Question
According to Newton, the viscous force acting between liquid layers of area $$A$$ and velocity gradient $$\frac{{\Delta v}}{{\Delta z}}$$ is given by $$F = - \eta A\frac{{dv}}{{dz}},$$ where $$\eta $$ is constant called
A.
$$\left[ {M{L^{ - 2}}{T^{ - 2}}} \right]$$
B.
$$\left[ {{M^0}{L^0}{T^0}} \right]$$
C.
$$\left[ {M{L^2}{T^{ - 2}}} \right]$$
D.
$$\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$$
Answer :
$$\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$$
Solution :
$$\eqalign{
& {\text{As }}F = - \eta A\frac{{dv}}{{dz}} \Rightarrow \eta = - \frac{F}{{A\left( {\frac{{dv}}{{dz}}} \right)}} \cr
& {\text{As}}\,F = \left[ {ML{T^{ - 2}}} \right],A = \left[ {{L^2}} \right] \cr
& dv = \left[ {L{T^{ - 1}}} \right],dz = \left[ L \right] \cr
& \therefore \eta = \frac{{\left[ {ML{T^{ - 2}}} \right]\left[ L \right]}}{{\left[ {{L^2}} \right]\left[ {L{T^{ - 1}}} \right]}} = \left[ {M{L^{ - 1}}\;{T^{ - 1}}} \right] \cr} $$