Q. What is velocity $v$ of a metallic ball of radius $r$ falling in a tank of liquid at the instant when its acceleration is one-half that of the freely falling body? (The densities of metal and of liquid are$\rho$ and $\sigma$ respectively, and the viscosity of the liquid is ɳ)
A) $\frac{r^{2}g}{9ɳ}(\rho - 2\sigma)$
B) $\frac{r^{2}g}{9ɳ}(2\rho - \sigma)$
C) $\frac{r^{2}g}{9ɳ}(\rho - \sigma)$
D) $\frac{{2r}^{2}g}{9ɳ}(\rho - \sigma)$
Explanation
(a) Net force on the ball = downward force$\ –\ $upward force
$= \frac{mg}{2}$
$\frac{3}{4}\pi r^{3}(\rho - \sigma)g - 6\pi ɳrv = \frac{mg}{2}$
$\frac{4}{3}\pi r^{3}(\rho - \sigma)g - 6\pi ɳrv = \frac{1}{2}\left( \frac{4}{3}\pi r^{3}\rho \right)g$
$r^{2}(\rho - \sigma)g - \frac{9}{2}ɳv = \frac{1}{2}r^{2}\rho g$
$\frac{9}{2}ɳv = r^{2}(\rho - \sigma)g - \frac{1}{2}r^{2}\rho g$
$= \frac{1}{2}r^{2}\rho g - r^{2}\sigma g$
$\frac{9}{2}ɳv = \frac{1}{2}r^{2}g(\rho - 2\sigma)$
$v = \frac{r^{2}g}{9ɳ}(\rho - 2\sigma)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.