An air bubble of diameter 6 mm rises steadily through a solution of density 1750kg / m3 at the rate of 0.35 cm/s. The co-efficient of viscosity of the solution (neglect density of air) is ____Pas (given, g = 10ms-2)
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Stokes’ Law relates the drag force to viscosity and terminal velocity
Correct Answer: 10
Solution and Explanation
Step 1: Apply Stokes’ Law for terminal velocity.- For uniform velocity, net force = 0.- Buoyant force = 6πηrv.
$\frac{4}{3}\pi r^3 \rho g = 6 \pi \eta r v$.
Step 2: Solve for η.
$\eta = \frac{2r^2 \rho g}{9v}$ .- Given r = $\frac{6}{2} \times 10^{-3}$ = 3 × 10-3m, ρ = 1750 kg/m3, v = 0.35 × 10-2 m/s:
$\eta = \frac{2 \cdot (3 \times 10^{-3})^2 \cdot 1750 \cdot 10}{9 \cdot 0.35 \times 10^{-2}}$.
η = 10 Pas.
Final Answer: The coefficient of viscosity is 10 Pas
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