Q157 / 180 · COMEDK 2025 (10 May Shift 2) · Physics
Water from a tap emerges vertically downwards with an initial speed of $1.0 \mathrm{~ms}^{-1}$. The cross-sectional area of the tap is $1 \mathrm{~cm}^2$. 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 cm below the tap is $\left(g=10 \mathrm{~ms}^{-2}\right)$
- A$0.05 \mathrm{~m}^2$
- B$5 \mathrm{~m}^2$
- Correct answerC$0.5 \mathrm{~cm}^2$
- D$0.2 \mathrm{~cm}^2$
Solution
Let the initial velocity at the tap be $v_1 = 1.0 \text{ ms}^{-1}$ and the cross-sectional area be $A_1 = 1 \text{ cm}^2$.
Let the velocity at a depth $h = 15 \text{ cm} = 0.15 \text{ m}$ be $v_2$ and the cross-sectional area be $A_2$.
Using the equation of motion $v_2^2 = v_1^2 + 2gh$, we have:
$v_2^2 = (1.0)^2 + 2 \times 10 \times 0.15 = 1 + 3 = 4$
$v_2 = 2 \text{ ms}^{-1}$
According to the equation of continuity for steady flow, $A_1 v_1 = A_2 v_2$.
$1 \text{ cm}^2 \times 1.0 \text{ ms}^{-1} = A_2 \times 2.0 \text{ ms}^{-1}$
$A_2 = \dfrac{1}{2} \text{ cm}^2 = 0.5 \text{ cm}^2$
Answer: $0.5 \mathrm{~cm}^2$
Let the velocity at a depth $h = 15 \text{ cm} = 0.15 \text{ m}$ be $v_2$ and the cross-sectional area be $A_2$.
Using the equation of motion $v_2^2 = v_1^2 + 2gh$, we have:
$v_2^2 = (1.0)^2 + 2 \times 10 \times 0.15 = 1 + 3 = 4$
$v_2 = 2 \text{ ms}^{-1}$
According to the equation of continuity for steady flow, $A_1 v_1 = A_2 v_2$.
$1 \text{ cm}^2 \times 1.0 \text{ ms}^{-1} = A_2 \times 2.0 \text{ ms}^{-1}$
$A_2 = \dfrac{1}{2} \text{ cm}^2 = 0.5 \text{ cm}^2$
Answer: $0.5 \mathrm{~cm}^2$
Hence, the correct answer is option: 'C'.