AravindG
A cylindrical tank of height 0.4 m is open at the top and has a diameter 0.16 m.Water is filled in it upto a height of 0.16 m .Calculate the time taken to empty the tank through a hole of radius 5x10^-3m in its bottom?
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AravindG
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@Vincent-Lyon.Fr , @ash2326 , @waterineyes
Vincent-Lyon.Fr
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First, use Torricelli's law to determine speed of water in the hole and rate of flow.
\(v=\sqrt{2gh}\)
AravindG
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then?
Vincent-Lyon.Fr
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Is the answer 45.80 s ?
AravindG
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i didnt get tht :O
AravindG
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i got 2.8 :P
Vincent-Lyon.Fr
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How?
AravindG
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wait
RaphaelFilgueiras
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h is variable
AravindG
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hmm
AravindG
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then how ?
Vincent-Lyon.Fr
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Total volume is 3.217 L, right?
AravindG
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i got 3.14x10^-4
Vincent-Lyon.Fr
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Try again. radius is 0.08m and height is 0.16m
AravindG
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3.21x10^-3
Vincent-Lyon.Fr
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ok
AravindG
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so?
Vincent-Lyon.Fr
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now escape velocity is \(v=\sqrt{2gh}\)
Vincent-Lyon.Fr
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and discharge is v multiplied by area of hole.
AravindG
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v is velocity ryt?
Vincent-Lyon.Fr
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yep
AravindG
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so how do we bring in volume?
Vincent-Lyon.Fr
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time taken is: volume / discharge
you should find 22.9s
AravindG
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hmm
AravindG
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but isnt h varying?
Vincent-Lyon.Fr
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Yes, but we can get round this difficulty
Vincent-Lyon.Fr
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Do you find the same values?
AravindG
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how/?
Vincent-Lyon.Fr
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initial escape velocity is v=root(2g hmax)
discharge is v multiplied by area of hole.
time taken is: volume / discharge
you should find 22.9s
This is the time it would take to empty the tank if initial discharge was constant.
Real duration, when h varies, is twice that time.