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Water from a tap emerges vertically downward with an initial speed of 1m/s. The cross-sectional area of tap is 10^–4 m². Assume that the pressure is constant throughout the stream of water and that the flow is steady the cross - sectional area of the stream 0.15 m below the tap is (g = 10m/s²)

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10^–5 m²

4 x 10^–5 m²

2 x 10^–5 m²

5 x 10^–5 m²

answer is D.

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Detailed Solution

$A_{1} V_{1} = A_{2} \sqrt{V_{1}^{2} + 2 gh}$

$(10^{- 4}) (1) = A_{2} (\sqrt{1^{2} + (2 \times 10 \times 0.15)}) \Rightarrow A_{2} = 5 \times 10^{- 5} m^{2}$

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