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.
$m_1=m\text{ or }{m}_2=8m$
Moreover $m=$ volume $\times$ density $=\frac{4}{3}\pi r^3\times d$
$\therefore m_1=\frac{4}{3}\pi r_1^{3}d,\Rightarrow m_2=\frac{4}{3}\pi r_2^{3}d$
$\frac{m_1}{m_2}=\frac{m}{8m}=\frac{\frac{4}{3}\pi r_1^{3}d}{\frac{4}{3}\pi r_2^{3}d}\Rightarrow \frac{1}{8}=\frac{r_1^3}{r_2^3}$
$\frac{r_1}{r_2}=\frac{1}{r}\Rightarrow r_2=2r_1$
$\frac{v_1}{v_2}=\frac{r_1^2}{r_2^2}$
$\Rightarrow \frac{v}{v_2}={\left(\frac{r_1}{2r_1}\right)}^2\text{ or }{v}_2=4v$