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 o and coefficient of viscosity t). It attains a terminal velocity due to two force; 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,   \[v=-2\times {{10}^{-3}}m/s,\,\,r={{10}^{-2}}m\] \[\rho =0,\,\sigma =1.5\times {{10}^{3}}kg/{{m}^{3}},g=10\,m/{{s}^{2}}\] \[\therefore \] \[\eta =\frac{2\times {{({{10}^{-2}})}^{2}}\times (-1.5\times {{10}^{3}})\times 10}{9\times (-2\times {{10}^{-3}})}\]                 \[=\frac{3}{18\times {{10}^{-3}}}=\frac{1}{6}\times {{10}^{3}}\] \[=0.167\times {{10}^{3}}=167\] units.