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The diameter of an air bubble which was initially $2 \mathrm{~mm}$ , rises steadily through a solution of density $1750 \mathrm{~kg} \mathrm{~m}^{-3}$ at the rate of $0.35 \: \mathrm{cms}^{-1}$ . The coefficient of viscosity of the solution is_________ poise (in nearest integer). (the density of air is negligible).
Option: 1
11

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

If it rises steadily that implies the net force on the bubble is zero
$\mathrm{mg=0}\mathrm{\left ( air\: density\: is\: negligible \right )}$

$\mathrm{F_{v}=viscous\: drag\: force }$

$\mathrm{=6\pi \eta rv.3\pi \eta dv}$

$\mathrm{d\rightarrow diameter\: of\: bubble}$

$\mathrm{B=mg+F_{v}}$

$\mathrm{B =m g+F_v }$

$\mathrm{\frac{4}{3} \pi r^3 g \rho =0+3 \pi \eta d v }$

$\mathrm{r =\frac{d}{2} }$

$\mathrm{\frac{\pi d^3 g \rho }{6} =3 \pi \eta d v }$

$\mathrm{\frac{d^2 \rho g}{18} =\eta v }$

$\mathrm{\eta =\frac{d^2 \rho 9}{18 V} }$

$\mathrm{=\frac{\left(2 \times 10^{-3}\right)^2 \times 1750 \times 10}{18 \times 35 \times 10^{-4}} }$

$\mathrm{=\frac{2000}{18} \times 10^{-1} \frac{\mathrm{NS}}{\mathrm{m}^2} }$

$\mathrm{\eta =11\frac{\mathrm{Ns}}{\mathrm{m}^2}}$

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