Question

Water from a tap emerges vertically downwards with an initial speed of \( 1.0 \, \text{ms}^{-1} \). The cross-sectional area of the tap is \( 1 \, \text{cm}^2 \). Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream \( 15 \, \text{cm} \) below the tap is \( (g = 10 \, \text{ms}^{-2}) \)

• A. \( 0.5 \, \text{cm}^2 \)
• B. \( 5 \, \text{m}^2 \)
• C. \( 0.2 \, \text{cm}^2 \)
• D. \( 0.05 \, \text{m}^2 \)

Hint:

For steady flow, use the continuity equation \( A_1v_1 = A_2v_2 \). As water falls down, its speed increases and cross-sectional area decreases.

Answer 1

The Correct Option is A

Step 1: Write the given values.
Initial speed of water is:
\[ v_1 = 1.0 \, \text{ms}^{-1} \]
Initial area of tap is:
\[ A_1 = 1 \, \text{cm}^2 \]
Distance below the tap is:
\[ h = 15 \, \text{cm} = 0.15 \, \text{m} \]

Step 2: Find the speed of water after falling through \( 0.15 \, \text{m} \).
Using equation of motion:
\[ v_2^2 = v_1^2 + 2gh \]

Step 3: Substitute the values.
\[ v_2^2 = (1)^2 + 2(10)(0.15) \]
\[ v_2^2 = 1 + 3 \]
\[ v_2^2 = 4 \]

Step 4: Calculate final speed.
\[ v_2 = 2 \, \text{ms}^{-1} \]

Step 5: Apply equation of continuity.
For steady flow of water:
\[ A_1v_1 = A_2v_2 \]

Step 6: Calculate the cross-sectional area below the tap.
\[ A_2 = \frac{A_1v_1}{v_2} \]
\[ A_2 = \frac{1 \times 1}{2} \]
\[ A_2 = 0.5 \, \text{cm}^2 \]

Step 7: Final Answer.
Therefore, the cross-sectional area of the stream is:
\[ \boxed{0.5 \, \text{cm}^2} \]