## AravindG 3 years ago 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?

1. AravindG

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

2. Vincent-Lyon.Fr

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

3. AravindG

then?

4. Vincent-Lyon.Fr

Is the answer 45.80 s ?

5. AravindG

i didnt get tht :O

6. AravindG

i got 2.8 :P

7. Vincent-Lyon.Fr

How?

8. AravindG

wait

9. RaphaelFilgueiras

h is variable

10. AravindG

hmm

11. AravindG

then how ?

12. Vincent-Lyon.Fr

Total volume is 3.217 L, right?

13. AravindG

i got 3.14x10^-4

14. Vincent-Lyon.Fr

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

15. AravindG

3.21x10^-3

16. Vincent-Lyon.Fr

ok

17. AravindG

so?

18. Vincent-Lyon.Fr

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

19. Vincent-Lyon.Fr

and discharge is v multiplied by area of hole.

20. AravindG

v is velocity ryt?

21. Vincent-Lyon.Fr

yep

22. AravindG

so how do we bring in volume?

23. Vincent-Lyon.Fr

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

24. AravindG

hmm

25. AravindG

but isnt h varying?

26. Vincent-Lyon.Fr

Yes, but we can get round this difficulty

27. Vincent-Lyon.Fr

Do you find the same values?

28. AravindG

how/?

29. Vincent-Lyon.Fr

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.