question_answer

An air bubble of radius \[{{10}^{-2}}\] m is rising up at steady rate of \[2\times {{10}^{-3}}m{{s}^{-1}}\] through a liquid of density \[1.5\times {{10}^{3}}\] \[kg{{m}^{-3}}\], the coefficient of viscosity neglecting the density of air, will be (Take \[g=10\,m{{s}^{-2}}\])

A)  \[23.2\] units                

B)  \[83.5\] units

C)  334 units                     

D)  167 units

Correct Answer: D

Solution :

Let a bubble of radius r and density p is rising up in a liquid whose density of a and coefficient of viscosity \[\eta \]. Since, bubble is moving up with constant velocity v, there is no acceleration in it, the net force acting on it must be zero. \[h=\frac{2}{9}\,\frac{{{r}^{2}}(\rho -\sigma )}{v}g\] Given \[v=-2\times {{10}^{3}}\,m{{s}^{-1}},\,\,r={{10}^{-2}}\,m\]             \[\rho =0,\,\,\sigma =1.5\,\times {{10}^{3}}\,k{{g}^{-3}},\,\,g=10\,m{{s}^{-2}}\] \[\eta \,=\frac{2\times {{({{10}^{-2}})}^{2}}\,\times (-1.5\times {{10}^{3}})\times 10}{9\times (-2\times {{10}^{-3}})}\] \[=\frac{3}{18\times {{10}^{-3}}\,}=\frac{1}{6}\times {{10}^{3}}\,=0.167\,\times {{10}^{3}}\]167 units