AP EAMCET · PHYSICS · Mechanical Properties of Fluids
An air bubble of radius \(1 \mathrm{~cm}\) rises from the bottom portion through a liquid of density \(1.5 \mathrm{~g} / \mathrm{cc}\) at a constant speed of \(0.25 \mathrm{~cm} \mathrm{~s}^{-1}\). If the density of air is neglected, the coefficient of viscosity of the liquid is approximately, (In Pas) :
- A 13000
- B 1300
- C 130
- D 13
Answer & Solution
Correct Answer
(C) 130
Step-by-step Solution
Detailed explanation
\(v=\frac{2}{9} \frac{r^2 \rho g}{\eta}\) \(\Rightarrow \eta=\frac{2}{9} \cdot \frac{r^2 \rho g}{v}\) \(=\frac{2}{9} \frac{\left(1 \times 10^{-2}\right)^2 \times\left(1.5 \times 10^3\right) \times 9.8}{0.25 \times 10^{-2}}\) \(=130\) pa-s.
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