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NVAEasyJEE 2023Viscosity & Stoke's Law

JEE Physics 2023 Question with Solution

An air bubble of diameter 6 mm6\,mm rises steadily through a solution of density 1750 kg/m31750\,kg/m^3 at the rate of 0.35 cm/s0.35\,cm/s. The coefficient of viscosity of the solution (neglect density of air) is _____,Pas (given, g=10 m/s2g = 10\,m/s^2)):

Answer

Correct answer:10

Step-by-step solution

Standard Method

Given: diameter of the air bubble =6 mm= 6\,mm, so radius r=3×10−3 mr = 3 \times 10^{-3}\,m; density of solution ρ=1750 kg/m3\rho = 1750\,kg/m^3; terminal speed v=0.35×10−2 m/sv = 0.35 \times 10^{-2}\,m/s; acceleration due to gravity g=10 m/s2g = 10\,m/s^2.

Find: The coefficient of viscosity η\eta.

Using Stokes’ Law for terminal velocity, for uniform velocity net force is zero. The effective weight of the bubble is balanced by viscous drag:

43πr3ρg=6πηrv\frac{4}{3}\pi r^3 \rho g = 6\pi \eta r v

Detailed Calculation

Now solve for η\eta:

η=2r2ρg9v\eta = \frac{2r^2 \rho g}{9v}

Substitute the given values:

η=2⋅(3×10−3)2⋅1750⋅109⋅0.35×10−2\eta = \frac{2 \cdot (3 \times 10^{-3})^2 \cdot 1750 \cdot 10}{9 \cdot 0.35 \times 10^{-2}}

On evaluation,

η=10 Pa s\eta = 10\,\text{Pa s}

Therefore, the coefficient of viscosity is 10 Pa s10\,\text{Pa s}.

Common mistakes

  • Using the diameter directly in Stokes’ formula instead of the radius. The formula requires rr, so use r=3×10−3 mr = 3 \times 10^{-3}\,m, not 6×10−3 m6 \times 10^{-3}\,m.

  • Not converting the speed from 0.35 cm/s0.35\,cm/s to SI units. It must be written as 0.35×10−2 m/s0.35 \times 10^{-2}\,m/s before substitution.

  • Ignoring that the bubble rises with terminal velocity and applying an acceleration-based equation. Here the speed is steady, so net force is zero and Stokes’ Law is used.

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