When tube is rotated, liquid starts to flow radially outward and air in sealed arm is compressed. Let the shift of liquid be x as shown in figure.
Let cross-sectional area of tube be S. Initial volume of air, V0= Sa and initial pressure P0=10500Nm−2 Final volume, V=S(a−x) ∴ Final pressure, P=
P0V0
V
=
P0⋅a
(a−x)
or Pressure at B,P2=P+xρg=
P0a
(a−x)
+xpg Centripetal force required for circular motion of vertical column of height x of liquid is provided by reaction of the tube while that to horizontal length (I−x) is provided by excess pressure at B. Force exerted by pressure difference is F1=(PB−PA)S=(P2−P0) S=[
P0x
(a−x)
+xρg] Mass of horizontal arm AB of liquid is, m=S(I−x)ρ Radius of circular path traced by its centre of mass is r=x+
ℓ−x
2
=(
ℓ+x
2
) ∴ Centripetal force, F2=mω02r But F2=F1 Sρ(I−x)}{
ℓ+x
2
}ω02={
P0x
(a−x)
+xpg}S or x=.01m=1cm ∴ Length of air column in sealed arm=(a−x) =5cm Ans