Length o f a horizontal arm o f a U -tube is L = 1 m and ends of both the vertical arms are open to atmospheric pressure P0=105N/m2. A liquid of density ρ=103kg/m3 is poured in the tube such that liquid just fills the horizontal part of the tube as shown in figure. 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 ω0=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  105N/m2

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 as 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=dmω2x=(ρSdx)ω2xdPS=ρω2SxdxHere the pressure difference between points A and B can be given by integrating the pressure difference across an element of width dx, which is given asdP=ρω2xdxNow integrating from A to B, we getPB−PA=∫vL ρω2xdx=ρω22L2−y2⇒pB=p0+ρω22L2−y2The pressure at point C can be given asPC=Pa−yρgAt point A, pressure is atmospheric . Thus, we havePC=ρω22L2−y2+P0−yρg=105N/m