JEE Advanced 2020 Full Test 2 Paper 2

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 ‌