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Water from a tap emerges vertically downwards with an initial speed of $5.0 {ms}^{- 1}$ . The cross-sectional area of tap is $10^{- 4} m^{2}$ . Assume that the pressure is constant throughout the stream of water and that the flow is steady, the cross- sectional area of stream 3.75 m below the tap is:

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

$1.0 \times 10^{- 4} m^{2}$

$5.0 \times 10^{- 3} m^{2}$

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

answer is C.

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Detailed Solution

$u = 5 .0 {ms}^{1}, v^{2} = u^{2} + 2 as$

$v^{2} = (5.0)^{2} + 2 (10) (3.75) = 25 + 75$

$v = 10 {ms}^{- 1}$

$A_{2} V = A_{1} u \Rightarrow A_{2} = 10^{- 4} \times \frac{5 .0}{10}$

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

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