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JEE Main 2023 (24 Jan, Shift 2) · Viscosity & Stokes' Law

Viscosity & Stokes' Law question from JEE Main 2023 (24 Jan, Shift 2)

Q17JEE Main 2023 · 24 JAN · SHIFT 2Medium

Viscosity & Stokes' Law

A Spherical ball of radius 1mm and density 10.5 g/cc is dropped in glycerine of coefficient of viscosity 9.8 poise and density 1.5 g/cc. Viscous force on the ball when it attains constant velocity is 3696×10−x3696 \times 10^{-x} N. The value of xx is ________.

(Given, g = 9.8 m/s2^{2} and π=227\pi = \frac{22}{7})

Numerical answer type. Work it out, then open the answer.

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Answer: 7

  1. At constant velocity: Fv=W−B=43πr3(ρ−σ)gF_v = W - B = \frac{4}{3}\pi r^3(\rho - \sigma)g.
  2. r=10−3r = 10^{-3} m, ρ−σ=9 g/cc=9000 kg/m3\rho - \sigma = 9\text{ g/cc} = 9000\text{ kg/m}^3.
  3. Fv=43⋅227⋅10−9⋅9000⋅9.8=3696×10−7F_v = \frac{4}{3}\cdot\frac{22}{7}\cdot10^{-9}\cdot9000\cdot9.8 = 3696\times10^{-7} N
  4. So x=7x = 7.
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