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 ${}^{\prime }8m^{\prime }$ is dropped in same liquid then the terminal velocity will be.

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

Answer 1

The correct option is C $4V$

Terminal velocity of the ball $V=\frac{2g{r}^2(\rho -\sigma )}{9\eta }$ ........(1)

where $\rho$ and $\sigma$ are the density of ball and liquid, respectively.

Mass of the metal ball $m=\rho \left(\frac{4\pi }{3}{r}^3\right)$

$⟹$ $r\propto m^{1/3}$

But from equation (1), we get $V\propto r^2$

$⟹$ $V\propto m^{2/3}$ ......(2)

Given : $V_1=V$ $m_1=m$ $m_2=8m$

From (2) we get $\frac{V_2}{V_1}=(\frac{m_2}{m_1}{)}^{2/3}$

Or $\frac{V_2}{V}=(\frac{8m}{m}{)}^{2/3}=8^{2/3}$

$⟹$ $V_2=(2^3{)}^{2/3}V=4V$