A

Step 1: Identify the forces acting on the slab.
The forces include the gravitational force, frictional force due to viscosity, and normal force.
The gravitational force acting down the slope can be calculated as:
$$ F_g = m imes g imes ext{sin}( heta) $$
where:
$m = 18 ext{ kg}$ (mass of the slab),
$g = 9.81 ext{ m/s}^2$ (acceleration due to gravity),
$ heta = 15^ ext{o}$ (angle of incline).
Step 2: Calculate the gravitational force:
$$ F_g = 18 imes 9.81 imes ext{sin}(15^ ext{o}) $$
Using a calculator, $$ ext{sin}(15^ ext{o}) ext{ is approximately } 0.2588 $$, so:
$$ F_g = 18 imes 9.81 imes 0.2588 ext{ N} ext{ (approximately)} $$
$$ F_g ext{ is approximately } 46.02 ext{ N} $$
Step 3: Calculate the viscous force using the formula for viscous drag:
$$ F_v = rac{6 imes ext{pi} imes ext{R}^2 imes ext{h} imes v}{d} $$
where:
$ ext{R} = ext{radius of the slab (half the width assumed for simplicity), or provided if necessary}$,
$d = 3 ext{ mm} = 0.003 m$ (thickness of oil),
$v$ is the terminal velocity, and $h$ is the height of the slab. Assuming a small plate, $h$ and $F_v$ will be tied to area $A$:
$$ F_v = rac{8 imes ext{viscosity} imes A imes ext{v}}{d} $$
Step 4: Set the net forces to zero for terminal velocity:
$$ F_g - F_v = 0 $$
or:
$$ F_v = F_g $$
$$ rac{8.14 imes 10^{-2} imes 0.3 imes v}{0.003} = 46.02 $$
Step 5: Solve for $v$:
$$ v = rac{46.02 imes 0.003}{8.14 imes 10^{-2} imes 0.3} $$
Calculate:
$$ v = rac{0.13806}{0.02442} $$
$$ v ext{ is approximately } 5.65 ext{ m/s} $$
Final Answer: The terminal velocity of the slab is approximately 5.65 m/s.