Question

The terminal velocity of a small ball falling in a viscous liquid depends upon
i) its mass $m$
ii) its radius $r$
iii) the coefficient of viscosity of the liquid $\eta$ and
iv) acceleration due to gravity.
Which of the following relations is dimensionally true for the terminal velocity?

• A. $V=\frac{Kmg}{\eta r}$
• B. $V=\frac{Kmgr}{\eta }$
• C. $V=\frac{Kmg\eta }{r}$
• D. $V=\frac{Kr\eta }{mg}$

Answer 1

The correct option is A $V=\frac{Kmg}{\eta r}$

For K being constant, $mg$ has units of $Kgm/s^2$

$\eta$ has units of $Kg/ms$

$r$ has units of $m$

$\therefore \frac{mg}{\eta r}$ has $m/s$, hence $V=\frac{Kmg}{\eta r}$ is dimensionally correct.