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Water from a tap emerges vertically down with an initial speed of ${1.0ms}^{- 1}$ . The cross sectional area of tap is $10 \times 10^{- 5} m^{2}$ . Assume that the pressure is constant throughout the stream of water, and that the flow is a steady, the cross-sectional area of the steam 0.15m below the tap is ______ (g = 10m/s²)

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5.0 x 10^-4 m²

1.0 x 10^-5 m²

5.0 x 10^-5 m²

2.0 x 10^-5 m²

answer is C.

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Detailed Solution

V² = u² + 2gh

V²=1²+2(10)(0.15)

v=2 m/s

and A₁V₁ = A₂V₂

$10 \times 10^{- 5} \times 1 = A_{2} \times 2 A_{2} = 5 \times 10^{- 5}$

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