Question

A small metal ball of mass ‘$m$’ is dropped in a liquid contained in a vessel, attains a terminal velocity ‘$v$’. If a metal ball of same material but of mass ‘$8m$’ is dropped in same liquid then the terminal velocity will be

• A. $V$
• B. $2V$
• C. $4V$
• D. $8V$

Answer 1

Answer: (C)

Thermal velocity
$v=\frac{2}{9} \frac{r^{2}\left(p-\rho_{0}\right) g}{\eta}$
In both cases all the variables will remain same except radius, density remains same as material is same.
$m_{1}=m \text { or } \,\,\, m_{2}=8 \,m$
Moreover $m=$ volume $\times$ density $=\frac{4}{3} \pi r^{3} \times d$
$\therefore \,\,\,m_{1}=\frac{4}{3} \pi r_{1}^{3} d, \Rightarrow m_{2}=\frac{4}{3} \pi r_{2}^{3} d$
$\frac{m_{1}}{m_{2}}=\frac{m}{8 m}=\frac{\frac{4}{3} \pi r_{1}^{3} d}{\frac{4}{3} \pi r_{2}^{3} d} \Rightarrow \frac{1}{8}=\frac{r_{1}^{3}}{r_{2}^{3}}$
$\frac{r_{1}}{r_{2}}=\frac{1}{r} \Rightarrow r_{2}=2 r_{1}$
$ \frac{v_{1}}{v_{2}}=\frac{r_{1}^{2}}{r_{2}^{2}} $
$\Rightarrow \frac{v}{v_{2}}=\left(\frac{r_{1}}{2 r_{1}}\right)^{2} \text { or } v_{2}=4 v$