Question

An air bubble of $1cm$ radius is rising at a steady rate of $2.00mm{s}^{-1}$ through a liquid of density $1.5gc{m}^{-3}$. Neglect density of air. If $g=1000cm{s}^{-2}$, then the coefficient of viscosity of the liquid is

• A. $16.6\times {10}^{3}poise$
  
• B. $166\times {10}^{3}poise$
  
• C. $0.166\times {10}^{3}poise$
  
• D. $1.66\times {10}^{3}poise$
  

Answer 1

The correct option is D $1.66\times {10}^{3}poise$


Given,
$r=1cm$
$v=2.00mm{s}^{-1}=0.2cm{s}^{-1}$
${\rho }_l=1.5gc{m}^{-3}$
$g=1000cm{s}^{-2}$

According to given condition,

$B=6\pi \eta rv$
$\Rightarrow \frac{4}{3}\pi r^3{\rho }_{l}g=6\pi \eta rv$

Put the values and get,

$\eta =1.66\times {10}^{3}poise$
Final Answer: $\left(b\right)$