Question
The force $$F$$ on a sphere of radius $$r$$ moving in a medium with velocity $$v$$ is given by $$F = 6\pi \eta rv.$$ The dimensions of $$\eta $$ are
A.
$$\left[ {M{L^-3}} \right]$$
B.
$$\left[ {ML{T^{ - 2}}} \right]$$
C.
$$\left[ {M{T^{ - 1}}} \right]$$
D.
$$\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$$
Answer :
$$\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$$
Solution :
Viscous force on a sphere of radius $$r$$ is
$$\eqalign{
& F = 6\pi \eta rv \Rightarrow \eta = \frac{F}{{6\pi rv}} \cr
& \left[ \eta \right] = \frac{{\left[ F \right]}}{{\left[ r \right]\left[ v \right]}} = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ L \right]\left[ {L{T^{ - 1}}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 1}}} \right] \cr} $$
NOTE
The above expression is obtained from Stokes’ law.