24Form here we get
PP=Po+12Ïω2(L/3)2=Po
=>ω=\frac{1}{L} \sqrt{\frac{18}{\rho }}
Simialrly form PQ
we get ω=\frac{1}{2L} \sqrt{\frac{18}{\rho }}
=> min ω=\frac{1}{L} \sqrt{\frac{18}{\rho }}
20 Answers
24
24@ philip..
excess pressure force due to rotation will balanec atmospheric pressure from top
1Eureka can you explain a situation when waters are only in the vertical tubes.....
what vertical force in the tubes will balance the pressure then ?
11This is the trouble with some FIITJEE qsns!
What's the trouble in clearly stating a qsn or illustrating the situation with a dia????
REALLY FRUSTRATING!
24same doubt as soumik in post #3...
question isnt clear...
see i can see two possible ways of solving it
1) Water separate means that we ahve water in vertical columns only..
we can find angular speed for this one by simply applying pressure variation eqn along rotating tube
2) Water separate means two separate films formed
we use surface tension eqn as used by karna......
11What do u mean actually by saying water separates out?
What's the situation? No water in the horizontal position or what......The qsn was not clear to me....
1oh no xyz .. not like this
we have to consider only centrifugal and pressure forces
1surface tension may be given .i may have missed some part.........anyone who gave ftse might help
24i think what xyz is doing is right.....thats the only visbile way to solve thsi one..
3there must be some force binding the water molecules in order to remain dat in liquid.Since no information abt the forces is given,i guessed so i hide it
1@ msp there is no viscousity................due to rotation water will experience a centrifugal force and then it will seperate but do not know how to proceed
3i think At minimum value the centripetal force just exceeds the viscous force.
24If water separates situation means that no water in horz tube..then ques is easy...
otherwise very tuff