IIT JEE , AIEEE Forum - Physics , Mechanics - fluids(help me outttttttt)-2

nick

now there is a U tube,both the arms are 10cm long,

the water at the bottom is 20cm(i.e. perhaps the length of the horizontal arm).

one end is closed with a cork,and it is rotated about the other end(open one) with ang. velocity w, and the water in the right arm rises to 5cm.

find w,g=9.8m/s^2 density of water is 10^3

the ans to this is 8.97 radians

guyssssssssss plss try this and tell all the forces,

also tell me how will water rise when it is raotated I mean even if u close one end water will rise !!! would'nt it???

elessar_iitkgp

Let the area of cross section be A
Consider a fluid element in the horizontal portion, ata distance x from the axis of rotation and of length dx. Then centrifugal force on the element is (

Adx)

² x.
Total centrifugal force on the horizontal fluid = ₀

^r

A

² xdx. =

A

² r²/2
where r= 20 cm

Now as the water rises to h (=5 cm), there is pressure imbalance on the horizontal section. This counterbalances the centrifugal force

A

² r²/2 =

ghA

= (1/r)

2gh

nick

hey ELESSAR gr8 work

but u have equated force to pressure for which u have to divide f by a,hence removing that term

also tell me what is r and then the ans. and i rate u

TRY Q2 AND Q3 OF MY OTHER TOPIC ON FLUID,PLSSS

and tell me wont the water rise if i only close one end????

elessar_iitkgp

Thanx for correcting me.
r = 20cm, the length of the horizontal section

On calculating, I am getting 3.5

2 = 4.949

nick

but the ans. is 8.97

it is a Q from BRILLIANTS TUTORIALS

plss try the other Q as well

joyfrancis

dy/dx = w²x/g (using tan

=a/g)

on integrating we get

y=w²x²/2g.

so,

0.05 = w²(0.2)²/2g

w = 5

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