Water from a tap emerges vertically downwards with initial velocity `4ms^(-1)`. The cross-sectional area of the tap is A. The flow is steady and pressure is constant throughout the stream of water. The distance `h` vertically below the tap, where the cross-sectional area of the stream becomes `((2)/(3))A` is `(g=10m//s^(2))`

Options

A. 2 m
B. 1 m
C. 0.5 m
D. 4 m

Explanation

To solve the problem, we will follow these steps:

### Step 1: Apply the Equation of Continuity
The equation of continuity states that the product of the cross-sectional area and the velocity of the fluid remains constant along a streamline. Mathematically, it can be expressed as:

\[ A_1 v_1 = A_2 v_2 \]

Given:
- Initial cross-sectional area \( A_1 = A \)
- Initial velocity \( v_1 = 4 \, \text{m/s} \)
- Final cross-sectional area \( A_2 = \frac{2}{3}A \)
- Final velocity \( v_2 \) (to be determined)

Substituting the known values into the equation:

\[ A \cdot 4 = \frac{2}{3}A \cdot v_2 \]

### Step 2: Solve for \( v_2 \)
We can cancel \( A \) from both sides (assuming \( A \neq 0 \)):

\[ 4 = \frac{2}{3} v_2 \]

Now, multiply both sides by \( \frac{3}{2} \):

\[ v_2 = 4 \cdot \frac{3}{2} = 6 \, \text{m/s} \]

### Step 3: Apply Bernoulli's Equation
Bernoulli's equation relates the pressure, height, and velocity at two points along a streamline. The equation is given by:

\[ P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2 \]

Since the pressure is constant throughout the stream, we can simplify the equation:

\[ \rho g h_1 + \frac{1}{2} \rho v_1^2 = \rho g h_2 + \frac{1}{2} \rho v_2^2 \]

### Step 4: Rearranging the Equation
Let \( h = h_1 - h_2 \). Thus, we can express the equation as:

\[ g h = \frac{1}{2} (v_2^2 - v_1^2) \]

### Step 5: Substitute Known Values
Substituting \( v_1 = 4 \, \text{m/s} \) and \( v_2 = 6 \, \text{m/s} \):

\[ g h = \frac{1}{2} (6^2 - 4^2) \]

Calculating the squares:

\[ g h = \frac{1}{2} (36 - 16) \]

\[ g h = \frac{1}{2} \cdot 20 = 10 \]

### Step 6: Solve for \( h \)
Now substituting \( g = 10 \, \text{m/s}^2 \):

\[ 10 h = 10 \]

Dividing both sides by 10:

\[ h = 1 \, \text{m} \]

### Final Answer
The distance \( h \) vertically below the tap, where the cross-sectional area of the stream becomes \( \frac{2}{3}A \), is:

\[ \boxed{1 \, \text{m}} \]

---