I don't whether it's solvable or not. It came as a thought to me while I was filling water in bottles.
So Here I " Go " :
A cylindrical tank of radius " r " contains water upto height " H " . There is a hose at height " h " from the bottom. At what angle should the tank be inclined so as to double the water output from the hose?
N.B. Assume necessary simplifications as I don't know what else this problem may require. Answer Unknown!
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4 Answers
49waah bhai waah...waah bhai waah!! :D :D :D
easy hai...ho jayega! :D :D
Suppositions: radius r..
Water present till height h
Total height H
hole present at distance L from the bottom.
CONSIDERING A DIAMETRICAL SLICE, the area covered by water must be equal...
net area in fig.1 = 2rh
area of triangular part = 2r2tan@
2rx+2r2tan@=2rh
or, x = h - r tan@
so, length of liquid in contact wid wall on the hole side = h + r tan@
so, length of liquid in contact wid wall above orifice = h + r tan@ - L
so, heigth of water above orifice = (h + r tan@ - L) cos@
Initial velocity of efflux(v1) = √2g(h-L)
final velocity of efflux(v2) = √2g(h + r tan@ - L)cos@
given v2=2v1
(h + r tan@ - L) cos@=4(h-L)
HO GAYA!!!!!!!!!!!!!!!!
HIP HIP HURRAH!! :D :D
49aur bhi suppose kiya hoon: orifice ka area bahut kam hai w.r.t. πr2.
Non-viscous flow.
Stream-lined Flow..!
Real life situation solve karunga agar engineering me sikhaye toh! :P :P :P
71Yeah.. I knew it was solvable but at Physics forums (PF Mentor said that it will have negligible effect and the other blame he made at me that it is a homework problem)
71Waise meri geometry thodi weak hai isliye nai kar paya!!