A tank containing water has an orifice in one vertical side. If the centre of orifice is 4.9 m below the surface level in the tank, the velocity of discharge is:

Options

A. 4.9 metre/second
B. 9.8 metre/second
C. 2.45 metre/second
D. zero

Answer

To find the velocity of discharge from an orifice in a tank of water, we can use Bernoulli's equation. Here’s a step-by-step solution:

### Step 1: Understand the Problem
We have a tank of water with an orifice located 4.9 meters below the surface of the water. We need to find the velocity of water discharging from the orifice.

### Step 2: Identify Relevant Parameters
- Depth of the orifice below the water surface, \( h = 4.9 \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 3: Apply Bernoulli's Equation
Bernoulli's equation relates the pressure, velocity, and height at two points in a fluid flow. For our case, we can simplify it as follows:

\[ P_1 + \rho gh_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho gh_2 + \frac{1}{2} \rho v_2^2 \]

Where:
- \( P_1 \) and \( P_2 \) are the pressures at points 1 and 2.
- \( \rho \) is the density of the fluid (water).
- \( h_1 \) and \( h_2 \) are the heights at points 1 and 2.
- \( v_1 \) and \( v_2 \) are the velocities at points 1 and 2.

### Step 4: Set Up the Equation
At the surface of the water (point 1):
- The pressure \( P_1 = P_0 \) (atmospheric pressure)
- The height \( h_1 = 4.9 \, \text{m} \)
- The velocity \( v_1 = 0 \) (water is still)

At the orifice (point 2):
- The pressure \( P_2 = P_0 \) (atmospheric pressure)
- The height \( h_2 = 0 \) (reference level at the orifice)
- The velocity \( v_2 = V \) (unknown discharge velocity)

### Step 5: Simplify the Equation
Substituting the known values into Bernoulli's equation:

\[ P_0 + \rho g h_1 + 0 = P_0 + 0 + \frac{1}{2} \rho V^2 \]

Cancelling \( P_0 \) and rearranging gives:

\[ \rho g h_1 = \frac{1}{2} \rho V^2 \]

### Step 6: Cancel Out the Density
Since \( \rho \) appears on both sides, we can cancel it out:

\[ g h_1 = \frac{1}{2} V^2 \]

### Step 7: Solve for Velocity \( V \)
Rearranging gives:

\[ V^2 = 2gh_1 \]

Taking the square root:

\[ V = \sqrt{2gh_1} \]

### Step 8: Substitute the Values
Now substituting \( g = 9.8 \, \text{m/s}^2 \) and \( h_1 = 4.9 \, \text{m} \):

\[ V = \sqrt{2 \times 9.8 \times 4.9} \]

Calculating:

\[ V = \sqrt{96.04} = 9.8 \, \text{m/s} \]

### Final Answer
The velocity of discharge is \( V = 9.8 \, \text{m/s} \).

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