PrepRung
Sign inSign up free

JEE Advanced 2024 Paper 1 · Physics · Fluid Mechanics, Elasticity and Surface Tension · hard

A vertical open cylindrical vessel of inner radius 10 cm and height 31 cm contains water to a depth of 20 cm. It is rotated about its vertical axis…

One or more options correct

A vertical open cylindrical vessel of inner radius 10 cm10\ \mathrm{cm} and height 31 cm31\ \mathrm{cm} contains water to a depth of 20 cm20\ \mathrm{cm}. It is rotated about its vertical axis at a steady angular speed ω\omega that is increased slowly from zero. Take g=9.8 m s−2g=9.8\ \mathrm{m\,s^{-2}} and water density 1000 kg m−31000\ \mathrm{kg\,m^{-3}}. Which of the following statements is/are correct?
  1. A. At ω=20 rad s−1\omega=20\ \mathrm{rad\,s^{-1}} the water level on the axis is 10.2 cm10.2\ \mathrm{cm} above the base.
  2. B. At ω=20 rad s−1\omega=20\ \mathrm{rad\,s^{-1}} the pressure at the base at the wall exceeds that on the axis by 4000 Pa4000\ \mathrm{Pa}.
  3. C. Water begins to spill over the rim at ω≈20.8 rad s−1\omega\approx20.8\ \mathrm{rad\,s^{-1}}.
  4. D. The free surface just touches the base on the axis at ω=28 rad s−1\omega=28\ \mathrm{rad\,s^{-1}}.
Show answer and solution

Answer: C

In steady rotation the free surface is a paraboloid h(r)=hc+ω2r22gh(r)=h_c+\dfrac{\omega^2r^2}{2g} and volume conservation puts the mean level at the original depth, so before spilling hc=h0−ω2R24gh_c=h_0-\dfrac{\omega^2R^2}{4g} and hrim=h0+ω2R24gh_{rim}=h_0+\dfrac{\omega^2R^2}{4g}. At ω=20\omega=20: ω2R2/4g=0.102 m\omega^2R^2/4g=0.102\ \mathrm{m}, so the axis level is 9.80 cm9.80\ \mathrm{cm} (A wrong, 10.210.2 cm is the drop) and the rim level 30.2 cm<31 cm30.2\ \mathrm{cm}<31\ \mathrm{cm} (no spill). The base pressure difference is ρg(hrim−hc)=ρω2R2/2=2000 Pa\rho g(h_{rim}-h_c)=\rho\omega^2R^2/2=2000\ \mathrm{Pa}, not 40004000 (B wrong). Spilling starts when hrim=0.31h_{rim}=0.31: ω2R2/4g=0.11 m\omega^2R^2/4g=0.11\ \mathrm{m}, ω=4(9.8)(0.11)/0.01=20.8 rad s−1\omega=\sqrt{4(9.8)(0.11)/0.01}=20.8\ \mathrm{rad\,s^{-1}} (C correct). Beyond this the volume is not conserved, so the axis touches the base when the rim stays at 0.31 m0.31\ \mathrm{m}: ω2R2/2g=0.31\omega^2R^2/2g=0.31, giving ω=24.7 rad s−1\omega=24.7\ \mathrm{rad\,s^{-1}}. The value 2828 comes from ignoring the spill (D wrong).
Report a mistake

Original question written in the JEE Advanced 2024 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake