Question

A small steel ball of radius $r$ is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity $\eta$. After some time the velocity of the ball attains a constant value known as terminal velocity $v_T.$ The terminal velocity depends on $\left(i\right)$ the mass of the ball $m$, $\left(ii\right)$ $\eta$, $\left(iii\right)$ $r$, and $\left(iv\right)$ acceleration due to gravity $g$. Which of the following relations is dimensionally correct?

• A. $v_T\propto \frac{mg}{\eta r}$
• B. $v_T\propto \frac{\eta r}{mg}$
• C. $v_T\propto \eta rmg$
• D. $v_T\propto \frac{mgr}{\eta }$

Answer 1

The correct option is A $v_T\propto \frac{mg}{\eta r}$

The dimension of terminal velocity $v_T$ is,
$\left[v_T\right]=\left[L{T}^{-1}\right]$ ................(1)
From the formula of drag force,
$F_D=6\pi \eta r{v}_T$
$\Rightarrow \left[\eta \right]=\frac{\left[F_D\right]}{\left[6\pi r{v}_T\right]}$
$\Rightarrow \left[\eta \right]=\frac{\left[ML{T}^{-2}\right]}{\left[L\right]\left[L{T}^{-1}\right]}=\left[M{L}^{-1}{T}^{-1}\right]$
Option $\left(a\right)$,
$\frac{\left[mg\right]}{\left[\eta r\right]}=\frac{\left[ML{T}^{-2}\right]}{[M{L}^{-1}{T}^{-1}\times L]}=\left[L{T}^{-1}\right]$
From (1), this option matches the dimension of $v_T$. Therefore, option $\left(a\right)$ is correct.