question_answer

A sphere of mass M and radius R is falling in a viscous fluid. The terminal velocity attained by the falling object will be proportional to:

A) \[{{R}^{2}}\]            

B)       R                                            

C) \[1/R\] 

D)                                       \[1/{{R}^{2}}\]

Correct Answer: A

Solution :

                                Key Idea: Since sphere is moving with constant velocity, there is no acceleration in it. When the sphere of radius^ is falling in a liquid of density a and coefficient of viscosity \[\eta \] it attains a terminal velocity v, under two forces (i) Effective force acting downward \[=V(\rho -\sigma )g=\frac{4}{3}\pi {{R}^{3}}(\rho -\sigma )g\] where \[\rho \] is density of sphere. (ii) Viscous forces acting upwards \[=6\pi \eta Rv\] Since the sphere is moving with a constant velocity v, there is no acceleration in it, the net force acting on it must be zero. That is \[6\pi \eta Rv=\frac{4}{3}\pi {{R}^{3}}(\rho -\sigma )g\] \[\Rightarrow \] \[v=\frac{2}{9}\frac{{{R}^{2}}(\rho -\sigma )g}{\eta }\] \[\Rightarrow \]   \[v\propto {{R}^{2}}\] Thus, terminal velocity is proportional to the square of its radius.