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NEET 2023 · Viscosity & Stokes' Law

Viscosity & Stokes' Law question from NEET 2023

Q13NEET 2023Easy

Viscosity & Stokes' Law

The viscous drag acting on a metal sphere of diameter 1 mm1\text{ mm}, falling through a fluid of viscosity 0.8 Pa0.8\text{ Pa} s with a velocity of 2 m s−12\text{ m s}^{-1} is equal to:

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Answer: (a)

  1. Stokes' law: F=6πηrvF = 6\pi\eta r v, with r=0.5 mm=5×10−4r = 0.5\text{ mm} = 5\times10^{-4} m.
  2. F=6π(0.8)(5×10−4)(2)=4.8π×10−3F = 6\pi(0.8)(5\times10^{-4})(2) = 4.8\pi\times10^{-3}
  3. F≈15×10−3F \approx 15\times10^{-3} N
  4. So the answer is (a) 15×10−315\times10^{-3} N.
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