c
Using the relation $\tau=\mu \frac{d u}{d y} .$ Here $\tau$ is the shear stress between the layers of the liquid which is equal to $\frac{F}{A} . d u$ is the change of velocity $=u-0=u$ and $d y$ is the clearance which is equal to $y$
Thus we get
$\frac{F}{A}=\mu \frac{u}{y}$
or
$F=\frac{A \mu u}{y} \propto u$
Thus we get
$\frac{F_{1}}{u_{1}}=\frac{F_{2}}{u_{2}}$
Substituting the values we get
$\frac{800}{1.5}=\frac{2400}{u_{2}}$
Thus we get $u_{2}=4.5 \mathrm{cm} / \mathrm{s}$