Question

The expression for viscous force $F$ acting on a tiny steel ball of radius $r$ moving in a viscous liquid of viscosity $\eta$ with a constant speed $v$ with the help of the method of dimensional analysis is

• A. $kr\eta v$
• B. $k{r}^2\eta v$
• C. $kr\eta v^{3/2}$
• D. $kr{\eta }^{2}v$

Answer 1

Answer: (A)

It is given here that, the viscous force $F$ depends on
(i) radius $r$ of steel ball (ii) coefficient of viscosity $\eta$ of viscous liquid and (iii) speed $v$ of the ball.
Let the relationship can be written as $F=k{r}^a{\eta }^b{v}^c$
where, $k$ is a dimensionless constant.
As dimensional formula of force $F=[ML{T}^{-2}$ ],
radius $r=[L]$, coefficient of viscosity,
$\eta =\left[M^1{L}^{-1}{T}^{-1}\right]$
and speed, $v=\left[L{T}^{-1}\right]$, hence we get
$\left[ML{T}^{-2}\right]=[L{]}^a{\left[M{L}^{-1}{T}^{-1}\right]}^b{\left[L{T}^{-1}\right]}^c$
$=\left[M^b{L}^{a-b+c}{T}^{-b-c}\right]$
Comparing powers of M, L and $T$ on either side of the equation, we get
$b=1$ .....(i)
$a-b+c=1$ .....(ii)
and $-b-c=-2$ ......(iii)
On solving these three equations, we get
$a=1,b=1$ and $c=1$
Hence, the relation becomes $F=kr\eta v$.