- Home
- Standard 11
- Physics
9-1.Fluid Mechanics
medium
The diameter of an air bubble which was initially $2\,mm$, rises steadily through a solution of density $1750\,kg\,m\,m ^{-3}$ at the rate of $0.35\,cms ^{-1}$. The coefficient of viscosity of the solution is poise (in nearest integer). (the density of air is negligible).
A
$12$
B
$11$
C
$10$
D
$8$
(JEE MAIN-2022)
Solution
As the bubble is rising steadily the net force acting on it will be zero
(Because of density of air the value of mg can be neglected)
So $B = F \Rightarrow \frac{4 \pi}{3} R ^{3} \rho g =6 \pi \eta Rv$
Putting $R =1\,mm =10^{-3}\,m$
$\rho=1.75 \times 10^{3}\,kg / m ^{3}$
$g =10\,m / s ^{2}$
$v =0.35 \times 10^{-2}\,m / s$
$\eta=\frac{10}{9}=1.11$ SI unit $=11$ poise $( CGS )$
Standard 11
Physics
