Water from a tap emerges vertically downwards with an initial speed of \(1.0\,ms^{-1}.\) The cross-sectional area of the tap is \(10^{-4}\,m^2.\) Assume that the pressure is constant throughout the stream of water and that flow is streamlined. The cross-sectional area of the stream, \(0.15\,m\) below the tap would be: (take \(g = 10\,ms^{-2}\) )

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Correct Answer

(B) \(5\times 10^{-5}\,m^2\)

Step-by-step Solution

Detailed explanation

\({10^{ - 4}}\, \times \,1\, = \sqrt {{{(1)}^2} + 2 \times 10 \times 0.15 \times } A\)

\(A = \frac{{{{10}^{ - 4}}}}{2} = 5 \times {10^{ - 5}}\)

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