Question

What is the 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 a freely falling body? (The densities of metal and of liquid are $\rho$ and $\sigma$ respectively, and the viscosity. of the liquid is $\eta$ ):

• A. $\frac{r^{2}g}{9\eta }(\rho -2\sigma )$
• B. $\frac{r^{2}g}{9\eta }(2\rho -\sigma )$
• C. $\frac{r^{2}g}{9\eta }(\rho -\sigma )$
• D. $\frac{2r^{2}g}{9\eta }(\rho -\sigma )$
• E. $\frac{r^{2}g}{18\eta }(\rho -2\sigma )$

Answer 1

Answer: (A)

Net force on the ball = downward force - upward force $=\frac{mg}{2}$
$\frac{4}{3}\pi r^3(\rho -\sigma )-6\pi \eta rv=\frac{mg}{2}$
$\frac{4}{3}\pi r^3(\rho -\sigma )g-6\pi \eta rv=\frac{1}{2}\left(\frac{4}{3}\pi r^3\rho \right)g$
$r^2(\rho -\sigma )g-\frac{9}{2}\eta v=\frac{1}{2}{r}^2\rho g$
$\frac{9}{2}v\eta =r^2(\rho -\sigma )g-\frac{1}{2}{r}^2\rho g$
$=\frac{1}{2}{r}^2\rho g-r^2\sigma g$
$\frac{9}{2}v\eta =\frac{1}{2}{r}^{2}g(\rho -2\sigma )$
$v=\frac{r^{2}g}{9\eta }(\rho -2\sigma )$