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

 

 

 

 

  • Option 1)

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

  • Option 2)

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

  • Option 3)

    5\times 10^{-4}m^{2}

  • Option 4)

    1\times 10^{-5}m^{2}

Answers (1)

best_answer

 

using Bernoulli's theorem we get

Velocity at point 2

V_{2}^{2}-U_{1}^{2}=2gh

V_{2}=\sqrt{V_{1}^{2}+2gh}

at\, \, \, \, h= 0.15m

V_{2}=2m/s

now use equation of continuity

A_{1}V_{_{1}}=A_{2}V_{2}

\left ( 10^{-4} \right )\left ( 1 \right )=A_{2}\times 2

A_{2}=.5\times 10^{-4}

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


Option 1)

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

Option 2)

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

Option 3)

5\times 10^{-4}m^{2}

Option 4)

1\times 10^{-5}m^{2}

Posted by

Plabita

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