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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 $1 0^{- 4} m^{2}$ . Assume that the pressure is constant throughout the stream of water and that the flow is steady. The cross-sectional area of the stream 0.15 m below the tap is:

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$2 . 0 \times 1 0^{- 5} m^{2}$

$5 . 0 \times 1 0^{- 4} m^{2}$

$1.0 \times 1 0^{- 5} m^{2}$

$5.0 \times 1 0^{- 5} m^{2}$

answer is C.

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Detailed Solution

From Bernoulli's principle,

$Δ P = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$

$\Rightarrow ρgh = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2})$

$\Rightarrow 10^{3} \times 10 \times 0.15 = \frac{1}{2} \times 10^{3} (v^{2} - 1)$

$\Rightarrow v = 2 m / s$

Using equation of continuity,

$A_{1} v_{1} = A_{2} v_{2}$

$\Rightarrow 10^{- 4} \times 1 = A_{2} \times 2$

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

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