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JEE Main 2018 · Physics · Fluid Mechanics, Elasticity and Surface Tension · easy

A cube of side 10 cm and density 880 kg m^-3 floats in a tank holding water (1000 kg m^-3) with a layer of oil (800 kg m^-3) on top that is deep…

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A cube of side 10 cm10\ \mathrm{cm} and density 880 kg m−3880\ \mathrm{kg\,m^{-3}} floats in a tank holding water (1000 kg m−31000\ \mathrm{kg\,m^{-3}}) with a layer of oil (800 kg m−3800\ \mathrm{kg\,m^{-3}}) on top that is deep enough to cover the cube completely. The depth of the cube that lies in the water is
  1. A. 8 cm8\ \mathrm{cm}
  2. B. 2 cm2\ \mathrm{cm}
  3. C. 4 cm4\ \mathrm{cm}
  4. D. 6 cm6\ \mathrm{cm}
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Answer: (C) 4 cm4\ \mathrm{cm}

Let the depth in water be xx and in oil (10−x)(10-x) cm; the cross-section AA cancels. Equilibrium: weight == total buoyancy, so 880(10)=1000 x+800(10−x)880(10)=1000\,x+800(10-x) in consistent units. This gives 8800=8000+200x8800=8000+200x, hence x=4 cmx=4\ \mathrm{cm}. Equivalently, the fraction in water is ρc−ρoilρw−ρoil=80200=0.4\dfrac{\rho_c-\rho_{oil}}{\rho_w-\rho_{oil}}=\dfrac{80}{200}=0.4. Treating the cube as floating in water alone would give 8.8 cm8.8\ \mathrm{cm}.
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