Q135 / 160 · AP EAMCET 2006 · Physics

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
  • Correct answer
    C
    130
  • D
    13

Solution

$v=\frac{2}{9} \frac{r^2 \rho g}{\eta}$ $\Rightarrow \eta=\frac{2}{9} \cdot \frac{r^2 \rho g}{v}$ $=\frac{2}{9} \frac{\left(1 \times 10^{-2}\right)^2 \times\left(1.5 \times 10^3\right) \times 9.8}{0.25 \times 10^{-2}}$ $=130$ pa-s.

Hence, the correct answer is option: 'C'.

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