Length of a horizontal arm of ends of both the vertical arms pressure `P_(0) = 10^(5) N//m^(2)`. A liquid of density `rho = 10^(3) kg //m^(3)` is poured in the tube such that liquid just fills the horizontal part of the tube as a shown in Fig. Now one end of the open ends is sealed and the tube is then rotated about a vertical axis passing through the other vertical arm with angular speed `omega_(0)=sqrt(40//3)rad//sec`. If length of each vertical arm is `a = 1 m` and in the sealed end liquid rises to a height `y = 1//2 m`, find pressure in the sealed tube during rotation. (in `xx 10^(5) N//m^(2))`

Answer

1

Explanation

When the tube is rotated, the liquid starts to flow radially outwards and air in sealed arm is compressed. Let the shift of the liquid be `y` shown in figure.

Let the cross sectional area of the tube be `S`. Here element of length `dx` is rotating with the tube. From the free body diagram of the element we can write
`(P+dP)S-PS=dmomega^(2)x=(rhoSdx)omega^(2)x`
`dPS=rhoomega^(2)Sxdx`
Here the pressure diference beween points `A` and `B` can be given by integrating the pressure difference across an element of width `dx`, which is given as
`dP=rhoomega^(2)xdx`
Now integrating from `A` to `B` we get
`P_(B)-P_(A)=int_(v)^(L)rhoomega^(2)xdx=(rhoomega^(2))/2(L^(2)-y^(2))`
`impliesP_(B)=p_(0)+(rhoomega^(2))/2(L^(2)-=y^(2))`
The pressure at point `C` can be given as
`P_(C)=P_(a)-y_(rho)g`
At point `A`, pressure is atmosphere. Thus, we have
`P_(C)=(rhoomega^(2))/2(L^(2)-y^(2))+P_(0)-y_(r)hog=10^(5)N//m`