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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?

Physics
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First, use Torricelli's law to determine speed of water in the hole and rate of flow. \(v=\sqrt{2gh}\)
then?

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Other answers:

Is the answer 45.80 s ?
i didnt get tht :O
i got 2.8 :P
How?
wait
h is variable
hmm
then how ?
Total volume is 3.217 L, right?
i got 3.14x10^-4
Try again. radius is 0.08m and height is 0.16m
3.21x10^-3
ok
so?
now escape velocity is \(v=\sqrt{2gh}\)
and discharge is v multiplied by area of hole.
v is velocity ryt?
yep
so how do we bring in volume?
time taken is: volume / discharge you should find 22.9s
hmm
but isnt h varying?
Yes, but we can get round this difficulty
Do you find the same values?
how/?
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.

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