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

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 $F=\left[ M L T ^{-2}\right.$ ],
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[ M L 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 $F=k r \eta v$.