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JEE Main 2025 (22 Jan, Shift 2) · Viscosity & Stokes' Law

Viscosity & Stokes' Law question from JEE Main 2025 (22 Jan, Shift 2)

Q4JEE Main 2025 · 22 JAN · SHIFT 2Easy

Viscosity & Stokes' Law

A small rigid spherical ball of mass M is dropped in a long vertical tube containing glycerine. The velocity of the ball becomes constant after some time. If the density of glycerine is half of the density of the ball, then the viscous force acting on the ball will be (consider g as acceleration due to gravity)

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Answer: (c)

MgBF (viscous)constant v: F + B = Mg
  1. At constant velocity the net force is zero: Fv+B=MgF_v + B = Mg.
  2. Buoyancy B=VσgB = V\sigma g with σ=ρ/2\sigma = \rho/2, so B=Vρg2=Mg2B = \frac{V\rho g}{2} = \frac{Mg}{2}.
  3. Fv=Mg−Mg2=Mg2F_v = Mg - \frac{Mg}{2} = \frac{Mg}{2}
  4. So the answer is (c) Mg2\frac{Mg}{2}.
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