Turpentine oil is flowing through a tube of length $$l$$ and radius $$r.$$ The pressure difference between the two ends of the tube is $$p.$$ The viscosity of oil is given by
$$\eta = \frac{{p\left( {{r^2} - {x^2}} \right)}}{{4vl}}$$
where, $$v$$ is the velocity of oil at distance $$x$$ from the axis of the tube. The dimensions of $$\eta $$ are

Solution :
$$\eqalign{ & {\text{Pressure}}\,\left( p \right) = \frac{{{\text{Force}}}}{{{\text{Area}}}} = \frac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right] \cr & {\text{Velocity,}}\,v = \left[ {L{T^{ - 1}}} \right] \cr} $$
From principle of homogeneity, the dimensions of $${r^2}$$ and $${x^2}$$ are same.
So, the dimensions of viscosity,
$$\eqalign{ & \eta = \frac{{\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]\left[ {{L^2}} \right]}}{{\left[ {L{T^{ - 1}}} \right]\left[ L \right]}} \cr & = \left[ {M{L^{ - 1}}{T^{ - 1}}} \right] \cr} $$