Get Answers to all your Questions

header-bg qa

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}}









 

Posted by

Shailly goel

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE