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MHT CET 2026 (11 Apr, Shift 1) · Viscosity & Stokes' Law

Viscosity & Stokes' Law question from MHT CET 2026 (11 Apr, Shift 1)

Q2MHT CET 2026 · 11 APR · SHIFT 1Medium

Viscosity & Stokes' Law

A spherical ball of radius 11 mm and density 10.510.5 g/cc is dropped in glycerine of coefficient of viscosity 9.89.8 poise and density 1.51.5 g/cc. Viscous force on the ball when it attains constant velocity is 3696×10−x3696 \times 10^{-x} N. The value of is

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

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Answer: (c)

  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=10.5−1.5=9 g/cc=9000 kg/m3\rho - \sigma = 10.5 - 1.5 = 9\text{ g/cc} = 9000\text{ kg/m}^3.
  3. Fv=43⋅227⋅10−9⋅9000⋅9.8=3.696×10−4=3696×10−7F_v = \frac{4}{3}\cdot\frac{22}{7}\cdot10^{-9}\cdot9000\cdot9.8 = 3.696\times10^{-4} = 3696\times10^{-7} N
  4. So x=7x = 7, answer (c) 77.
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