A area of cross section of a large tank is `0.5m^2`. It has an opening near the bottom having area of cross section `1cm^2`. A load of 20 kg is applied on the water at the top. Find the velocity of the water coming out of the opening at the time when the height of water level is 50 cm above the bottom. Take `g=10ms^-2`

Explanation

To solve the problem step by step, we will follow the principles of fluid mechanics, particularly Bernoulli's equation.

### Step 1: Understand the Problem
We have a large tank with an area of cross-section \( A_t = 0.5 \, m^2 \) and an opening at the bottom with an area of cross-section \( A_o = 1 \, cm^2 = 0.0001 \, m^2 \). A load of \( 20 \, kg \) is placed on the water, and we need to find the velocity of water exiting the opening when the height of the water is \( h = 50 \, cm = 0.5 \, m \).

### Step 2: Calculate the Pressure at the Opening
The pressure at the water surface (point A) consists of the atmospheric pressure and the pressure due to the load. The pressure due to the load can be calculated using the formula:

\[ P = P_0 + \frac{mg}{A_t} \]

Where:
- \( P_0 \) is the atmospheric pressure (which we will assume cancels out later).
- \( m = 20 \, kg \)
- \( g = 10 \, m/s^2 \)
- \( A_t = 0.5 \, m^2 \)

Calculating the pressure due to the load:

\[ P = P_0 + \frac{20 \times 10}{0.5} = P_0 + 400 \, N/m^2 \]

### Step 3: Apply Bernoulli's Equation
Using Bernoulli's equation between point A (water surface) and point B (opening):

\[ P_A + \rho g h + \frac{1}{2} \rho v_A^2 = P_B + \rho g h_B + \frac{1}{2} \rho v_B^2 \]

At point A, the velocity \( v_A \) is negligible due to the large area compared to point B. Thus, we can simplify the equation:

\[ P_A + \rho g h = P_B + \frac{1}{2} \rho v_B^2 \]

### Step 4: Substitute Known Values
Assuming \( P_B \) is atmospheric pressure \( P_0 \), we can write:

\[ P_0 + 400 + \rho g h = P_0 + \frac{1}{2} \rho v_B^2 \]

Canceling \( P_0 \) from both sides:

\[ 400 + \rho g h = \frac{1}{2} \rho v_B^2 \]

Where \( \rho \) (density of water) is approximately \( 1000 \, kg/m^3 \) and \( h = 0.5 \, m \):

\[ 400 + 1000 \times 10 \times 0.5 = \frac{1}{2} \times 1000 \times v_B^2 \]

Calculating the left side:

\[ 400 + 5000 = \frac{1}{2} \times 1000 \times v_B^2 \]

\[ 5400 = 500 \times v_B^2 \]

### Step 5: Solve for \( v_B \)
Rearranging gives:

\[ v_B^2 = \frac{5400}{500} = 10.8 \]

Taking the square root:

\[ v_B = \sqrt{10.8} \approx 3.29 \, m/s \]

### Final Answer
The velocity of the water coming out of the opening is approximately \( 3.29 \, m/s \).

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