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{2{{r}^{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 ) $