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JEE Advanced 2023 Paper 1 · Physics · Fluid Mechanics, Elasticity and Surface Tension · medium

A uniform solid cylinder of mass 0.80 kg and cross-sectional area 40 cm^2 floats upright in water (density 1000 kg m^-3) in a wide tank. A light…

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A uniform solid cylinder of mass 0.80 kg0.80\ \mathrm{kg} and cross-sectional area 40 cm240\ \mathrm{cm^2} floats upright in water (density 1000 kg m−31000\ \mathrm{kg\,m^{-3}}) in a wide tank. A light vertical spring of force constant 60 N m−160\ \mathrm{N\,m^{-1}} joins the bottom of the cylinder to the floor of the tank; in equilibrium the spring is stretched and the cylinder is partly submerged. The cylinder is pushed slightly down and released. Take g=10 m s−2g=10\ \mathrm{m\,s^{-2}} and neglect the motion of the water and viscosity. The period of the small vertical oscillations is closest to
  1. A. 0.56 s0.56\ \mathrm{s}
  2. B. 0.89 s0.89\ \mathrm{s}
  3. C. 0.73 s0.73\ \mathrm{s}
  4. D. 1.15 s1.15\ \mathrm{s}
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Answer: (A) 0.56 s0.56\ \mathrm{s}

The spring joins the cylinder to the floor and is stretched, so it pulls the cylinder down. Displace the cylinder downward by xx: the spring is stretched less, so its downward pull falls by kxkx (an upward change in the net force), and the extra submerged length raises the buoyant force by ρgAx\rho gAx. Both changes are upward and proportional to xx, so the net restoring constant is keff=k+ρgA=60+1000×10×4×10−3=100 N m−1k_{eff}=k+\rho gA=60+1000\times10\times4\times10^{-3}=100\ \mathrm{N\,m^{-1}}. Hence T=2πm/keff=2π0.80/100=0.56 sT=2\pi\sqrt{m/k_{eff}}=2\pi\sqrt{0.80/100}=0.56\ \mathrm{s}. The wrong values come from omitting the spring (0.89 s0.89\ \mathrm{s}, only buoyancy), omitting buoyancy (0.73 s0.73\ \mathrm{s}, only the spring) or combining the two stiffnesses as if in series (1.15 s1.15\ \mathrm{s}); the two restoring effects act on the same body in the same direction, so they add.
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