What would be the rotaion rate (in rad/sec) if during rotation the water in the right leg (with a dimeter of d) decrease a distance h1=0.05m

 Q:-  What would be the rotaion rate (in rad/sec) if during rotation the water in the right leg (with a dimeter of d) decrease a distance h1=0.05m

Answer - 

From the figure :-

Rise of liquid in left leg is "y"

also,  Vloume of  water rise in left leg = Volume of liquid fall in right leg

                                       π(2d)^2 (y)/4    =   π(d)^2 (h1)/4

                                                   (2d)^2 y =  (d)^2 (0.05)

                                                               y = 0.0125 m

So,         Z2   =  0.2 + y = 0.2 + 0.0125 = 0.2125 m

and,        Z1  =  0.2 - 0.05   =  0.15 m

Also, We know that 

      Z2 - Z1  =  w^2[ (0.4)^2 -  (0.2)^2 ]/2g

     0.2125 - 0.15  = w^2[ 0.12]/2(9.81)

                       w = 3.19667 rad/sec

Hence , the  rotation rate is 3.19667 radian per second.