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 (i) the mass of the ball $m$, (ii) $\eta$, (iii) $r$ and (iv) 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

Answer: (A)

$v_{T} \propto \frac{mg}{\eta r}$

Answer 2

By substituting dimension of each quantity in R.H.S. of option (1) we get $\left\lbrack \frac{mg}{\eta r} \right\rbrack\mspace{6mu} = \mspace{6mu}\left\lbrack \frac{M \times LT^{- 2}}{ML^{- 1}T^{- 1} \times L} \right\rbrack$

=$\lbrack LT^{- 1}\rbrack$.

This option gives the dimension of velocity.

$\left\lbrack \frac{mg}{\eta r} \right\rbrack\mspace{6mu} = \mspace{6mu}\left\lbrack \frac{M \times LT^{- 2}}{ML^{- 1}T^{- 1} \times L} \right\rbrack$=$\lbrack LT^{- 1}\rbrack$