \( H \). At the bottom of the tank, there is an orifice with a cross-sectional area of \( \frac{A}{3} \).

### Step 2: Apply Bernoulli’s Equation
We will apply Bernoulli’s equation between two points:
1. At the top surface of the liquid (Point 1).
2. At the orifice (Point 2).

The Bernoulli's equation states:
\[ P_1 + \rho g H + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2 \]
Where:
- \( P_1 \) and \( P_2 \) are the pressures at points 1 and 2 respectively.
- \( \rho \) is the density of the liquid.
- \( g \) is the acceleration due to gravity.
- \( H \) is the height of the liquid column.
- \( v_1 \) is the velocity of the liquid at the top surface.
- \( v_2 \) is the velocity of the liquid exiting the orifice.

### Step 3: Define Pressures and Velocities
At the top surface of the liquid (Point 1):
- The pressure \( P_1 = P_0 \) (atmospheric pressure).
- The height \( H \) is the height of the liquid column.
- The velocity \( v_1 \) is what we want to find.

At the orifice (Point 2):
- The pressure \( P_2 = P_0 \) (atmospheric pressure).
- The height \( h_2 = 0 \) (since it is at the base).
- The velocity \( v_2 \) can be related to \( v_1 \) using the continuity equation.

### Step 4: Apply the Continuity Equation
The continuity equation states:
\[ A_1 v_1 = A_2 v_2 \]
Where:
- \( A_1 = A \) (cross-sectional area of the tank).
- \( A_2 = \frac{A}{3} \) (cross-sectional area of the orifice).

From this, we can express \( v_2 \):
\[ v_2 = \frac{A}{A/3} v_1 = 3 v_1 \]

### Step 5: Substitute into Bernoulli’s Equation
Substituting the values into Bernoulli’s equation:
\[ P_0 + \rho g H + \frac{1}{2} \rho v_1^2 = P_0 + 0 + \frac{1}{2} \rho (3 v_1)^2 \]

### Step 6: Simplify the Equation
Cancelling \( P_0 \) from both sides and simplifying gives:
\[ \rho g H + \frac{1}{2} \rho v_1^2 = \frac{1}{2} \rho (9 v_1^2) \]
\[ \rho g H = \frac{1}{2} \rho (9 v_1^2 - v_1^2) \]
\[ \rho g H = \frac{1}{2} \rho (8 v_1^2) \]
Cancelling \( \rho \) from both sides:
\[ g H = 4 v_1^2 \]

### Step 7: Solve for \( v_1 \)
Rearranging gives:
\[ v_1^2 = \frac{g H}{4} \]
Taking the square root:
\[ v_1 = \sqrt{\frac{g H}{4}} = \frac{1}{2} \sqrt{g H} \]

### Final Answer
The initial velocity of the top surface of the liquid is:
\[ v_1 = \frac{1}{2} \sqrt{g H} \]