First slide

Fluid dynamics

Question

Water from a tap emerges vertically downwards with an initial speed of 1.0 m s^-1. The cross-sectional area of the 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:

Easy

A

$5.0 \times 10^{- 4} m^{2}$
B

$1.0 \times 10^{- 5} m^{2}$
C

$5.0 \times 10^{- 5} m^{2}$
D

$2.0 \times 10^{- 5} m^{2}$

Solution

Velocity of stream at 0.15 m below the tap
$u^{2} = u^{2} + 2 gh = {(1)}^{2} + 2 \times 10 \times 0.15 = 4 \Rightarrow u = 2 m / s .$
From equation of continuity $A_{1} u_{1} = A_{2} u_{2} \Rightarrow A_{2} = A_{1} (\frac{u_{1}}{u_{2}}) = 10^{- 4} (\frac{1}{2}) = 5 \times 10^{- 5} m^{2} .$

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