question_answer

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

A)  23.2 units           

B)  83.5 units           

C)         334 units       

D)         167 units

Correct Answer: D

Solution :

Key Idea: Since air bubble is moving up with a constant velocity, there is no acceleration in it. Let bubble of radius r and density p is falling in a liquid whose density is a and coefficient of viscosity \[\eta .\] It attains a terminal velocity due to two forces effective force acting downward                 \[=V(\rho -\sigma )g=\frac{4}{3}\pi {{r}^{3}}(\rho -\sigma )g\] Viscous force acting upward \[=6\pi \eta \,rv.\] Since, ball is moving up with constant velocity v, there is no acceleration in it, the net force acting on it must be zero. \[\therefore \]  \[6\pi \,\eta rv=\frac{4}{3}\pi {{r}^{3}}(\rho -\sigma )g\]                 \[\eta =\frac{2}{9}\frac{{{r}^{2}}(\rho -\sigma )}{v}g\] Given, \[v=-2\times {{10}^{-3}}m/s,\,r={{10}^{-2}}m\] \[\rho =0,\,\sigma =1.5\times {{10}^{3}}kg/{{m}^{3}},g=10\,m/{{s}^{2}}\] \[\therefore \]  \[\eta =\frac{2\times {{({{10}^{-2}})}^{2}}\times (-1.5\times {{10}^{3}})\times 10}{9\times (-2\times {{10}^{-3}})}\] \[=\frac{3}{18\times {{10}^{-3}}}=\frac{1}{6}\times {{10}^{3}}\] \[=0.167\times {{10}^{3}}=167\,\text{units}\]