Question

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?

Answer 1

@Vincent-Lyon.Fr , @ash2326 , @waterineyes

Answer 2

First, use Torricelli's law to determine speed of water in the hole and rate of flow.

\(v=\sqrt{2gh}\)

Answer 3

then?

Answer 4

Is the answer 45.80 s ?

Answer 5

i didnt get tht :O

Answer 6

i got 2.8 :P

Answer 7

How?

Answer 8

wait

Answer 9

h is variable

Answer 10

hmm

Answer 11

then how ?

Answer 12

Total volume is 3.217 L, right?

Answer 13

i got 3.14x10^-4

Answer 14

Try again. radius is 0.08m and height is 0.16m

Answer 15

3.21x10^-3

Answer 16

ok

Answer 17

so?

Answer 18

now escape velocity is \(v=\sqrt{2gh}\)

Answer 19

and discharge is v multiplied by area of hole.

Answer 20

v is velocity ryt?

Answer 21

yep

Answer 22

so how do we bring in volume?

Answer 23

time taken is: volume / discharge
you should find 22.9s

Answer 24

hmm

Answer 25

but isnt h varying?

Answer 26

Yes, but we can get round this difficulty

Answer 27

Do you find the same values?

Answer 28

how/?

Answer 29

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.