Q. A small spherical ball falling through a viscous medium of negligible density has terminal velocity $v$. Another ball of the same mass but of radius twice that of the earlier falling through the same viscous medium will have terminal velocity
Explanation
(c) Terminal velocity of the ball through a viscous medium
$v = \frac{2}{9} \times \frac{g}{ɳ}(\rho - \sigma)r^{2}$
$v = \frac{2}{9} \times \frac{g}{ɳ}(\rho)r^{2}$
because viscous medium of negligible density $(\sigma = 0)$
$v = \frac{2}{9} \times \frac{g}{ɳ} \times \frac{m}{\frac{4}{3}{\pi r}^{3}} \times r^{2\ \ \ \ \ \ \ \ \ }\ \ \ \ \left( \therefore e = \frac{m}{\frac{4}{3}{\pi r}^{3}} \right)$
$or\ \ v = \frac{2}{9} \times \frac{g}{ɳ} \times \frac{m}{\frac{4}{3}{\pi r}^{3}}$
$\Longrightarrow \ \ \ \ v = \frac{1}{r}$
For the second ball
$v \propto \frac{1}{2r}$
Because radius of second ball is twice that of the first ball
$\frac{v}{v} = \frac{2r}{r}\ or\ v^{'} = \frac{v}{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.